1 Introduction
The Yoneda characterization of
${\operatorname{Ext}}$
is familiar to most students of homological algebra. Let
$A,B$
be two objects in an abelian category
$\cal{A}$
. Then
${\operatorname{Ext}}_{\cal{A}}(A,B)$
is the set of all equivalence classes of sequences of the form
$0\to B\to C\to A\to 0$
, where two sequences
$\alpha ,\beta $
are equivalent if we have the following commutative diagram:
Now,
${\operatorname{Ext}}_{\cal{A}}(A,B)$
can be given an abelian group structure by the well-known Baer sum as described in, for instance, [38, Tag 010I]. This consideration can be carried out more generally in any exact category (see [Reference Dräxler, Reiten, Smalø and Solberg14, §1.2]).
The purpose of this note is to study the following rather natural questions: what if we place additional constraints on the short exact sequences? When do we get a subfunctor of
${\operatorname{Ext}}^1$
? Can one apply such functors to study ring and module theory, similar to the ways homological algebra has been very successfully applied in the last decades?
Let us elucidate our goals with a concrete example. Let
$(R,\mathfrak {m},k)$
be a Noetherian local ring. Let
$A,B$
be finitely generated R-modules. We consider exact sequences
$0\to B\to C\to A\to 0$
of R-modules, with the added condition that
$\mu (C)=\mu (A)+\mu (B)$
, where
$\mu (-)$
denotes the minimal number of generators. As we shall see later, the equivalence classes of such sequences do form a subfunctor of
${\operatorname{Ext}}^1_R$
, which we denote by
${\operatorname{Ext}}^1_R(-,-)^{\mu }$
. More surprisingly, the vanishing of a single module
${\operatorname{Ext}}^1_R(-,-)^{\mu }$
can be used to characterize the regularity of R, a feature that is lacking with the classical
${\operatorname{Ext}}^1$
.
Analogous versions of classical homological functors have been studied by various authors, notably starting with Hochschild [Reference Hochschild23] who studied relative
${\operatorname{Ext}}$
and
$\operatorname{Tor}$
groups of modules over a ring with respect to subrings of the original ring. This work is further developed by Butler–Horrocks as well as Auslander–Solberg, where the point of view is switched to allowable exact sequences that give rise to sub(bi)functors of
${\operatorname{Ext}}^1$
. The whole circle of ideas is now thriving on its own under the name relative homological algebra, with exact structures playing a fundamental role (see [Reference Brüstle, Hassoun, Langford and Roy4], [Reference Buan5], [Reference Crivei8], [Reference Dräxler, Reiten, Smalø and Solberg14], [Reference Enochs and Jenda16]–[Reference Enomoto18], [Reference Matsui28], [Reference Rump33], [Reference Zhu and Zhuang40]] for an incomplete list of literature and [Reference Bühler6], [Reference Solberg34] for some excellent surveys). In commutative algebra, as far as we know, this line of inquiry has not been exploited thoroughly; however, traces of it can be found in [Reference Puthenpurakal31], [Reference Striuli35], and [Reference Eisenbud and Schreyer15, §§1 and 2].
Although the existing literature provides excellent starting points and inspiring ideas for this present work, it is not always easy to extract the precise results needed for our intended applications. For instance, while the connections between subfunctors of
${\operatorname{Ext}}^1$
and certain sub-exact structures on a fixed category are well known [Reference Dräxler, Reiten, Smalø and Solberg14, §1.2], checking the conditions of substructures in each case can be time-consuming.
We are able to find criteria that can be applied in broad settings to identify exact subcategories and subfunctors. Here is a sample result applicable to our motivating example above, which follows from Theorem 4.8 and Proposition 3.8.
Theorem 1.1. Let
$({\cal{A}},{\cal{E}})$
be an exact category. Let
$\phi :{\cal{A}} \to \mathbb Z$
be a function such that
$\phi $
is constant on isomorphism classes of objects in
$\cal{A}$
,
$\phi $
is additive on finite biproducts, and
$\phi $
is sub-additive on kernel–cokernel pairs in
$\cal{E}$
(i.e., if
is in
$\cal{E}$
, then
$\phi (N)\le \phi (M)+\phi (L)$
). Set
${\cal{E}}^{\phi }:=\{\text {kernel--cokernel pairs in } {\cal{E}} \text { on which } \phi \text { is additive} \}$
. Then
${\cal{E}}^{\phi }$
gives rise, via the Yoneda construction, to a subfunctor of
${\operatorname{Ext}}_{\cal{E}}(-,-)$
.
Our above theorem is partly motivated by, and can be used to recover and extend, recent interesting work of Puthenpurakal in [Reference Puthenpurakal31] (see Theorem 4.17).
Another situation we would like to have convenient criteria for subfunctors is when one restricts to a certain subcategory. A concrete example we have in mind concerns Ulrich modules and their recent generalizations. These modules form a subcategory of Cohen–Macaulay modules over a commutative ring and have been receiving increasing attention over the years due to very interesting and useful algebraic and geometric properties that their existence or abundance imply. We are able to show that even the generalized notion of I-Ulrich modules, recently introduced in [Reference Dao, Maitra and Sridhar11], induces subfunctors of
${\operatorname{Ext}}^1_R(-,-)$
. See Proposition 4.1 and Corollary 4.7 in this regard.
While this work is mainly concerned with foundational results, we also study the properties of some chosen new subfunctors, just to see if they are worth our efforts to show their existence! The early returns seem promising: these functors can be used to detect a wide range of ring and module-theoretic properties. Below we shall describe the organization of the paper and describe the most interesting findings in more detail.
Section 2 is devoted to preliminary results on subfunctors of additive functors. While these results are perhaps not new, we were unable to locate convenient references, hence their inclusion.
Section 3 establishes various foundational results on exact subcategories, which form the cornerstone of the theory. As mentioned above, our applications require some extra care in preparation, and we try to give complete proofs whenever possible.
Section 4 concerns our first main application. We study sub-additive numerical functions
$\phi $
on an exact category and show that under mild conditions they induce exact subcategories and hence subfunctors of
${\operatorname{Ext}}^1(-,-)$
, which we denote by
${\operatorname{Ext}}^1(-,-)^{\phi }$
. See Theorem 4.8. A key consequence, Theorem 4.17, is inspired by, as well as extends, [Reference Puthenpurakal31, Ths. 3.11 and 3.13]. We also give similar results about subfunctors of
${\operatorname{Ext}}^1$
induced by half-exact functors, in the spirit of [Reference Auslander and Solberg1] (see Corollaries 4.13 and 4.14).
In §5, we focus on two special types of subfunctors, which arise from simple applications of previous sections. Already these cases appear to be interesting and useful. The first one is
${\operatorname{Ext}}^1_R(-,-)^{\mu }$
, where
$\mu $
is the minimal number of generators function mentioned above. We compute the values of this subfunctor on all (pair of) finitely generated modules over a discrete valuation ring, i.e., DVR (Corollary 5.1.14), as well as for certain pairs of modules over a Cohen–Macaulay ring of minimal multiplicity (Proposition 5.1.22).
Using this subfunctor, we prove the following characterization of the regularity of local rings, which is the combination of Theorems 5.1.1 and 5.1.12, Proposition 5.1.14, and Corollary 5.2.7. Note that the regularity can be detected by the vanishing of a single
${\operatorname{Ext}}^1_R(-,-)^{\mu }$
-module. In the following statement, we mention that for a finitely generated module M over the local ring R,
$\Omega ^i_R M$
denotes the ith syzygy in a minimal free resolution of M.
Theorem 1.2. Let
$(R,\mathfrak {m},k)$
be a local ring of depth
$t>0$
. Then, the following are equivalent:
-
(1) R is regular.
-
(2)
${\operatorname{Ext}}^1_R(k,R/(x_1,...,x_{t-1})R)^{\mu }=0$
for some R-regular sequence
$x_1,...,x_{t-1}$
. -
(3)
${\operatorname{Ext}}^1_R(k,M)^{\mu }=0$
for some finitely generated R-module M of projective dimension
$t-1$
. -
(4) R is Cohen–Macaulay, and
${\operatorname{Ext}}^1_R(\Omega ^{t-1}_R k,N)^{\mu }=0$
for some finitely generated nonzero R-module N of finite injective dimension.Moreover, if
$t=1$
, then the above are also equivalent to each of the following. -
(5)
${\operatorname{Ext}}^1_R(M,N)^{\mu }=\mathfrak {m}{\operatorname{Ext}}^1_R(M,N)$
for all finitely generated R-modules M and N. -
(6) R is Cohen–Macaulay, and there exist Ulrich modules
$M,N$
such that N is faithful,
$M\ne 0$
, and for every R-module X that fits into a short exact sequence
$0\to N\to X\to M \to 0 $
, one has X is also an Ulrich module.
We note here that the usual
${\operatorname{Ext}}$
-modules (without the
$\mu $
) in (2)–(4) of the above theorem are always nonzero. Moreover, the statement of part (6) of Theorem 1.2 apparently has nothing to do with subfunctor of
${\operatorname{Ext}}^1$
, but we do not know a proof of
$(1)\iff (6)$
(which is contained in Corollary 5.2.7) without resorting to
${\operatorname{Ext}}^1_{\operatorname{Ul}(R)}(-,-): \operatorname{Ul}(R)^{op}\times \operatorname{Ul}(R)\to {\operatorname{mod}}\ R$
.
It is worth mentioning that one can also use
${\operatorname{Ext}}^1_R(-,-)^{\mu }$
to detect the property of R being a hypersurface of minimal multiplicity (Corollary 5.1.23) or the weak
$\mathfrak {m}$
-fullness of a submodule (Proposition 5.1.11).
The second type arises from I-Ulrich modules, where I is any
$\mathfrak {m}$
-primary ideal in a Noetherian local ring
$(R,\mathfrak {m})$
. See Definition 4.5 and Corollary 4.7 for the definition of I-Ulrich modules and the fact that they form an exact category. In Theorem 5.2.14, using a subfunctor of
${\operatorname{Ext}}^1$
corresponding to Ulrich modules over one-dimensional local Cohen–Macaulay rings, we give some characterizations of modules belonging to
$\operatorname{add}_R(B(\mathfrak {m}))$
and also characterize when
$B(\mathfrak {m})$
is a Gorenstein ring in terms of annihilator of
${\operatorname{Ext}}^1_R$
of Ulrich modules. Here,
$B(-)$
denotes the blow-up. We give some applications of Theorem 5.2.14, one of which relates annihilation of
${\operatorname{Ext}}^1_R(\operatorname{Ul}(R),\mathfrak {m})$
with that of
${\operatorname{Ext}}^1_R\left (\operatorname{Ul}_{\omega }(R),B(\omega )\right )$
(see Corollary 5.2.17).
Finally, we should mention that one of the main applications of our results has appeared in a separate work, where we study the splitting of short exact sequences of Ulrich modules and connections to other properties of singularities (see [Reference Dao, Dey and Dutta9]).
2 Preliminaries on subfunctors of additive functors
Unless otherwise stated, all rings in this paper are assumed to be commutative, Noetherian, and with unity. For a ring R,
$Q(R)$
will denote its total ring of fractions. For an R-module M,
$\lambda _R(M)$
will denote its length. For a finitely generated R-module M, and
$i\geq 1$
, by
$\Omega ^i_R M$
we mean
$\text {Im} f_{i}$
, where we have an exact sequence
$F_{i}\xrightarrow {f_{i}} F_{i-1}\xrightarrow {f_{i-1}}\cdots \to F_0\to M \to 0$
, with each
$F_j$
being a finitely generated projective R-module. When R is moreover local, we choose this so that
$\text {Im}(f_j)\subseteq \mathfrak {m} F_{j-1}$
for each j (see [Reference Bruns and Herzog3, Prop. 1.3.1]).
For definitions, and basic properties of additive categories, additive functors, R-linear categories, and R-linear functors, we refer the reader to [38, Tags 09SE, 010M, and 09MI]. We now recall the definition of subfunctors as in [29].
Definition 2.1. Let
${\cal{A}},{\cal{B}}$
be two categories. Let
$F: {\cal{A}} \to {\cal{B}}$
be a covariant (resp. contravariant) functor. A covariant (resp. contravariant) functor
$G: {\cal{A}} \to {\cal{B}}$
is called a subfunctor of F if for every
$M\in {\cal{A}}$
, there exists a monomorphism
$j_M:G(M)\to F(M)$
, and moreover, for every
$M,N\in {\cal{A}}$
, and
$f\in \operatorname{Mor}_{\cal{A}}(M,N)$
, the following diagrams are commutative, where the left one stands for the covariant case and the right one for the contravariant case:
2.2. For our purposes, we will always take
$\cal{B}$
to be either the category of abelian groups
$\mathbf {Ab}$
,
${\operatorname{Mod}}\ R$
or
${\operatorname{mod}}\ R$
(hence monomorphisms are just injective morphisms) for some commutative ring R, and
$j_M$
will usually be just the inclusion map.
The following lemma is probably well known, but we could not find an appropriate reference; hence, we include a proof. This will be used throughout the remainder of the article, possibly without further reference.
Lemma 2.3.
-
(1) Subfunctor of an additive functor, between two additive categories, is additive.
-
(2) Subfunctor of an R-linear functor, between two R-linear categories, is R-linear.
Proof. We will only prove the covariant case of both, since the contravariant case is similar.
(1) Let
${\cal{A}},{\cal{B}}$
be two additive categories, and let
$F:{\cal{A}}\to {\cal{B}}$
be an additive functor. Also, let
$G: {\cal{A}} \to {\cal{B}}$
be a subfunctor of F. Then we have to show that the map
$G:\operatorname{Mor}_{\cal{A}}(X,Y)\to \operatorname{Mor}_{\cal{B}}(G(X),G(Y))$
is a homomorphism of abelian groups for all
$X,Y\in {\cal{A}}$
. Fix two objects
$X,Y\in {\cal{A}}$
. Then we need to prove that
$G(f+g)=G(f)+G(g)$
for all
$f,g\in \operatorname{Mor}_{\cal{A}}(X,Y)$
. Since G is a subfunctor of F, we have the following commutative diagrams:
Then we get
$$ \begin{align*} j_Y\circ G(f+g) &=F(f+g)\circ j_X\\ &= (F(f)+F(g))\circ j_X\text{ [Since }F\text{ is additive]}\\ &=F(f)\circ j_X+F(g)\circ j_X\\ &=j_Y\circ G(f)+j_Y\circ G(g)\\ &=j_Y\circ (G(f)+G(g)), \end{align*} $$
where the third and fifth equalities follow from the fact that
$\cal{B}$
is an additive category. Now, since
$j_Y$
is a monomorphism, so
$j_Y\circ G(f+g)=j_Y\circ (G(f)+G(g))$
implies that
$G(f+g)=G(f)+G(g)$
. Hence, G is additive.
(2) Let
${\cal{A}},{\cal{B}}$
be two R-linear categories, and let
$F:{\cal{A}}\to {\cal{B}}$
be an R-linear functor. Also, let
$G: {\cal{A}} \to {\cal{B}}$
be a subfunctor of F. Then we have to show that the map
$G:\operatorname{Mor}_{\cal{A}}(X,Y)\to \operatorname{Mor}_{\cal{B}}(G(X),G(Y))$
is an R-linear map for all
$X,Y\in {\cal{A}}$
. Fix two objects
$X, Y\in {\cal{A}}$
. Then, by part (1), we already have the additivity of G, so we only need to prove that
$G(rf)=rG(f)$
for all
$f\in \operatorname{Mor}_{\cal{A}}(X,Y)$
and for all
$r\in R$
. Since G is a subfunctor of F, we have the following commutative diagrams:
Then we get
$$ \begin{align*} j_Y\circ G(rf) &=F(rf)\circ j_X\\ &= (rF(f))\circ j_X\text{ [Since }F\text{ is }R\text{-linear]}\\ &=r(F(f)\circ j_X)\\ &=r(j_Y\circ G(f))\\ &=j_Y\circ (rG(f)), \end{align*} $$
where the third and fifth equalities follow from the fact that
$\cal{B}$
is an R-linear category. Now, since
$j_Y$
is a monomorphism, so
$j_Y\circ G(rf)=j_Y\circ (rG(f))$
implies that
$G(rf)=rG(f)$
. Hence, G is R-linear.
We finish this section with a submodule inclusion result relating subfunctors of R-linear functors which will be applied in §5.
Lemma 2.4. Let
$\cal{A}$
be an additive R-linear category, and let
$G:{\cal{A}}\to {\operatorname{mod}}\ R$
be a subfunctor of an R-linear functor
$F:{\cal{A}} \to {\operatorname{Mod}}\ R$
, and for every object
$A\in {\cal{A}}$
, let
$j_A:G(A)\to F(A)$
be the monomorphism as in the definition of a subfunctor. Let I be an ideal of R. Let
$\{A_i\}_{i=1}^n$
be objects in
$\cal{A}$
such that
$j_{A_i}\left (G(A_i)\right )\subseteq IF(A_i)$
(resp.
$j_{A_i}\left (G(A_i)\right )\supseteq IF(A_i)$
) for all
$i=1,...,n$
. Let
$X:=\oplus _{i=1}^nA_i$
. Then, it holds that
$j_{X} \left (G(X)\right )\subseteq IF(X)$
(resp.
$j_{X} \left (G(X)\right )\supseteq IF(X)$
).
Proof. We will only do the covariant case, the contravariant case being similar. For every i, we have
$\pi _i:X \to A_i$
, which gives rise to the following commutative diagram:
Consequently, we get the following commutative diagram:
where the horizontal arrows are isomorphisms, since F and consequently G are additive functors (see Lemma 2.3). Call the top horizontal map
$\theta $
, and the bottom one
$\alpha $
, so that
$\alpha \circ j_X=(\oplus _{i=1}^nj_{A_i})\circ \theta $
; hence,
$j_X\circ \theta ^{-1}=\alpha ^{-1}\circ (\oplus _{i=1}^nj_{A_i})$
. So, now, we get
$$ \begin{align*} j_X(G(X))=(j_X \circ \theta^{-1}) (\theta(G(X)))=(j_X\circ \theta^{-1})(\oplus_{i=1}^nG(A_i)) &=(\alpha^{-1}\circ (\oplus_{i=1}^nj_{A_i}\kern-1pt))(\oplus_{i=1}^nG(A_i)\kern-1pt)\\ &=\alpha^{-1}(\oplus_{i=1}^n j_{A_i}(G(A_i))). \end{align*} $$
So, if
$j_{A_i}\left (G(A_i)\right )\subseteq IF(A_i)$
(resp.
$j_{A_i}\left (G(A_i)\right )\supseteq IF(A_i)$
) for all
$i=1,...,n$
, then
$$ \begin{align*}j_X(G(X))\subseteq \alpha^{-1}(\oplus_{i=1}^n IF(A_i))=\alpha^{-1}(I\left(\oplus_{i=1}^n F(A_i)\right))=I\alpha^{-1}(\oplus_{i=1}^n F(A_i))=IF(X)\end{align*} $$
(resp.
$j_X(G(X))\supseteq \alpha ^{-1}(\oplus _{i=1}^n IF(A_i))=\alpha ^{-1}(I\left (\oplus _{i=1}^n F(A_i)\right ))=I\alpha ^{-1}(\oplus _{i=1}^n F(A_i))=IF(X)$
),
where we have used
$\alpha ^{-1}(IM)=I\alpha ^{-1}(M)$
, since
$\alpha $
is an R-linear map.
3 Some generalities about exact subcategories
In this section, we record some generalities about exact subcategories of an exact category that we will later use for subcategories of
${\operatorname{mod}}\ R$
when R is a commutative Noetherian ring. All our subcategories are strict (closed under isomorphism classes) and full, and we often abbreviate this as strictly full. We will follow the definition of an exact category described in [Reference Bühler6, Def. 2.1]. We try to provide complete proofs whenever possible.
Given an exact category
$({\cal{A}},{\cal{E}})$
, we call a monomorphism
$X\xrightarrow {i} Y$
to be an
$\cal{E}$
-inflation (also, called an admissible monic) if it is the part of a kernel–cokernel pair
$X\xrightarrow {i} Y\xrightarrow {} Z$
, which lies in
$\cal{E}$
. Dually, we call an epimorphism
$Y\xrightarrow {p} Z$
to be an
$\cal{E}$
-deflation (also, called an admissible epic) if it is the part of a kernel–cokernel pair
$X\xrightarrow {} Y\xrightarrow {p} Z$
, which lies in
$\cal{E}$
. We will often denote an admissible monic (resp. an admissible epic) by
(resp.
).
We begin by stating a lemma on morphisms and kernel–cokernel pairs, which we will use frequently while proving that certain structures are closed under isomorphism classes of kernel–cokernel pairs. This should be standard and well known, but we could not find an appropriate reference; hence, we include a proof.
Lemma 3.1. Let
$\cal{A}$
be an additive category. Let
$M,N,L\in {\cal{A}}$
be such that
$M\xrightarrow {i} N\xrightarrow {d} L$
is a kernel–cokernel pair in
$\cal{A}$
. Also, let
$M',N',L'\in {\cal{A}}$
. Consider morphisms
$M'\xrightarrow {i'} N'$
,
$N'\xrightarrow {d'} L'$
. If we have the following diagram with commutative squares:
where the vertical arrows
$\phi _1,\phi _2,\phi _3$
are isomorphisms, then
$M'\xrightarrow {i'} N'\xrightarrow {d'} L'$
is a kernel–cokernel pair in
$\cal{A}$
.
Proof. We will only prove that
$\ker (d')=i'$
, since the proof of
$\operatorname{Coker}(i')=d'$
can be given by a dual argument. From the above commutative diagram, we have
$$ \begin{align} \phi_2\circ i=i'\circ\phi_1,\quad\phi_3\circ d=d'\circ\phi_2. \end{align} $$
Since
$M\xrightarrow {i} N\xrightarrow {d} L$
is a kernel–cokernel pair in
$\cal{A}$
, we get
$d\circ i=0$
. Then, by Equation (3.1), we have
$d'\circ i'=(\phi _3\circ d\circ \phi _2^{-1})\circ (\phi _2\circ i\circ \phi _1^{-1})=\phi _3\circ d\circ i\circ \phi _1^{-1}=\phi _3\circ 0\circ \phi _1^{-1}=0$
. Now, let
$K\in {\cal{A}}$
and consider a morphism
$f':K\to N'$
such that
$d'\circ f'=0$
. Define
$f:=\phi _2^{-1}\circ f':K\to N$
. Then, by Equation (3.1), we get that
$d\circ f=d\circ (\phi _2^{-1}\circ f')=(d\circ \phi _2^{-1})\circ f'=(\phi _3^{-1}\circ d')\circ f'=\phi _3^{-1}\circ (d'\circ f')=\phi _3^{-1}\circ 0=0$
. Now, since
$\ker (d)=i$
, by the universal property of a kernel of a map, there exists a morphism
$u:K\to M$
such that
$i\circ u=f$
. Now, define
$u':=\phi _1\circ u:K\to M'$
. Then, by Equation (3.1), we get that
$i'\circ u'=i'\circ (\phi _1\circ u)=(i'\circ \phi _1)\circ u=(\phi _2\circ i)\circ u=\phi _2\circ f=\phi _2\circ (\phi _2^{-1}\circ f')=f'$
. This implies that
$\ker (d')=i'$
.
We now record a lemma on the intersection of exact subcategories. Note that this is slightly different (and in view of [Reference Rump33, Cor. 2], more general) than [Reference Brüstle, Hassoun, Langford and Roy4, Lem. 5.2].
Lemma 3.2. Let
$({\cal{A}},{\cal{E}})$
be an exact category. Also, let
$({\cal{A}}_{\lambda },{\cal{E}}_{\lambda })$
be an arbitrary family of exact subcategories of
$({\cal{A}},{\cal{E}})$
. Then
$(\cap _{\lambda } {\cal{A}}_{\lambda },\cap _{\lambda }{\cal{E}}_{\lambda })$
is exact subcategory of
$({\cal{A}},{\cal{E}})$
.
Proof. Clearly,
$(\cap _{\lambda } {\cal{A}}_{\lambda },\cap _{\lambda }{\cal{E}}_{\lambda })$
is a strictly full additive subcategory of
$({\cal{A}},{\cal{E}})$
. By [Reference Bühler6, Def. 2.1] and [Reference Bühler6, Rem. 2.4], it is enough to show that
$\cap _{\lambda }{\cal{E}}_{\lambda }$
is closed under isomorphisms of kernel–cokernel pairs and
$(\cap _{\lambda } {\cal{A}}_{\lambda },\cap _{\lambda }{\cal{E}}_{\lambda })$
satisfies the axioms [E0], [E0
$^{\text {op}}$
], [E1
$^{\text {op}}$
], [E2], and [E2
$^{\text {op}}$
] of [Reference Bühler6, Def. 2.1]. First, we will show that
$\cap _{\lambda }{\cal{E}}_{\lambda }$
is closed under isomorphisms of kernel–cokernel pairs. Let
be a kernel–cokernel pair in
$\cap _{\lambda }{\cal{E}}_{\lambda }$
, so
$M,N,L\in {\cal{A}}_{\lambda }$
and
is a kernel–cokernel pair in
${\cal{A}}_{\lambda }$
for all
$\lambda $
. Also, let
$M'\to N'\to L'$
be a kernel–cokernel pair in
$\cap _{\lambda }{\cal{A}}_{\lambda }$
such that it is isomorphic to
. So,
$M',N',L'\in {\cal{A}}_{\lambda }$
and
$M'\to N'$
,
$N'\to L'$
are two morphisms in
${\cal{A}}_{\lambda }$
for all
$\lambda $
. Hence, by Lemma 3.1, we get that
$M'\to N'\to L'$
is a kernel–cokernel pair in
${\cal{A}}_{\lambda }$
for all
$\lambda $
. Since
$({\cal{A}}_{\lambda },{\cal{E}}_{\lambda })$
is an exact category and
is in
${\cal{E}}_{\lambda }$
for all
$\lambda $
, we have
is in
${\cal{E}}_{\lambda }$
for all
$\lambda $
. So,
is in
$\cap _{\lambda }{\cal{E}}_{\lambda }$
. Thus,
$\cap _{\lambda }{\cal{E}}_{\lambda }$
is closed under isomorphisms. Next, we will show that
$(\cap _{\lambda } {\cal{A}}_{\lambda },\cap _{\lambda }{\cal{E}}_{\lambda })$
satisfies the axiom [E0]. Let
$A\in \cap _{\lambda } {\cal{A}}_{\lambda }$
. Since
$({\cal{A}}_{\lambda }, {\cal{E}}_{\lambda })$
is an exact category for all
$\lambda $
, by [Reference Bühler6, Lem. 2.7], we get that
is in
${\cal{E}}_{\lambda }$
for all
$\lambda $
. As
${\cal{A}}_{\lambda }$
is an additive subcategory of
$\cal{A}$
for all
$\lambda $
, so
$0_{\cal{A}}=0_{{\cal{A}}_{\lambda }}$
for all
$\lambda $
. Hence, by definition,
is in
$\cap _{\lambda }{\cal{E}}_{\lambda }$
, which implies that
$1_A$
is an admissible monic in
$\cap _{\lambda }{\cal{E}}_{\lambda }$
. So,
$(\cap _{\lambda } {\cal{A}}_{\lambda },\cap _{\lambda }{\cal{E}}_{\lambda })$
satisfies the axiom [E0]. Next, we will show that
$(\cap _{\lambda } {\cal{A}}_{\lambda },\cap _{\lambda }{\cal{E}}_{\lambda })$
satisfies the axiom [E0
$^{\text {op}}$
]. Let
$A\in \cap _{\lambda } {\cal{A}}_{\lambda }$
. Since
$({\cal{A}}_{\lambda }, {\cal{E}}_{\lambda })$
is an exact category for all
$\lambda $
, by [Reference Bühler6, Lem. 2.7], we get that
is in
${\cal{E}}_{\lambda }$
for all
$\lambda $
. As
${\cal{A}}_{\lambda }$
is an additive subcategory of
$\cal{A}$
for all
$\lambda $
, so
$0_{\cal{A}}=0_{{\cal{A}}_{\lambda }}$
for all
$\lambda $
. Hence, by definition
is in
$\cap _{\lambda }{\cal{E}}_{\lambda }$
, which implies that
$1_A$
is an admissible epic in
$\cap _{\lambda }{\cal{E}}_{\lambda }$
. So,
$(\cap _{\lambda } {\cal{A}}_{\lambda },\cap _{\lambda }{\cal{E}}_{\lambda })$
satisfies the axiom [E0
$^{\text {op}}$
]. Next, we will show that
$(\cap _{\lambda } {\cal{A}}_{\lambda },\cap _{\lambda }{\cal{E}}_{\lambda })$
satisfies the axiom [E1
$^{\text {op}}$
]. Let
and
be two admissible epics in
$\cap _{\lambda }{\cal{E}}_{\lambda }$
. Then we will show that
is an admissible epic in
$\cap _{\lambda }{\cal{E}}_{\lambda }$
. Now,
and
are admissible epics in
${\cal{E}}_{\lambda }$
for all
$\lambda $
. Since
$({\cal{A}}_{\lambda },{\cal{E}}_{\lambda })$
is an exact category for all
$\lambda $
, we get
is an admissible epic in
${\cal{E}}_{\lambda }$
for all
$\lambda $
. Hence, there exist objects
$D_{\lambda }\in {\cal{A}}_{\lambda }$
and kernel–cokernel pairs
in
${\cal{E}}_{\lambda }$
for all
$\lambda $
. So,
is in
$\cal{E}$
for all
$\lambda $
. Hence,
$\sigma _{\lambda }$
is a kernel–cokernel pair in
$\cal{A}$
for all
$\lambda $
, so by the universal property of kernels, we get that the kernel–cokernel pairs
$\sigma _{\lambda }$
’s are all isomorphic to each other. Hence, all the
$D_{\lambda }$
’s are isomorphic to each other. Now, fix a
$\lambda $
, say
$\lambda _0$
. Then
$D_{\lambda _0}\cong D_{\lambda }$
for all
$\lambda $
. Since
${\cal{A}}_{\lambda }$
is a strict subcategory of
$\cal{A}$
for all
$\lambda $
, we have
$D_{\lambda _0}\in \cap _{\lambda }{\cal{A}}_{\lambda }$
. Now,
${\cal{A}}_{\lambda }$
is a strictly full subcategory of
$\cal{A}$
for all
$\lambda $
, so
$D_{\lambda _0}\in {\cal{A}}_{\lambda }$
implies that
$\sigma _{\lambda _0}$
is a kernel–cokernel pair in
${\cal{A}}_{\lambda }$
for all
$\lambda $
. Since
${\cal{E}}_{\lambda }$
is closed under isomorphisms of kernel–cokernel pairs in
${\cal{A}}_{\lambda }$
and
$\sigma _{\lambda }$
’s are all isomorphic to each other, we get
$\sigma _{\lambda _0}\in {\cal{E}}_{\lambda }$
for all
$\lambda $
. So,
$\sigma _{\lambda _0}\in \cap _{\lambda }{\cal{E}}_{\lambda }$
. Hence,
is in
$\cap _{\lambda }{\cal{E}}_{\lambda }$
, so
is an admissible epic in
$\cap _{\lambda }{\cal{E}}_{\lambda }$
. Thus,
$(\cap _{\lambda }{\cal{A}}_{\lambda },\cap _{\lambda }{\cal{E}}_{\lambda })$
satisfies the axiom [E1
$^{\text {op}}$
]. Now, we will show that
$(\cap _{\lambda }{\cal{A}}_{\lambda },\cap _{\lambda }{\cal{E}}_{\lambda })$
satisfies the axiom [E2]. Let
be an admissible monic in
$\cap _{\lambda }{\cal{E}}_{\lambda }$
, and let
$A\xrightarrow {f}A'$
be an arbitrary morphism in
$\cap _{\lambda }{\cal{A}}_{\lambda }$
. Now,
is an admissible monic in
${\cal{E}}_{\lambda }$
for all
$\lambda $
. Since
$({\cal{A}}_{\lambda },{\cal{E}}_{\lambda })$
is an exact category for all
$\lambda $
, by [Reference Bühler6, Prop. 2.12(iv)], we have the following pushout commutative diagram with rows being kernel–cokernel pairs in
${\cal{E}}_{\lambda }$
:
Since
$({\cal{A}}_{\lambda },{\cal{E}}_{\lambda })$
is an exact subcategory of
$({\cal{A}},{\cal{E}})$
, by [Reference Bühler6, Prop. 5.2], we get that the square
is a pushout square in
$\cal{E}$
for all
$\lambda $
. Then, by the universal property of pushout, we get that the kernel–cokernel pairs
$\beta _{\lambda }$
’s are all isomorphic to each other, so all the
$B^{\prime }_{\lambda }$
’s are isomorphic to each other. Now, fix a
$\lambda $
, say
$\lambda _0$
. Then
$B^{\prime }_{\lambda _0}\cong B^{\prime }_{\lambda }$
for all
$\lambda $
. Since
${\cal{A}}_{\lambda }$
is a strict subcategory of
$\cal{A}$
for all
$\lambda $
, we have
$B^{\prime }_{\lambda _0}\in \cap _{\lambda }{\cal{A}}_{\lambda }$
. Now,
${\cal{A}}_{\lambda }$
is a strictly full subcategory of
$\cal{A}$
for all
$\lambda $
, so
$B^{\prime }_{\lambda _0}\in {\cal{A}}_{\lambda }$
implies that
$\beta _{\lambda _0}$
is a kernel–cokernel pair in
${\cal{A}}_{\lambda }$
for all
$\lambda $
. Since
${\cal{E}}_{\lambda }$
is closed under isomorphisms of kernel–cokernel pairs in
${\cal{A}}_{\lambda }$
and
$\beta _{\lambda }$
’s are all isomorphic to each other, we have
$\beta _{\lambda _0}\in {\cal{E}}_{\lambda }$
for all
$\lambda $
. So,
$\beta _{\lambda _0}\in \cap _{\lambda }{\cal{E}}_{\lambda }$
. Hence,
is in
$\cap _{\lambda }{\cal{E}}_{\lambda }$
, so
is an admissible monic in
$\cap _{\lambda }{\cal{E}}_{\lambda }$
. Thus,
$(\cap _{\lambda }{\cal{A}}_{\lambda },\cap _{\lambda }{\cal{E}}_{\lambda })$
satisfies the axiom [E2]. A dual argument will show that
$(\cap _{\lambda }{\cal{A}}_{\lambda },\cap _{\lambda }{\cal{E}}_{\lambda })$
satisfies the axiom [E2
$^{\text {op}}$
]. Hence,
$(\cap _{\lambda }{\cal{A}}_{\lambda },\cap _{\lambda }{\cal{E}}_{\lambda })$
is an exact subcategory of
$({\cal{A}},{\cal{E}})$
.
Next, we record a useful lemma for proving the exactness of certain structures on additive subcategories of a given exact category.
Lemma 3.3. Let
$({\cal{A}},{\cal{E}})$
be an exact category. Let
${\cal{A}}'$
be a strictly full additive subcategory of
$\cal{A}$
. Define
. If the pullback (resp. pushout) in
$\cal{A}$
of every
$\mathcal E$
-deflation (resp. inflation) of sequences in
$\mathcal E|_{\cal{A}'}$
along every morphism in
$\mathcal A'$
is again in
$\mathcal A'$
, then
$({\cal{A}}',{\cal{E}}|_{\cal{A}'})$
is an exact subcategory of
$({\cal{A}},{\cal{E}})$
.
The proof of Lemma 3.3 depends on Lemma 3.1 and the following proposition.
Proposition 3.4. Let
$({\cal{A}},{\cal{E}})$
be an exact category. Suppose that we have the following diagram:
where the square commutative diagram is a pullback diagram and
$e,p$
are admissible epics and i is a kernel of p. Then
$i'$
is an admissible monic with a cokernel given by
.
Proof. The proof follows from the construction in the proof of [Reference Bühler6, Prop. 2.15]. The only missing point in the proof of [Reference Bühler6, Prop. 2.15], toward showing
$i'$
is a kernel of
$p\circ e$
, is the following: it was not shown that
$(p\circ e)\circ i'=0$
. This can be easily checked as follows:
$(p\circ e)\circ i'=p\circ (e\circ i')=p\circ (i\circ e')=(p\circ i)\circ e'=0\circ e'=0$
, since i is a kernel of p.
Proof of Lemma 3.3
Clearly, each kernel–cokernel pair in
${\cal{E}}|_{\cal{A}'}$
is a kernel–cokernel pair in
$\cal{A}$
, hence also a kernel–cokernel pair in
$\cal{A}'$
. Thus,
${\cal{E}}|_{\cal{A}'}$
consists of kernel–cokernel pairs in
$\cal{A}'$
. Now, we will show that
$({\cal{A}}',{\cal{E}}|_{\cal{A}'})$
satisfies the axiom [E2]. Let
$A,B,A'\in {\cal{A}}'$
, let
be an admissible monic in
${\cal{E}}|_{\cal{A}'}$
, and let
$A\xrightarrow {f}A'$
be an arbitrary morphism in
${\cal{A}}'$
. Now,
is an admissible monic in
$\cal{E}$
. Since
$({\cal{A}},{\cal{E}})$
is an exact category, by [Reference Bühler6, Prop. 2.12(iv)], we have the following pushout commutative diagram with rows being kernel–cokernel pairs in
$\cal{E}$
:
Now, by the assumption, we have
$B'\in {\cal{A}}'$
. Since
is an admissible monic in
${\cal{E}}|_{\cal{A}'}$
, we have
$C\in {\cal{A}}'$
. Thus,
$\beta \in {\cal{E}}|_{\cal{A}'}$
. Since
$\cal{A}'$
is a strictly full subcategory of
$\cal{A}$
, the above diagram is a pushout diagram in
$\cal{A}'$
as well. Thus,
$({\cal{A}}',{\cal{E}}|_{\cal{A}'})$
satisfies the axiom [E2]. Similarly,
$({\cal{A}}',{\cal{E}}|_{\cal{A}'})$
satisfies the axiom [E2
$^{\text {op}}$
]. Now, by [Reference Bühler6, Def. 2.1] and [Reference Bühler6, Rem. 2.4], it is enough to show that
${\cal{E}}|_{\cal{A}'}$
is closed under isomorphisms and
$({\cal{A}}',{\cal{E}}|_{\cal{A}'})$
satisfies the axioms [E0], [E0
$^{\text {op}}$
], and [E1
$^{\text {op}}$
]. First, we will show that
${\cal{E}}|_{\cal{A}'}$
is closed under isomorphisms. Let
be a kernel–cokernel pair in
${\cal{E}}|_{\cal{A}'}$
, so
$M,N,L\in {\cal{A}}'$
and
is a kernel–cokernel pair in
$\cal{A}$
. Also, let
$M'\to N'\to L'$
be a kernel–cokernel pair in
$\cal{A}'$
such that it is isomorphic to
. Hence, by Lemma 3.1, we get that
$M'\to N'\to L'$
is a kernel–cokernel pair in
$\cal{A}$
. Since
$({\cal{A}},{\cal{E}})$
is an exact category, we have
is in
${\cal{E}}$
. Hence, by definition,
is in
${\cal{E}}|_{\cal{A}'}$
. Thus,
${\cal{E}}|_{\cal{A}'}$
is closed under isomorphisms. Next, we will show that
$({\cal{A}}',{\cal{E}}|_{\cal{A}'})$
satisfies the axiom [E0]. Let
$A\in {\cal{A}}'$
. Since
$({\cal{A}},{\cal{E}})$
is an exact category, by [Reference Bühler6, Lem. 2.7], we get that
is in
${\cal{E}}$
. As
$\cal{A}'$
is an additive subcategory of
$\cal{A}$
, so
$0_{\cal{A}}=0_{\cal{A}'}$
. Hence, by definition,
is in
${\cal{E}}|_{\cal{A}'}$
, which implies that
$1_A$
is an admissible monic in
${\cal{E}}|_{\cal{A}'}$
. So,
$({\cal{A}}',{\cal{E}}|_{\cal{A}'})$
satisfies the axiom [E0]. Next, we will show that
$({\cal{A}}',{\cal{E}}|_{\cal{A}'})$
satisfies the axiom [E0
$^{\text {op}}$
]. Let
$A\in {\cal{A}}'$
. Since
$({\cal{A}},{\cal{E}})$
is an exact category, by [Reference Bühler6, Lem. 2.7], we get that
is in
${\cal{E}}$
. As
$\cal{A}'$
is an additive subcategory of
$\cal{A}$
, so
$0_{\cal{A}}=0_{\cal{A}'}$
. Hence, by definition,
is in
${\cal{E}}|_{\cal{A}'}$
, which implies that
$1_A$
is an admissible epic in
${\cal{E}}|_{\cal{A}'}$
. So,
$({\cal{A}}',{\cal{E}}|_{\cal{A}'})$
satisfies the axiom [E0
$^{\text {op}}$
]. Next, we will show that
$({\cal{A}}',{\cal{E}}|_{\cal{A}'})$
satisfies the axiom [E1
$^{\text {op}}$
]. Let
and
be two admissible epics in
$\cal{E}$
such that they are in
${\cal{E}}|_{\cal{A}'}$
, which means that there exist two kernel–cokernel pairs
and
in
$\cal{E}$
such that
$A,B,C,D,B'\in {\cal{A}}'$
. Then we will show that
is an admissible epic in
${\cal{E}}|_{\cal{A}'}$
. From Proposition 3.4, we get that
is a kernel–cokernel pair in
$\cal{E}$
. By the hypothesis of Lemma 3.3 and the diagram of Proposition 3.4, we get that
$A'\in {\cal{A}}'$
. Hence,
is an admissible epic in
${\cal{E}}|_{\cal{A}'}$
. Thus,
$({\cal{A}}',{\cal{E}}|_{\cal{A}'})$
satisfies the axiom [E1
$^{\text {op}}$
]. So,
$({\cal{A}}',{\cal{E}}|_{\cal{A}'})$
is an exact subcategory of
$({\cal{A}},{\cal{E}})$
.
From now on, given a subcategory
$\cal{A}'$
of an exact category
$({\cal{A}},{\cal{E}})$
, the notation
${\cal{E}}|_{\cal{A}'}$
will stand for as defined in Lemma 3.3. Using Lemma 3.3, we now record two quick consequences, which give a sufficient condition on a subcategory
$\cal{A}'$
such that
$({\cal{A}}',{\cal{E}}|_{\cal{A}'})$
is an exact subcategory of
$({\cal{A}},{\cal{E}})$
. The first of which we state below now is well known (see, e.g., [Reference Bühler6, Lem. 10.20]). However, due to the absence of a proof in [Reference Bühler6, Lem. 10.20], we give a proof using our Lemma 3.3.
Proposition 3.5. (cf. [Reference Bühler6, Lem. 10.20]) Let
$({\cal{A}},{\cal{E}})$
be an exact category. Let
$\cal{A}'$
be a strictly full additive subcategory of
$\cal{A}$
. Assume that for every
$X\to Y\to Z$
in
${\cal{E}}$
, if
$X,Z\in {\cal{A}}'$
, then
$Y\in {\cal{A}}'$
(i.e.,
$\cal{A}'$
is closed under extensions). Then
$({\cal{A}}',{\cal{E}}|_{\cal{A}'})$
is an exact subcategory of
$({\cal{A}},{\cal{E}})$
.
Proof. By Lemma 3.3, it is enough to show that the pullback (resp. pushout) in
$\cal{A}$
of every
$\mathcal E$
-deflation (resp. inflation) of sequences in
$\mathcal E|_{\cal{A}'}$
along every morphism in
$\mathcal A'$
is again in
$\mathcal A'$
. Let
be an inflation in
$\mathcal E|_{\cal{A}'}$
, and let
$f:A\to A'$
be a morphism in
$\mathcal A'$
. Then, by [Reference Bühler6, Prop. 2.12(iv)], we get the following pushout commutative diagram:
Since
$A',C\in {\cal{A}}'$
, by assumption and the bottom row of the above diagram, we get that
$B'\in {\cal{A}}'$
. The pullback case follows by a dual argument. Hence,
$({\cal{A}}',{\cal{E}}|_{\cal{A}'})$
is an exact subcategory of
$({\cal{A}},{\cal{E}})$
.
Theorem 3.6. Let
$({\cal{A}},{\cal{E}})$
be an exact category. Let
$\cal{A}'$
be a strictly full additive subcategory of
$\cal{A}$
. Assume that
$\cal{A}'$
is closed under kernels and co-kernels of admissible epics and monics in
$\cal{E}$
, respectively. Then,
$({\cal{A}}',{\cal{E}}|_{\cal{A}'})$
is an exact subcategory of
$({\cal{A}},{\cal{E}})$
.
Proof. By Lemma 3.3, it is enough to show that the pullback (resp. pushout) in
$\cal{A}$
of every
$\mathcal E$
-deflation (resp. inflation) of sequences in
$\mathcal E|_{\cal{A}'}$
along every morphism in
$\mathcal A'$
is again in
$\mathcal A'$
. Let
be an inflation in
$\mathcal E|_{\cal{A}'}$
, and let
$f:A\to A'$
be a morphism in
$\mathcal A'$
. Now, we have the following pushout commutative diagram in
$(A,{\cal{E}})$
:
Then, by [Reference Bühler6, Prop. 2.12(ii)], we have the following kernel–cokernel pair in
${\cal{E}}$
:
. As
$B,A'\in {\cal{A}}'$
and
$\cal{A}'$
is additive, so
$B\oplus A'\in {\cal{A}}'$
. Since
is an admissible monic in
$\cal{E}$
, we get the cokernel
$B'\in {\cal{A}}'$
. The pullback case follows by a dual argument. Hence,
$({\cal{A}}',{\cal{E}}|_{\cal{A}'})$
is an exact subcategory of
$({\cal{A}},{\cal{E}})$
.
Given any exact category
$({\cal{A}},{\cal{E}})$
and
$C,A\in {\cal{A}}$
, one can define the Yoneda Ext group
${\operatorname{Ext}}_{\cal{E}}(C,A)$
, which has an abelian group structure by Baer sum (see the beginning of [Reference Dräxler, Reiten, Smalø and Solberg14, §1.2] for a description of
${\operatorname{Ext}}_{\cal{E}}(-,-)$
). When
$({\cal{A}},{\cal{E}})$
is moreover an R-linear category, then
${\operatorname{Ext}}_{\cal{E}}(C,A)$
can be given an R-linear structure via either of the following constructions, both of which yield equivalent triples in
${\operatorname{Ext}}_{\cal{E}}(C,A)$
:
Given a kernel–cokernel pair
in
${\operatorname{Ext}}_{\cal{E}}(C,A)$
, the multiplication
$r\cdot \sigma $
is either given by the following pullback diagram:
or by the following pushout diagram:
That both of these yield equivalent triplet in
${\operatorname{Ext}}_{\cal{E}}(-,-)$
follows from [Reference Bühler6, Prop. 3.1]. Moreover, this makes
${\operatorname{Ext}}_{\cal{E}}(-,-)$
into an R-linear functor in each component.
Now, we show that if
$({\cal{A}}',{\cal{E}'})$
is a strictly full exact subcategory of an exact category
$({\cal{A}},{\cal{E}})$
, then
${\operatorname{Ext}}_{\cal{E}'}(-,-):{\cal{A}}^{\prime \operatorname{op}}\times {\cal{A}}'\to \mathbf {Ab}$
is naturally a subfunctor of
${\operatorname{Ext}}_{\cal{E}}(-,-)|_{{\cal{A}}^{\prime \operatorname{op}}\times {\cal{A}}'}:{\cal{A}}^{\prime \operatorname{op}}\times {\cal{A}}'\to \mathbf {Ab}$
. For this, we first record a remark.
Remark 3.7. Let
$({\cal{A}}',{\cal{E}} ')$
be a strictly full exact subcategory of
$(\cal{A},{\cal{E}})$
. Then, for
$\sigma \in {\cal{E}} ' (\subseteq {\cal{E}})$
, it is actually true that
$[\sigma ]_{\cal{E} '}=[\sigma ]_{\cal{E}}$
; hence, the map
${\operatorname{Ext}}_{\cal{E} '}(-,-)\xrightarrow {[\sigma ]_{\cal{E} '}\mapsto [\sigma ]_{\cal{E}}} {\operatorname{Ext}}_{\cal{E}}(-,-)$
is the natural inclusion map. Indeed, to see
$[\sigma ]_{\cal{E} '}=[\sigma ]_{\cal{E}}$
: let
$\sigma $
be a kernel–cokernel pair
in
${\cal{E}}'$
, so
$A,B,C\in {\cal{A}}'$
. Let
$\beta \in [\sigma ]_{\cal{E}}$
be the kernel–cokernel pair
in
$\cal{E}$
, so
$B'\in {\cal{A}}$
. Hence, there exists
$f\in \operatorname{Mor}_{\cal{A}}(B,B')$
such that we have the following commutative diagram in
$\cal{A}$
:
By [Reference Bühler6, Cor. 3.2], we get that f is an isomorphism. Since
$B\in {\cal{A}}'$
,
$B'\in {\cal{A}}$
, and
$\cal{A}'$
is a strict subcategory of
$\cal{A}$
,
$B\cong B'$
implies that
$B'\in {\cal{A}}'$
. Since
$\cal{A}'$
is full,
$f\in \operatorname{Mor}_{\cal{A}'}(B,B')$
. Hence, the above commutative diagram is in
$\cal{A}'$
. Since
$({\cal{A}}',{\cal{E}}')$
is an exact category,
${\cal{E}}'$
is closed under isomorphisms. Hence,
$\beta \in {\cal{E}}'$
and
$\beta \in [\sigma ]_{\cal{E}'}$
. Thus,
$[\sigma ]_{\cal{E}}\subseteq [\sigma ]_{\cal{E}'}$
. Now,
$({\cal{A}}',{\cal{E}}')$
is a subcategory of
$({\cal{A}},{\cal{E}})$
, so
$[\sigma ]_{\cal{E}'}\subseteq [\sigma ]_{\cal{E}}$
as well. Hence,
$[\sigma ]_{\cal{E}'}=[\sigma ]_{\cal{E}}$
.
Proposition 3.8. Let
$({\cal{A}},{\cal{E}})$
be an exact category. Let
$({\cal{A}}',{\cal{E}}')$
be a strictly full exact subcategory of
$({\cal{A}},{\cal{E}})$
. Then
${\operatorname{Ext}}_{\cal{E}'}(-,-):{\cal{A}}^{\prime \operatorname{op}}\times {\cal{A}}'\to \mathbf {Ab}$
is a subfunctor of
${\operatorname{Ext}}_{\cal{E}}(-,-)|_{{\cal{A}}^{\prime \operatorname{op}}\times {\cal{A}}'}:{\cal{A}}^{\prime \operatorname{op}}\times {\cal{A}}'\to \mathbf {Ab}$
, where for every
$C,A\in {\cal{A}}'$
, the natural inclusion map
${\operatorname{Ext}}_{\cal{E}'}(C,A)\to {\operatorname{Ext}}_{\cal{E}}(C,A)$
is given by
$[\sigma ]_{\cal{E}'} \mapsto [\sigma ]_{\cal{E}}$
. If
${\cal{A}}$
is moreover R-linear, then the natural inclusion map is also R-linear; hence,
${\operatorname{Ext}}_{\cal{E}'}(-,-):{\cal{A}'}^{\operatorname{op}}\times {\cal{A}'}\to {\operatorname{mod}}\ R$
is a subfunctor of
${\operatorname{Ext}}_{\cal{E}}(-,-)|_{{\cal{A}'}^{\operatorname{op}}\times {\cal{A}}'}:{\cal{A}'}^{\operatorname{op}}\times {\cal{A}}'\to {\operatorname{mod}}\ R$
.
Proof. Let
$C,A\in {\cal{A}}'$
. Define
$\phi _{C,A}:{\operatorname{Ext}}_{\cal{E}'}(C,A)\to {\operatorname{Ext}}_{\cal{E}}(C,A)$
by
$\phi _{C,A}([\sigma ]_{\cal{E}'})=[\sigma ]_{\cal{E}}$
. From now on, we will call this map
$\phi $
. The well-definedness and injectivity of
$\phi $
follow from Remark 3.7. Next, we will show that
$\phi $
is a group homomorphism. Let
$[\sigma ]_{\cal{E}'},[\beta ]_{\cal{E}'}\in {\operatorname{Ext}}_{\cal{E}'}(C,A)$
, where
and
for some
$B,B'\in {\cal{A}}'$
. Then
$[\sigma ]_{\cal{E}'}+[\beta ]_{\cal{E}'}=[\alpha ]_{\cal{E}'}$
is given by the following commutative diagram in
$({\cal{A}}',{\cal{E}}')$
:
We know that
$\gamma $
is the pushout of
$\sigma \oplus \beta $
in
$\cal{E}'$
by the sum map
$A\oplus A\xrightarrow {\Sigma }A$
. Since
$({\cal{A}}',{\cal{E}}')$
is an exact subcategory of
$({\cal{A}},{\cal{E}})$
, by [Reference Bühler6, Prop. 5.2], we get that
$\gamma $
is also the pushout of
$\sigma \oplus \beta $
in
${\cal{E}}$
by the sum map
$A\oplus A\xrightarrow {\Sigma }A$
. Next,
$\alpha $
is the pullback of
$\gamma $
in
$\cal{E}'$
by the diagonal map
$C\xrightarrow {\Delta } C\oplus C$
, so by [Reference Bühler6, Prop. 5.2] we get that
$\alpha $
is also the pullback of
$\gamma $
in
$\cal{E}$
by the diagonal map
$C\xrightarrow {\Delta } C\oplus C$
. This implies that
$\alpha $
is a representative of
$[\sigma ]_{\cal{E}}+[\beta ]_{\cal{E}}$
in
${\operatorname{Ext}}_{\cal{E}}(C,A)$
, that is,
$[\alpha ]_{\cal{E}}=[\sigma ]_{\cal{E}}+[\beta ]_{\cal{E}}$
. Since
$\phi ([\sigma ]_{\cal{E}'}+[\beta ]_{\cal{E}'})=\phi ([\alpha ]_{\cal{E}'})=[\alpha ]_{\cal{E}}$
and
$\phi ([\sigma ]_{\cal{E}'})+\phi ([\beta ]_{\cal{E}'})=[\sigma ]_{\cal{E}}+[\beta ]_{\cal{E}}$
, we get
$\phi $
is a group homomorphism.
Let
$A,B,C\in {\cal{A}}'$
, and let
$f\in \operatorname{Mor}_{\cal{A}'}(A,B)$
. First, we will show that
${\operatorname{Ext}}_{\cal{E}'}(C,-)$
is a subfunctor of
${\operatorname{Ext}}_{\cal{E}}(C,-)$
, so we need to prove that the following diagram commutes:
where for every
$[\sigma ]_{\cal{E}} \in {\operatorname{Ext}}_{\cal{E}}(C,A)$
, a representative of
$f^*([\sigma ]_{\cal{E}})$
is the pushout of
$\sigma $
by f in
$\cal{E}$
and for every
$[\beta ]_{\cal{E}'} \in {\operatorname{Ext}}_{\cal{E}'}(C,A)$
, a representative of
$\tilde {f}([\beta ]_{\cal{E}'})$
is the pushout of
$\beta $
by f in
$\cal{E}'$
. Let
$[\sigma ]_{\cal{E}'}\in {\operatorname{Ext}}_{\cal{E}'}(C,A)$
. Now,
$\phi _{C,A}([\sigma ]_{\cal{E}'})=[\sigma ]_{\cal{E}}$
and
$f^*([\sigma ]_{\cal{E}})=[\gamma ]_{\cal{E}}$
, where
$\gamma $
is the pushout of
$\sigma $
by f in
$\cal{E}$
. Next, we have
$\tilde {f}([\sigma ]_{\cal{E}'})=[\alpha ]_{\cal{E}'}$
, where
$\alpha $
is the pushout of
$\sigma $
by f in
$\cal{E}'$
. Hence, by [Reference Bühler6, Prop. 5.2], we get that
$\alpha $
is also the pushout of
$\sigma $
by f in
$\cal{E}$
. Hence,
$[\alpha ]_{\cal{E}}=[\gamma ]_{\cal{E}}$
, so
$\phi _{C,B}(\tilde {f}([\sigma ]_{\cal{E}'}))=\phi _{C,B}([\alpha ]_{\cal{E}'})=[\alpha ]_{\cal{E}}=[\gamma ]_{\cal{E}}=f^*([\sigma ]_{\cal{E}})=f^*(\phi _{C,A}([\sigma ]_{\cal{E}'}))$
, and therefore the above diagram commutes. Hence,
${\operatorname{Ext}}_{\cal{E}'}(C,-)$
is a subfunctor of
${\operatorname{Ext}}_{\cal{E}}(C,-)$
for any
$C\in {\cal{A}}'$
. Similarly,
${\operatorname{Ext}}_{\cal{E}'}(-,C)$
is a subfunctor of
${\operatorname{Ext}}_{\cal{E}}(-,C)$
for any
$C\in {\cal{A}}'$
. Thus,
${\operatorname{Ext}}_{\cal{E}'}(-,-):{\cal{A}}^{\prime \operatorname{op}}\times {\cal{A}}'\to \mathbf {Ab}$
is a subfunctor of
${\operatorname{Ext}}_{\cal{E}}(-,-)|_{{\cal{A}}^{\prime \operatorname{op}}\times {\cal{A}}'}:{\cal{A}}^{\prime \operatorname{op}}\times {\cal{A}}'\to \mathbf {Ab}$
. Now, let
$\cal{A}$
be R-linear; hence, so is
$\cal{A}'$
because it is a full subcategory of
$\cal{A}$
. Let
$C,A\in {\cal{A}}'$
, and let
$[\sigma ]_{\cal{E}'}\in {\operatorname{Ext}}_{\cal{E}'}(C,A)$
. Let
$[\gamma ]_{\cal{E}'}=r\cdot [\sigma ]_{\cal{E}'}$
, so
$\gamma $
is obtained from
$\sigma $
by pullback in
$\cal{E}'$
along the map
$C\xrightarrow {r\cdot \text {id}_C} C$
. Now,
$({\cal{A}}',{\cal{E}}')$
is an exact subcategory of
$({\cal{A}},{\cal{E}})$
; hence, this is also the pullback in
$\cal{E}$
along the map
$C\xrightarrow {r\cdot \text {id}_C} C$
(see [Reference Bühler6, Prop. 5.2]). So,
$[\gamma ]_{\cal{E}}=r\cdot [\sigma ]_{\cal{E}}=r\phi ([\sigma ]_{\cal{E}'})$
. Also,
$[\gamma ]_{\cal{E}}=\phi ([\gamma ]_{\cal{E}'})=\phi (r\cdot [\sigma ]_{\cal{E}'})$
. This shows that the natural inclusion map
$\phi $
is R-linear.
In the following corollary, we denote
$\operatorname{Hom}_{\cal{A}}(-,-)$
just by
$\operatorname{Hom}(-,-)$
(we completely ignore the subcategory, since all our subcategories are full). In view of Definition 4.10, and the discussion following 4.11, the following corollary is crucial for recovering [Reference Striuli35, Props. 1.37 and 1.38].
Corollary 3.9. Let
$({\cal{A}},{\cal{E}})$
be an exact category. Let
$({\cal{A}}',{\cal{E}}')$
be a strictly full exact subcategory of
$({\cal{A}},{\cal{E}})$
. Let
be a kernel–cokernel pair in
$\cal{E}'$
. Then, for every
$A\in {\cal{A}}'$
, we have the following commutative diagrams of long exact sequences:
and
Proof. We will only prove the covariant version (the first diagram above), since the proof of the contravariant version is given by the dual argument. In the first diagram above, the commutativity of the first two squares is obvious. Also, the commutativity of the last two squares follows directly from the proof of Proposition 3.8 (i.e.,
${\operatorname{Ext}}_{\cal{E}'}(-,-)$
is a subfunctor of
${\operatorname{Ext}}_{\cal{E}}(-,-)$
). So, it is enough to prove that the following square is commutative:
Now,
$f(g)=[\gamma ]_{\cal{E}}$
, where
$\gamma $
is the pullback of
$\sigma $
by g in
$\cal{E}$
and
$f^*(g)=[\alpha ]_{\cal{E}'}$
, where
$\alpha $
is the pullback of
$\sigma $
by g in
$\cal{E}'$
. Since
$({\cal{A}}',{\cal{E}}')$
is an exact subcategory of
$({\cal{A}},{\cal{E}})$
, by [Reference Bühler6, Prop. 5.2], we get that
$[\gamma ]_{\cal{E}}=[\alpha ]_{\cal{E}}$
. Also, by definition,
$\phi _{A,M}([\alpha ]_{\cal{E}'})=[\alpha ]_{\cal{E}}$
. Thus,
$f(g)=\phi _{A,M}(f^*(g))$
, so the above square commutes.
Let
$({\cal{A}},{\cal{E}})$
be an exact category. Let
$\cal{A}'$
be a strictly full subcategory of
$\cal{A}$
. If
and
are two kernel–cokernel pairs in
$\cal{E}$
belonging to the same class in
${\operatorname{Ext}}_{\cal{E}}(A_2,A_1)$
, then
$B\cong C$
(by short five lemma); hence,
$B\in {\cal{A}}'$
if and only if
$C\in {\cal{A}}'$
. Hence, for every
$M,N\in {\cal{A}}'$
, the collection
$\{[\sigma ]_{\cal{E}}\in {\operatorname{Ext}}_{\cal{E}}(M,N) : \text {the middle object of } \sigma \text { is in } {\cal{A}}'\}$
is well defined.
Proposition 3.10. Let
$({\cal{A}},{\cal{E}})$
be an exact category. Let
$\cal{A}'$
be a strictly full additive subcategory of
$\cal{A}$
. For every
$M,N\in {\cal{A}}'$
, define
$F(M,N):=\{[\sigma ]_{\cal{E}}\in {\operatorname{Ext}}_{\cal{E}}(M,N) : \text { the middle object of } \sigma \text { is in } {\cal{A}}'\}$
. If
$F(-,-)$
is a subfunctor of
${\operatorname{Ext}}_{\cal{E}}(-,-)|_{{\cal{A}}^{\prime \operatorname{op}}\times {\cal{A}}'}:{\cal{A}}^{\prime \operatorname{op}}\times {\cal{A}}'\to \mathbf {Ab}$
via the natural inclusion maps, then
$({\cal{A}}',{\cal{E}}|_{\cal{A}'})$
is an exact subcategory of
$({\cal{A}},{\cal{E}})$
.
Proof. By Lemma 3.3, it is enough to show that the pullback (resp. pushout) in
$\cal{A}$
of every
$\cal{E}$
-deflation (resp. inflation) of sequences in
${\cal{E}}|_{\cal{A}'}$
along every morphism in
$\cal{A}'$
is again in
$\cal{A}'$
. We will only prove the pullback case, since the proof of the pushout case can be given by a similar argument. Let
be an admissible epic in
$\cal{E}$
such that it is in
${\cal{E}}|_{\cal{A}'}$
, which means that there exist a kernel–cokernel pair
in
$\cal{E}$
such that
$A, B,C\in {\cal{A}}'$
. Also, let
$B'\xrightarrow {f} C$
be a morphism in
$\cal{A}'$
. Then we have a map
$f^*:{\operatorname{Ext}}_{\cal{E}}(C,A)\to {\operatorname{Ext}}_{\cal{E}}(B',A)$
defined as follows: for every
$[\sigma ]_{\cal{E}} \in {\operatorname{Ext}}_{\cal{E}}(C,A)$
, a representative of
$f^*([\sigma ]_{\cal{E}})$
is the pullback of
$\sigma $
by f in
$({\cal{A}},{\cal{E}})$
. Now, by definition of
$F(-,-)$
, it is clear that
$F(M,N)\subseteq {\operatorname{Ext}}_{\cal{E}}(M,N)$
for any
$M,N\in {\cal{A}}'$
. Since
$F(-,-)$
is a subfunctor of
${\operatorname{Ext}}_{\cal{E}}(-,-)|_{{\cal{A}}^{\prime \operatorname{op}}\times {\cal{A}}'}:{\cal{A}}^{\prime \operatorname{op}}\times {\cal{A}}'\to \mathbf {Ab}$
, we have the following commutative square:
where the columns are natural inclusion maps. So, for every
$[\beta ]_{\cal{E}} \in F(C,A)$
, a representative of
$\tilde {f}([\beta ]_{\cal{E}})$
is the pullback of
$\beta $
by f in
$({\cal{A}},{\cal{E}})$
. Now, consider the pullback of
$\gamma $
by f in
$({\cal{A}},{\cal{E}})$
as follows:
Now,
$[\gamma ]_{\cal{E}}\in F(C,A)$
, so
$\tilde {f}([\gamma ]_{\cal{E}})\in F(B',A)$
implies that
$A'\in {\cal{A}}'$
. Hence, the pullback in
$\cal{A}$
of every
$\cal{E}$
-deflation of sequences in
${\cal{E}}|_{\cal{A}'}$
along every morphism in
$\cal{A}'$
is again in
$\cal{A}'$
. Hence, by Lemma 3.3, we get that
$({\cal{A}}',{\cal{E}}|_{\cal{A}'})$
is an exact subcategory of
$({\cal{A}},{\cal{E}})$
.
4 Subcategories and subfunctors of
${\operatorname{Ext}}^1$
from numerical functions and applications to module categories
In this section, we present tools to identify exact subcategories of an exact category coming from certain numerical functions. Consequently, due to Proposition 3.8, we are also able to identify subfunctors of
${\operatorname{Ext}}^1$
associated with certain numerical functions. Our first result in this direction is an application of Theorem 3.6.
Proposition 4.1. Let
$({\cal{A}},{\cal{E}})$
be an exact category. Let
$\phi :{\cal{A}}\to \mathbb {Z}_{\le 0}$
be a function such that
$\phi $
is constant on isomorphism classes of objects in
$\cal{A}$
,
$\phi $
is additive on finite biproducts, and
$\phi $
is sub-additive on kernel–cokernel pairs in
$\cal{E}$
(i.e., if
is in
$\cal{E}$
, then
$\phi (N)\le \phi (M)+\phi (L)$
). Let
$\cal{A}'$
be the strictly full subcategory of
$\cal{A}$
, whose objects are given by
$\{M\in {\cal{A}}: \phi (M)=0\}$
. Then
$({\cal{A}}',{\cal{E}}|_{\cal{A}'})$
is an exact subcategory of
$({\cal{A}},{\cal{E}})$
.
Proof. Note that
$\cal{A}'$
is a strict subcategory of
$\cal{A}$
since
$\phi $
is constant on isomorphism classes of objects in
$\cal{A}$
. Since
$\phi $
is additive on finite biproduct,
$\phi (0_{\cal{A}})=\phi (0_{\cal{A}}\oplus 0_{\cal{A}})=\phi (0_{\cal{A}})+\phi (0_{\cal{A}})$
, and hence
$\phi (0_{\cal{A}})=0$
. Hence,
$0_{\cal{A}}\in {\cal{A}}'$
. Since
$\cal{A}$
is an additive category and
$\phi $
is additive on finite biproducts, by definition of
$\cal{A}'$
, we get that for every
$X,Y\in {\cal{A}}'$
, the biproduct of X and Y in
$\cal{A}$
also belongs to
$\cal{A}'$
. Hence,
$\cal{A}'$
is an additive subcategory of
$\cal{A}$
. Let
$B\in {\cal{A}}'$
and
be a kernel–cokernel pair in
$\mathcal E$
. We will show that
$A_1, A_2 \in {\cal{A}}'$
. Since
$\phi $
is subadditive on kernel–cokernel pairs in
$\cal{E}$
and
$B\in {\cal{A}}'$
, we get
$0=\phi (B)\leq \phi (A_1)+\phi (A_2)\le 0$
. Hence,
$\phi (A_1)+\phi (A_2)=0$
. Since we know
$\phi $
always takes non-positive values, we get
$\phi (A_1)=\phi (A_2)=0$
; hence,
$A_1,A_2\in {\cal{A}}'$
. Thus,
$\cal{A}'$
is closed under kernels and co-kernels of admissible epics and monics in
$\cal{E}$
, respectively. Then, by Theorem 3.6, we get that
$({\cal{A}}',{\cal{E}}|_{\cal{A}'})$
is an exact subcategory of
$({\cal{A}},{\cal{E}})$
.
We now proceed to give our main example of Proposition 4.1. First, we recall some terminologies and prove some preliminary lemmas. For any unexpected concepts and notations, we refer the reader to [Reference Bruns and Herzog3] and [Reference Leuschke and Wiegand27].
Let R be a commutative ring, and let
${\cal{S}}_R$
be the collection of all short exact sequences of R-modules, which gives the standard exact structure on
${\operatorname{mod}}\ R$
. When the ring in question is clear, we drop the suffix R and write only
$\cal{S}$
. Note that since
${\operatorname{mod}}\ R$
is extension closed in
${\operatorname{mod}}\ R$
, the collection of all short exact sequences in
${\operatorname{mod}}\ R$
gives an exact subcategory of
${\operatorname{mod}}\ R$
, and we take this as the standard exact structure of
${\operatorname{mod}}\ R$
. Note that if
$({\cal{X}},{\cal{E}})$
is an exact subcategory of
${\operatorname{mod}}\ R$
and
${\cal{X}}\subseteq {\operatorname{mod}}\ R$
, then
$({\cal{X}},{\cal{E}})$
is also an exact subcategory of
${\operatorname{mod}}\ R$
.
Now, for a Noetherian local ring
$(R,\mathfrak {m})$
of dimension d and for an integer
$s\ge 0$
, let
$\operatorname{CM}^s(R)$
denote the full subcategory of
${\operatorname{mod}}\ R$
consisting of the zero-module, and all Cohen–Macaulay R-modules [Reference Bruns and Herzog3, Def. 2.1.1] of dimension s. Note that when
$s=d$
,
$\operatorname{CM}^d(R)$
is just the category of all maximal Cohen–Macaulay modules, which we will also denote by
$\operatorname{CM}(R)$
.
4.2. We quickly note that for each
$s\ge 0$
,
$\operatorname{CM}^s(R)$
is closed under finite direct sums, direct summands, and closed under extensions in
${\operatorname{mod}}\ R$
. Indeed, let
$0\to L \to M \to N \to 0$
be a short exact sequence with
$L,N\in \operatorname{CM}^s(R)$
. If L or N is zero, then there is nothing to prove. So, assume
$L,N$
are nonzero. Now,
$M\in {\operatorname{mod}}\ R$
, and we also have the following calculation for
$\dim M$
:
$\dim M = \dim \operatorname{Supp}(M)=\dim \left (\operatorname{Supp}(L)\cup \operatorname{Supp}(N)\right )=\max \{\dim \operatorname{Supp}(L),\dim \operatorname{Supp}(N)\}=\max \{\dim L, \dim N\}=s$
. Consequently,
$s=\inf \{\operatorname{depth}_R L, \operatorname{depth}_R N\}\le \operatorname{depth}_R M \le \dim M=s$
, which shows
$M\in \operatorname{CM}^s(R)$
. Thus,
$\operatorname{CM}^s(R)$
is closed under extensions in
${\operatorname{mod}}\ R$
. To also show
$\operatorname{CM}^s(R)$
is closed under direct summands, let
$M,N$
be nonzero modules with
$M\oplus N\in \operatorname{CM}^s(R)$
. Then,
$M,N\in {\operatorname{mod}}\ R$
, and
$s=\operatorname{depth}_R(M\oplus N)=\inf \{\operatorname{depth}_R M,\operatorname{depth}_R N\}\le \operatorname{depth}_R M\le \dim M\le \dim (M\oplus N)=s$
, and thus
$M\in \operatorname{CM}^s(R)$
.
Hence, if
$\cal{S}$
is the standard exact structure on
${\operatorname{mod}}\ R$
, then
$(\operatorname{CM}^s(R),{\cal{S}}|_{\operatorname{CM}^s(R)})$
is an exact subcategory of
${\operatorname{mod}}\ R$
by Proposition 3.5. Now, let I be an
$\mathfrak {m}$
-primary ideal and let
$\phi _I:\operatorname{CM}^s(R)\to \mathbb Z$
be the function defined by
$\phi _I(M):=\lambda _R(M/IM)-e_R(I,M)$
, where
$e_R(I,M)$
is the multiplicity of M with respect to I (see [Reference Bruns and Herzog3, Def. 4.6.1]). We first show that
$\phi _I$
satisfy all the hypothesis of Proposition 4.1. To prove this, we will need to pass to the faithfully flat extension
$S=R[X]_{\mathfrak {m}[X]}$
whose unique maximal ideal is
$\mathfrak {m} S$
. That we can harmlessly pass to this extension, is discussed in the following.
4.3. Let
$(R,\mathfrak {m})$
be a local ring, and consider
$S=R[X]_{\mathfrak {m}[X]}$
, which is a faithfully flat extension of R and the unique maximal ideal of S is
$\mathfrak {m} S$
, whose residue field
$S/\mathfrak {m} S$
is infinite (see [Reference Huneke and Swanson24, §8.4]). Let
$M\in {\operatorname{mod}} (R)$
. By [Reference Bruns and Herzog3, Th. 2.1.7] (and the sentences following it), we have
$M\in \operatorname{CM}^s(R)$
if and only if
$M\otimes _R S \in \operatorname{CM}^s(S)$
. Let I be an
$\mathfrak {m}$
-primary ideal of R. As
$\mathfrak {m} S$
is the maximal ideal of S, so
$IS=I\otimes _R S$
is also primary to the maximal ideal of S. Now,
$S\otimes _R \dfrac {M}{IM}\cong \dfrac {S\otimes _R M}{S\otimes _R (IM)}=\dfrac {S\otimes _R M}{(IS)(S\otimes _R M)}$
by Lemma 5.2.5(1). Hence,
$\lambda _S\left (\dfrac {S\otimes _R M}{(IS)(S\otimes _R M)}\right )=\lambda _S\left (S\otimes _R \dfrac {M}{IM} \right )=\lambda _R\left (\dfrac {M}{IM}\right )$
, where the last equality is by [38, Tag 02M1] remembering that
$S/\mathfrak {m} S$
is the residue field of S. Also,
$e_R(I,M):=\lim _{n\to \infty }\dfrac {(\dim M)!}{n^{\dim M}}\lambda _R\left (\dfrac {M}{I^{n+1}M}\right )$
, and
$\dim _S(M\otimes _R S)=\dim _R(M)$
[Reference Bruns and Herzog3, Th. A.11(b)], so a similar argument as the previous one shows
$e_R(I,M)=e_S(IS, S\otimes _R M)$
. Thus,
$\phi _I(M)=\phi _{IS}(S\otimes _R M)$
.
Lemma 4.4. The function
$\phi _I:\operatorname{CM}^s(R)\to \mathbb Z$
satisfies
$\phi _I(M)\le 0$
, is constant on isomorphism classes of modules in
$\operatorname{CM}^s(R)$
, additive on finite biproducts, and subadditive on short exact sequences of modules in
$\operatorname{CM}^s(R)$
.
Proof. It is obvious that
$\phi _I$
is constant on isomorphism classes of modules in
$\operatorname{CM}^s(R)$
, and additive on finite biproducts (direct sums). Now,
$e_R(I,-)$
is additive on short exact sequences of modules in
$\operatorname{CM}^s(R)$
by [Reference Bruns and Herzog3, Cor. 4.7.7]. Since
$\lambda _R((-)\otimes _R R/I)$
is always subadditive on short exact sequences, this proves
$\phi _I$
is also subadditive on short exact sequences of modules in
$\operatorname{CM}^s(R)$
. Now, we finally prove that
$\phi _I(M)\le 0$
for all
$M\in \operatorname{CM}^s(R)$
. If
$\dim M=0$
, then
$e_R(I,M)=\lambda _R(M)\ge \lambda _R(M/IM)$
, so
$\phi _I(M)\le 0$
. Now, assume that
$s=\dim M>0$
. We may assume that R has infinite residue field due to 4.3. By [Reference Bruns and Herzog3, Cor. 4.6.10], we have
$e_R(I,M)=e_R((\mathbf x),M)$
for some system of parameters
$\mathbf x=x_1,...,x_s$
on M, which is a reduction of I with respect to M. Then,
$\mathbf x$
is an M-regular sequence, since M is Cohen–Macaulay (see [Reference Bruns and Herzog3, Th. 2.1.2(d)]). Let
$J=(\mathbf x)$
. Since
$\mathbf x$
is an M-regular sequence, by [Reference Bruns and Herzog3, Th. 1.1.8], we have
$J^nM/J^{n+1}M\cong (M/JM)^{\oplus \dfrac {(s+n-1)!}{n!(s-1)!}}$
. Hence,
$e_R(I,M)=e_R(J,M)=(s-1)!\lim _{n\to \infty }\dfrac {\lambda _R(J^nM/J^{n+1}M)}{n^{s-1}}=\lambda _R(M/JM)\ge \lambda _R(M/I M)$
(the last inequality follows by noticing that
$J\subseteq I$
). Hence,
$\phi _I(M)\le 0$
.
Definition 4.5. Let
$(R,\mathfrak {m})$
be a Noetherian local ring, and let I be an
$\mathfrak {m}$
-primary ideal. Let
$s\ge 0$
be an integer. We denote by
$\operatorname{Ul}^s_I(R)$
the full subcategory of all modules
$M\in \operatorname{CM}^s(R)$
such that
$\phi _I(M)=0$
, that is,
$\lambda _R(M/IM)=e_R(I,M)$
. When
$s=\dim R$
, we will denote this subcategory simply by
$\operatorname{Ul}_I(R)$
. When
$I=\mathfrak {m}$
,
$\operatorname{Ul}^s_{\mathfrak {m}}(R)$
will be denoted by
$\operatorname{Ul}^s(R)$
. We will also denote
$\operatorname{Ul}^{\dim R}_{\mathfrak {m}}(R)$
simply by
$\operatorname{Ul}(R)$
. Note that when
$s=\dim R=1$
and R is Cohen–Macaulay,
$\operatorname{Ul}_I(R)$
is exactly the collection of all I-Ulrich modules as defined in [Reference Dao, Maitra and Sridhar11, Def. 4.1].
Remark 4.6. In terms of [Reference Goto, Takahashi and Taniguchi21, Def. 2.1], a nonzero R-module M is Ulrich if
$M\in \operatorname{Ul}^s(R)$
for some
$s\ge 0$
in our notation. When R is Cohen–Macaulay, then the modules in
$\operatorname{Ul}(R)$
are simply the maximally generated modules as studied in [Reference Brennan, Herzog and Ulrich2].
Corollary 4.7. For each integer
$s\ge 0$
and
$\mathfrak {m}$
-primary ideal I,
$(\operatorname{Ul}^s_I(R), {\cal{S}}|_{{\operatorname{Ul}}^s_I(R)})$
is an exact subcategory of
${\operatorname{mod}}\ R$
(hence of
${\operatorname{mod}}\ R$
). Hence,
${\operatorname{Ext}}^1_{\operatorname{Ul}^s_I(R)}(-,-): \operatorname{Ul}^s_I(R)^{op}\times \operatorname{Ul}^s_I(R)\to {\operatorname{mod}}\ R$
is a subfunctor of
${\operatorname{Ext}}^1_{R}(-,-): \operatorname{Ul}^s_I(R)^{op}\times \operatorname{Ul}^s_I(R)\to {\operatorname{mod}}\ R$
.
Proof. Since it has been noticed that
$(\operatorname{CM}^s(R),{\cal{S}}|_{\operatorname{CM}^s(R)})$
is an exact subcategory of
${\operatorname{mod}}\ R$
, by Lemma 4.4 and Proposition 4.1, it follows that
$(\operatorname{Ul}^s_I(R), {\cal{S}}|_{\operatorname{Ul}^s_I(R)})$
is an exact subcategory of
$(\operatorname{CM}^s(R),{\cal{S}}|_{\operatorname{CM}^s(R)})$
, and hence of
${\operatorname{mod}}\ R$
. The subfunctor part now follows from Proposition 3.8.
Next, we record a proposition for constructing a special kind of exact substructure of an exact structure
$\cal{E}$
on a category
$\cal{A}$
, without shrinking the category, that also comes from certain kinds of numerical functions. This is used in the next section to recover and improve some results on module categories.
Theorem 4.8. Let
$({\cal{A}},{\cal{E}})$
be an exact category. Let
$\phi :{\cal{A}} \to \mathbb Z$
be a function such that
$\phi $
is constant on isomorphism classes of objects in
$\cal{A}$
,
$\phi $
is additive on finite biproducts, and
$\phi $
is subadditive on kernel–cokernel pairs in
$\cal{E}$
(i.e., if
is in
$\cal{E}$
, then
$\phi (N)\le \phi (M)+\phi (L)$
). Set
${\cal{E}}^{\phi }:=\{\text {kernel--cokernel pairs in } {\cal{E}} \text { on which } \phi \text { is additive} \}$
. Then,
$({\cal{A}},{\cal{E}}^{\phi })$
is an exact subcategory of
$({\cal{A}},{\cal{E}})$
.
Proof. By [Reference Bühler6, Def. 2.1] and [Reference Bühler6, Rem. 2.4], it is enough to show that
${\cal{E}}^{\phi }$
is closed under isomorphisms and
$({\cal{A}},{\cal{E}}^{\phi })$
satisfies the axioms [E0], [E0
$^{\text {op}}$
], [E1
$^{\text {op}}$
], [E2], and [E2
$^{\text {op}}$
]. First, we will show that
${\cal{E}}^{\phi }$
is closed under isomorphisms. Let
be a kernel–cokernel pair in
${\cal{E}}^{\phi }$
. Also, let
$M'\to N'\to L'$
be a kernel–cokernel pair in
$\cal{A}$
such that it is isomorphic to
, which implies that
is in
$\cal{E}$
and
$M\cong M'$
,
$N\cong N'$
,
$L\cong L'$
. Since
$\phi (N)=\phi (M)+\phi (L)$
and
$\phi $
is constant on isomorphism classes of objects in
$\cal{A}$
, we have
$\phi (N')=\phi (N)=\phi (M)+\phi (L)=\phi (M')+\phi (L')$
. So,
is in
${\cal{E}}^{\phi }$
. Hence,
${\cal{E}}^{\phi }$
is closed under isomorphisms. Next, we will show that
$({\cal{A}},{\cal{E}}^{\phi })$
satisfies the axiom [E0]. Let
$A\in {\cal{A}}$
. Since
$({\cal{A}},{\cal{E}})$
is an exact category, by [Reference Bühler6, Lem. 2.7], we get that
is in
$\cal{E}$
. Since
$\phi $
is additive on finite biproducts,
$\phi $
is additive on the kernel–cokernel pair
. Hence, by definition,
is in
${\cal{E}}^{\phi }$
, which implies that
$1_A$
is an admissible monic in
${\cal{E}}^{\phi }$
. So,
$({\cal{A}},{\cal{E}}^{\phi })$
satisfies the axiom [E0]. Next, we will show that
$({\cal{A}},{\cal{E}}^{\phi })$
satisfies the axiom [E0
$^{\text {op}}$
]. Let
$A\in {\cal{A}}$
. Since
$({\cal{A}},{\cal{E}})$
is an exact category, by [Reference Bühler6, Lem. 2.7], we get that
is in
$\cal{E}$
. Since
$\phi $
is additive on finite biproducts,
$\phi $
is additive on the kernel–cokernel pair
. Hence, by definition,
is in
${\cal{E}}^{\phi }$
, which implies that
$1_A$
is an admissible epic in
${\cal{E}}^{\phi }$
. So,
$({\cal{A}},{\cal{E}}^{\phi })$
satisfies the axiom [E0
$^{\text {op}}$
]. Next, we will show that
$({\cal{A}},{\cal{E}}^{\phi })$
satisfies the axiom [E1
$^{\text {op}}$
]. Let
and
be two admissible epics in
${\cal{E}}^{\phi }$
. Then we will show that
is an admissible epic in
${\cal{E}}^{\phi }$
. Let
be the complete kernel–cokernel pair of p, and consider a pullback square as in Proposition 3.4. From Proposition 3.4, we get that
is a kernel–cokernel pair in
$\cal{E}$
, so
$\phi (B')\leq \phi (C)+\phi (A')$
.
Now, consider the pullback square of Proposition 3.4
By the dual (pullback version) of [Reference Bühler6, Prop. 2.12(i)
$\Rightarrow$
(iv)] we get a commutative diagram in
$\cal{E}$
as follows:
The top row gives
$\phi (A')\leq \phi (A)+\phi (D)$
. Thus,
$\phi (C)+\phi (A')\leq \phi (A)+\phi (C)+\phi (D)$
. Since
is an admissible epic in
${\cal{E}}^{\phi }$
and
is in
$\cal{E}$
and
$\phi (-)$
is constant on isomorphism classes of objects, we have
$\phi (A)+\phi (C)=\phi (B)$
. Thus,
$\phi (C)+\phi (A')\leq \phi (A)+\phi (C)+\phi (D)$
implies
$\phi (C)+\phi (A')\leq \phi (B)+\phi (D)$
. Since
is an admissible epic in
${\cal{E}}^{\phi }$
, the bottom row of the above diagram gives
$\phi (B')=\phi (B)+\phi (D)$
. Hence,
$\phi (C)+\phi (A')\leq \phi (B')$
. Thus,
in
$\cal{E}$
is actually in
${\cal{E}}^{\phi }$
. This shows
$({\cal{A}},{\cal{E}}^{\phi })$
satisfies the axiom [E1
$^{\text {op}}$
].
Now, we will show that
$({\cal{A}},{\cal{E}}^{\phi })$
satisfies the axiom [E2]. Let
be an admissible monic in
${\cal{E}}^{\phi }$
, and let
$A\xrightarrow {f}A'$
be an arbitrary morphism in
$\cal{A}$
. Then we have the following pushout commutative square in
$\cal{E}$
:
Then, by [Reference Bühler6, Prop. 2.12(ii)], we have the following kernel–cokernel pair in
$\cal{E}$
:
, so
$\phi (B)+\phi (A')=\phi (B\oplus A')\leq \phi (A)+\phi (B')$
. Also, by [Reference Bühler6, Prop. 2.12(iv)], we have the following commutative diagram with rows being kernel–cokernel pairs in
$\cal{E}$
:
Since
is an admissible monic in
${\cal{E}}^{\phi }$
, we have
$\phi (B)=\phi (A)+\phi (C)$
. This implies
$\phi (A)+\phi (C)+\phi (A')\leq \phi (A)+\phi (B')$
, so
$\phi (C)+\phi (A')\leq \phi (B')$
. Also,
is a kernel–cokernel pair in
$\cal{E}$
, so
$\phi (B')\leq \phi (C)+\phi (A')$
. Thus,
$\phi (C)+\phi (A')=\phi (B')$
. Hence,
is a kernel–cokernel pair in
${\cal{E}}^{\phi }$
, so
is an admissible monic in
${\cal{E}}^{\phi }$
. Thus,
$({\cal{A}},{\cal{E}}^{\phi })$
satisfies the axiom [E2]. A dual argument will show that
$({\cal{A}},{\cal{E}}^{\phi })$
satisfies the axiom [E2
$^{\text {op}}$
]. Hence,
$({\cal{A}},{\cal{E}}^{\phi })$
is an exact subcategory of
$({\cal{A}},{\cal{E}})$
.
Moving forward, we record some applications of Theorem 4.8.
4.9. Let
$\cal{X}$
be a subcategory of
${\operatorname{Mod}} (R)$
such that
$({\cal{X}},{\cal{S}}|_{\cal{X}})$
is an exact subcategory of
${\operatorname{mod}}\ R$
. In this case, for
$M,N\in {\cal{X}}$
, by
${\operatorname{Ext}}^1_{\cal{X}}(M,N)$
, we will mean
${\operatorname{Ext}}_{{\cal{S}}|_{\cal{X}}}(M,N)$
, that is,
${\operatorname{Ext}}^1_{\cal{X}}(M,N)=\{[\sigma ]: \text {the middle object of }\sigma \text { is in } {\cal{X}}\}$
. (Note that it does not matter whether the equivalence class is taken in
${\cal{S}}|_{\cal{X}}$
or
$\cal{S}$
by Remark 3.7.) Note that
${\operatorname{Ext}}^1_{\cal{X}}(-,-): {\cal{X}}^{op}\times {\cal{X}}\to {\operatorname{mod}}\ R$
is a subfunctor of
${\operatorname{Ext}}^1_R(-,-): {\cal{X}}^{op}\times {\cal{X}}\to {\operatorname{mod}}\ R$
by Proposition 3.8. For example, we can apply this discussion to
${\cal{X}}=\operatorname{Ul}^s_I(R)$
due to Corollary 4.7.
Note that if
$\cal{X}$
is closed under taking extensions in
${\operatorname{mod}}\ R$
, then
$({\cal{X}},{\cal{S}}|_{\cal{X}})$
is exact subcategory of
${\operatorname{mod}}\ R$
by Proposition 3.5. In this case,
${\operatorname{Ext}}^1_{\cal{X}}(M,N)={\operatorname{Ext}}^1_{R}(M,N)$
for all
$M,N\in {\cal{X}}$
.
Definition 4.10. Let
$\cal{X}$
be a subcategory of
${\operatorname{mod}}\ R$
such that
$({\cal{X}},{\cal{S}}|_{\cal{X}})$
is an exact subcategory of
${\operatorname{mod}}\ R$
, and let
$\phi :{\cal{X}}\to \mathbb Z$
be a function satisfying the hypothesis of Theorem 4.8, that is,
$\phi $
is constant on isomorphism classes of modules in
$\cal{X}$
,
$\phi $
is additive on finite direct sums, and subadditive on short exact sequences of modules in
$\cal{X}$
. Then, in the notation of Theorem 4.8,
$({\cal{X}},{\cal{S}}|_{{\cal{X}}^{\phi }})$
is an exact subcategory of
$({\cal{X}},{\cal{S}}|_{\cal{X}})$
, hence an exact subcategory of
${\operatorname{mod}}\ R$
. For
$M,N\in {\cal{X}}$
, define
${\operatorname{Ext}}_{\cal{X}}^1(M,N)^{\phi }:={\operatorname{Ext}}_{{\cal{S}}|_{{\cal{X}}^{\phi }}}(M,N)=\{[\alpha ]\in {\operatorname{Ext}}_{\cal{X}}^1(M,N)\text { }:\text { }\phi \text { is additive on }\alpha \}$
. Again, we notice that if
${\cal{X}}\subseteq {\operatorname{mod}}\ R$
is closed under taking extensions, then
${\operatorname{Ext}}_{\cal{X}}^1(M,N)^{\phi } =\{[\alpha ]\in {\operatorname{Ext}}^1_{R}(M,N)\text { }:\text { }\phi \text { is additive on }\alpha \}$
, and in this case, we denote it just by
${\operatorname{Ext}}^1_R(M,N)^{\phi }$
.
4.11. Note that
${\operatorname{Ext}}_{\cal{X}}^1(-,-)^{\phi }:{\cal{X}}^{op}\times {\cal{X}}\to {\operatorname{mod}}\ R$
is a subfunctor of
${\operatorname{Ext}}_{\cal{X}}^1(-,-):{\cal{X}}^{op}\times {\cal{X}}\to {\operatorname{mod}}\ R$
, which in turn is a subfunctor of
${\operatorname{Ext}}^1_{R}(-,-):{\cal{X}}^{op}\times {\cal{X}}\to {\operatorname{mod}}\ R$
(Proposition 3.8) and we have corresponding commutative diagram of long exact sequences by Corollary 3.9.
With the extension-closed subcategory
${\cal{X}}={\operatorname{mod}}\ R$
, where
$(R,\mathfrak {m},k)$
is a Noetherian local ring, and the subadditive function
$\phi (-)=\mu (-):{\operatorname{mod}}\ R\to \mathbb Z$
being the number of generators function in Definition 4.10, we see that
${\operatorname{Ext}}^1_R(M,N)^{\mu }=\langle M, N\rangle $
in the notation of [Reference Striuli35, Def. 1.35]. So, by our discussion, Theorem 4.8, Proposition 3.8, and Corollary 3.9 recover [Reference Striuli35, Cor. 1.36 and Props. 1.37 and 1.38].
Before we apply our discussion to more certain special cases, we record one preliminary lemma.
Lemma 4.12. Let
$({\cal{X}},{\cal{E}})$
be an exact subcategory of
${\operatorname{mod}}\ R$
. Let
$G:{\cal{X}} \to \operatorname{fl}(R)$
be an additive half-exact functor, where
$\operatorname{fl}(R)$
denotes the full subcategory of
${\operatorname{mod}}\ R$
consisting of finite length modules. Then,
$\phi (-):=\lambda _R(G(-)): {\cal{X}} \to \mathbb Z$
satisfies the hypothesis of Theorem 4.8. Moreover, given a short exact sequence
$\sigma $
with objects in
$\cal{X}$
,
$G(\sigma )$
is a short exact sequence of modules if and only if
$\phi $
is additive on
$\sigma $
.
Proof. We will only prove the statement when G is a covariant functor, since the proof for a contravariant functor is similar. Since G is a functor and
$\lambda _R$
is constant on isomorphism classes of objects in
${\operatorname{mod}}\ R$
,
$\phi $
is constant on isomorphism classes of objects in
$\cal{X}$
. Let
$A,B\in {\cal{X}}$
. Then
$\phi (A\oplus B)=\lambda _R(G(A\oplus B))=\lambda _R(G(A)\oplus G(B))=\lambda _R(G(A))+\lambda _R(G(B))=\phi (A)+\phi (B)$
, so
$\phi $
is additive on finite biproducts. Next, let
$\sigma :0\to M\xrightarrow {f} N\xrightarrow {g} L\to 0$
be a short exact sequence with objects in
$\cal{X}$
. Then
$G(\sigma )$
is the exact sequence
$G(\sigma ):0\to K\to G(M)\xrightarrow {G(f)} G(N)\xrightarrow {G(g)} G(L)\to P\to 0$
, where
$K=\operatorname{Ker}(G(f))$
and
$P=\operatorname{Coker}(G(g))$
. Then we have
$\phi (M)+\phi (L)-\phi (N)=\lambda _R(G(M))+\lambda _R(G(L))-\lambda _R(G(N))=\lambda _R(K)+\lambda _R(P)\geq 0$
, so
$\phi (M)+\phi (L)\ge \phi (N)$
. Hence,
$\phi $
is subadditive on a short exact sequence with objects in
$\cal{X}$
. Thus,
$\phi $
satisfies the hypothesis of Proposition 4.8. Next, let
$\sigma :0\to M\xrightarrow {f} N\xrightarrow {g} L\to 0$
be a short exact sequence with objects in
$\cal{X}$
. Then
$G(\sigma )$
is short exact if and only if
$K=\operatorname{Ker}(G(f))=0$
and
$P=\operatorname{Coker}(G(g))=0$
if and only if
$\lambda _R(K)=\lambda _R(P)=0$
if and only if
$\phi (M)+\phi (L)-\phi (N)=\lambda _R(G(M))+\lambda _R(G(L))-\lambda _R(G(N))=\lambda _R(K)+\lambda _R(P)= 0$
if and only if
$\phi (M)+\phi (L)=\phi (N)$
if and only if
$\phi $
is additive on
$\sigma $
.
Corollary 4.13. Let
$({\cal{X}},{\cal{E}})$
be an exact subcategory of
${\operatorname{mod}}\ R$
. Let
$G:{\cal{X}} \to \operatorname{fl}(R)$
be an additive half-exact functor, where
$\operatorname{fl}(R)$
denotes the full subcategory of
${\operatorname{mod}}\ R$
consisting of finite length modules. Set
${\cal{E}}^G:=\{\sigma \in {\cal{E}}: G(\sigma ) \text { is a short exact sequence of modules}\}$
. Then,
$({\cal{X}}, {\cal{E}}^G)$
is an exact subcategory of
${\operatorname{mod}}\ R$
.
Proof. From Lemma 4.12, it follows that
${\cal{E}}^G={\cal{E}}^{\phi }$
, where
$\phi (-):=\lambda _R(G(-)): {\cal{X}} \to \mathbb Z$
. Hence, the claim now follows from Theorem 4.8.
Now again, let
${\cal{X}}={\operatorname{mod}}\ R$
, where R is Noetherian. Basic examples of additive half-exact functors
$G:{\cal{X}} \to \operatorname{fl}(R)$
are
$G(-):={\operatorname{Ext}}^i_R(-,X),{\operatorname{Ext}}^i_R(X,-), \operatorname{Tor}^R_i(X,-)$
, where X is a fixed R-module of finite length and
$i\ge 0$
is an integer.
Given a collection
${\cal{C}}\subseteq {\operatorname{mod}}\ R$
, in [Reference Auslander and Solberg1], the authors considered the following:
$$ \begin{align*} F_{\cal{C}}(C,A)&:=\{\sigma: 0\to A \to B \to C\to 0\text{ } |\text{ }\operatorname{Hom}_R(X,\sigma) \text{ is short exact for all } X \in {\cal{C}}\},\\ F^{\cal{C}}(C,A)&:=\{\sigma: 0\to A \to B \to C\to 0\text{ }|\text{ }\operatorname{Hom}_R(\sigma,X) \text{ is short exact for all } X \in {\cal{C}}\} \end{align*} $$
and showed in [Reference Auslander and Solberg1, Prop. 1.7] that these define subfunctors of
${\operatorname{Ext}}^1_R(-,-):{\operatorname{mod}}\ R^{op}\times {\operatorname{mod}}\ R \to \textbf {Ab}$
. Using Corollary 4.13, we recover this result when
${\cal{C}}\subseteq \operatorname{fl}(R)$
as follows.
Corollary 4.14. Let R be a commutative Noetherian ring. Let
${\cal{C}}\subseteq \operatorname{fl}(R)$
. Then,
$F_{\cal{C}}(-,-)$
and
$F^{\cal{C}}(-,-)$
are subfunctors of
${\operatorname{Ext}}^1_R(-,-):{\operatorname{mod}}\ R^{op}\times {\operatorname{mod}}\ R \to {\operatorname{mod}}\ R$
.
Proof. We only prove the case of
$F_{\cal{C}}(-,-)$
, since the proof of
$F^{\cal{C}}(-,-)$
is similar.
Let
${\cal{S}}^{\cal{C}}$
be the collection of all short exact sequences
$\sigma $
of finitely generated R-modules such that
$\operatorname{Hom}_R(X,\sigma )$
is short exact for all
$X \in {\cal{C}}$
. We first prove that
$({\operatorname{mod}}\ R,{\cal{S}}^{\cal{C}})$
is an exact subcategory of
${\operatorname{mod}}\ R$
. Since
${\cal{S}}^{\cal{C}}=\cap _{C \in {\cal{C}}} {\cal{S}}^{C}$
, it is enough to prove that for each
$C\in {\cal{C}}$
,
${\cal{S}}^C$
gives an exact substructure on
${\operatorname{mod}}\ R$
(intersection of exact substructures is again an exact substructure by Lemma 3.2). Now, note that
${\cal{S}}^C=\{\sigma : G(\sigma ) \text { is short exact}\}$
, where
$G:{\operatorname{mod}}\ R \to \operatorname{fl}(R)$
is given by
$G(-):=\operatorname{Hom}_R(C,-)$
, so by Corollary 4.13 we get that
$({\operatorname{mod}}\ R,{\cal{S}}^C)$
is an exact subcategory of
${\operatorname{mod}}\ R$
. Since
${\operatorname{Ext}}_{{\cal{S}}^{\cal{C}}}(-,-)=F_{\cal{C}}(-,-)$
, by Proposition 3.8 we are done.
Next, we recover and improve upon [Reference Puthenpurakal31, Ths. 3.11 and 3.13].
For a Noetherian local ring
$(R,\mathfrak {m})$
, let
$\operatorname{MD}(R)$
denote the full subcategory of
${\operatorname{mod}}\ R$
consisting of all modules M such that either
$M=0$
or
$\dim M=\dim R$
. We recall that if I is an
$\mathfrak {m}$
-primary ideal, then
$e_R(I,-)$
is additive on
$\operatorname{MD}(R)$
. We also recall that a finitely generated module M satisfies Serre’s condition
$(S_n)$
if
$\operatorname{depth}\ M_{\mathfrak {p}}\ge \inf \{n,\dim R_{\mathfrak {p}}\}$
for all
$\mathfrak {p}\in \operatorname{Spec} R$
. The full subcategory of
${\operatorname{mod}}\ R$
consisting of all modules satisfying
$(S_n)$
is denoted by
$S_n(R)$
. Note that
$S_n(R)$
is closed under extensions in
${\operatorname{mod}}\ R$
.
Lemma 4.15. Let R be an equidimensional local ring. Let
$S_1(R)$
be the collection of all modules in
${\operatorname{mod}}\ R$
satisfying Serre’s condition
$(S_1)$
. Then,
$S_1(R)\subseteq \operatorname{MD}(R)$
.
Proof. The hypothesis on R implies that
$\dim (R/\mathfrak {p})=\dim R$
for all
$\mathfrak {p} \in \operatorname{Min}(R)$
. So, if
$0\ne M$
satisfy
$(S_1)$
, then there exists
$\mathfrak {p} \in \operatorname{Ass}(M)\subseteq \operatorname{Min}(R)$
. Hence,
$\dim (M)\ge \dim (R/\mathfrak {p})=\dim R$
, and we are done.
Now, inspecting the proof of [Reference Puthenpurakal30, Prop. 17], the only place, where R is Cohen–Macaulay and M is maximal Cohen–Macaulay are required, is to ensure that
$e_R(I,-)$
is additive on
$0\to \Omega _R M \to F \to M \to 0$
and that
$M,\Omega _R M \in \operatorname{MD}(R)$
. But if we assume R is equidimensional and satisfy
$(S_1)$
, and
$M\in S_1(R)\subseteq \operatorname{MD}(R)$
, then since
$\Omega _R M\in S_1(R)\subseteq \operatorname{MD}(R)$
(Lemma 4.15) and
$e_R(I,-)$
is additive on short exact sequence of modules of the same dimension [Reference Bruns and Herzog3, Cor. 4.7.7]; hence, we get the following lemma by following the same proof as in [Reference Puthenpurakal30, Prop. 17].
Lemma 4.16. Let
$(R,\mathfrak {m})$
be an equidimensional local ring satisfying
$(S_1)$
. Let
$M\in S_1(R)$
. Let
$d=\dim R\ge 1$
, and let I be an
$\mathfrak {m}$
-primary ideal. Then, the function
$n \mapsto \lambda (\operatorname{Tor}^R_1(M,R/I^{n+1}))$
is given by a polynomial of degree
$\le d-1$
for
$n\gg 0$
. So, in particular, the limit
$\lim _{n\to \infty }\dfrac {\lambda (\operatorname{Tor}^R_1(M,R/I^{n+1}))}{n^{d-1}}$
exists.
If R and M are as in Lemma 4.16, then let us denote
$e^T_I(M):=(d-1)!\lim _{n\to \infty }\dfrac {\lambda (\operatorname{Tor}^R_1(M,R/I^{n+1}))}{n^{d-1}}$
. (What [Reference Puthenpurakal31] denotes by
$e^T_R(M)$
is exactly the same as
$e^T_{\mathfrak {m}}(M)$
in our notation.)
The following theorem generalizes [Reference Puthenpurakal31, Ths. 3.11 and 3.13] owing to the fact that a local Cohen–Macaulay ring R is equidimensional, satisfies
$(S_1)$
and
$\operatorname{CM}(R)\subseteq S_1(R)$
.
Theorem 4.17. Let
$(R,\mathfrak {m})$
be an equidimensional local ring satisfying
$(S_1)$
. Let
${\cal{X}}=S_1(R)$
be the subcategory of
${\operatorname{mod}}\ R$
consisting of all modules satisfying
$(S_1)$
. Let I be an
$\mathfrak {m}$
-primary ideal. Then,
${\operatorname{Ext}}^1_R(-,-)^{e_I^T}:S_1(R)^{op}\times S_1(R)\to {\operatorname{mod}}\ R$
is a subfunctor of
${\operatorname{Ext}}^1_R(-,-):S_1(R)^{op}\times S_1(R)\to {\operatorname{mod}}\ R$
.
Proof. Since
$S_1(R)$
is an extension closed subcategory of
${\operatorname{mod}}\ R$
, we have
$(S_1(R),{\cal{S}}|_{S_1(R)})$
is an exact subcategory of
${\operatorname{mod}}\ R$
. Since for each n, the function
${\operatorname{mod}}\ R \to \mathbb Z$
given by
$M\mapsto \lambda (\operatorname{Tor}^R_1(M,R/I^{n+1}))$
is subadditive on short exact sequences of
${\operatorname{mod}}\ R$
, we have
$e^T_I:S_1(R)\to \mathbb Z$
is also subadditive on short exact sequences of
$S_1(R)$
. Thus,
$(S_1(R),{\cal{S}}|_{S_1(R)}^{e^T_I})$
is an exact subcategory of
$(S_1(R),{\cal{S}}|_{S_1(R)})$
by Theorem 4.8. Hence, we are done by Proposition 3.8.
5 Special subfunctors of
${\operatorname{Ext}}^1$
and applications
Throughout this section, R will denote a Noetherian local ring with unique maximal ideal
$\mathfrak {m}$
and residue field k. We will denote by
$\mu _R(-)$
the minimal number of generators function for finitely generated modules, that is,
$\mu _R(M)=\lambda _R(M\otimes _R R/\mathfrak {m})$
for all
$M\in {\operatorname{mod}}\ R$
. We drop the subscript R when the ring in question is clear. We shall study properties and applications of a number of subfunctors of
${\operatorname{Ext}}^1$
, whose existence follow from results in previous sections.
5.1 Some computations and applications of the subfunctor
${\operatorname{Ext}}^1_R(-,-)^{\mu }$
We start this subsection with a characterization of regularity among Cohen–Macaulay rings of positive dimension d in terms of vanishing of certain
${\operatorname{Ext}}^1_R(-,-)^{\mu }$
(see Definition 4.10 and the discussion after 4.11 for notation).
Theorem 5.1.1. Let R be a local Cohen–Macaulay ring of dimension
$d\ge 1$
. Then, R is regular if and only if
${\operatorname{Ext}}^1_R(\Omega ^{d-1}_R k, N)^{\mu }=0$
for some finitely generated nonzero R-module N of finite injective dimension.
When R is regular, we show
${\operatorname{Ext}}^1_R(\Omega ^{d-1}_R k, R)^{\mu }=0$
, which shows one direction of Theorem 5.1.1. For this, we first need two preliminary lemmas.
Lemma 5.1.2. Let Y be a submodule of an R-module X such that
$\mathfrak {m} X\subseteq Y$
. If X is cyclic and
$Y\ne X$
, then
$Y=\mathfrak {m} X$
.
Proof. We prove that if
$Y\ne \mathfrak {m} X$
, then
$Y=X$
. If
$Y\ne \mathfrak {m} X$
, then pick
$y\in Y\setminus \mathfrak {m} X\subseteq X\setminus \mathfrak {m} X$
. Then, y is part of a minimal system of generators of X, but X is cyclic, so
$X=Ry$
. Also,
$Ry\subseteq Y\subseteq X$
. Hence,
$Y=X$
.
Lemma 5.1.3. Let M be a finitely generated R-module with the first Betti number
$1$
. Then,
${\operatorname{Ext}}^1_R(M,\Omega _R M)^{\mu }=\mathfrak {m} {\operatorname{Ext}}^1_R(M,\Omega _R M)$
.
Proof.
${\operatorname{Ext}}^1_R(M,\Omega _R M)^{\mu }$
is a submodule of
${\operatorname{Ext}}^1_R(M,\Omega _R M)$
such that
$\mathfrak {m} {\operatorname{Ext}}^1_R(M,\Omega _R M)\subseteq {\operatorname{Ext}}^1_R(M,\Omega _R M)^{\mu }$
by Lemma 5.2.1. By hypothesis, we have
$\Omega _R M\cong R/I$
for some ideal
$I\ne R$
, and we have an exact sequence
$ 0\to R/I \to F \to M \to 0$
, for some free R-module F. So, in particular,
$\mu (F)=\mu (M)\ne \mu (R/I)+\mu (M)$
. Hence, this exact sequence is not
$\mu $
-additive. So,
${\operatorname{Ext}}^1_R(M,\Omega _R M)^{\mu }\ne {\operatorname{Ext}}^1_R(M,\Omega _R M)$
. Also, applying
$\operatorname{Hom}_R(-,R/I)$
to the exact sequence
$0\to R/I \to F \to M \to 0$
, we get an exact sequence
$0\to \operatorname{Hom}_R(M,R/I)\to \operatorname{Hom}_R(F,R/I)\to \operatorname{Hom}_R(R/I,R/I)\cong R/I \to {\operatorname{Ext}}^1_R(M,R/I)\cong {\operatorname{Ext}}^1_R(M,\Omega _RM)\to 0$
. Hence,
${\operatorname{Ext}}^1_R(M,\Omega _R M)$
is a cyclic module. Now, applying Lemma 5.1.2 to
$Y={\operatorname{Ext}}^1_R(M,\Omega _R M)^{\mu }$
and
$X={\operatorname{Ext}}^1_R(M,\Omega _R M)$
, we get
${\operatorname{Ext}}^1_R(M,\Omega _R M)^{\mu }=\mathfrak {m}{\operatorname{Ext}}^1_R(M,\Omega _R M)$
.
Now, one direction of Theorem 5.1.1 is the following.
Corollary 5.1.4. Let
$(R,\mathfrak {m},k)$
be a regular local ring of dimension
$d\ge 1$
. Then,
${\operatorname{Ext}}^1_R(\Omega ^{d-1}_R k, R)^{\mu }=0$
.
Proof. We know
$\Omega _R \Omega ^{d-1}_R k\cong R$
by looking at the Koszul complex of
$R/\mathfrak {m}$
, so the first Betti number of
$\Omega ^{d-1}_R k$
is
$1$
. Hence, by Lemma 5.1.3, we get that
${\operatorname{Ext}}^1_R(\Omega ^{d-1}_Rk,R)^{\mu }=\mathfrak {m}{\operatorname{Ext}}^1_R(\Omega ^{d-1}_Rk,R)\cong \mathfrak {m}{\operatorname{Ext}}^d_R(k,R)\cong \mathfrak {m}\cdot k=0$
.
The other direction of Theorem 5.1.1 requires more work. First, we begin with the following setup.
5.1.5. Let
$(R,\mathfrak {m},k)$
be a local Cohen–Macaulay ring of dimension
$d\ge 1$
admitting a canonical module
$\omega _R$
. Then,
${\operatorname{Ext}}^1_R(\Omega ^{d-1}_Rk,\omega _R)\cong k$
. Hence, for any two non-split exact sequences
$\alpha , \beta \in {\operatorname{Ext}}^1_R(\Omega ^{d-1}_Rk,\omega _R)$
, we have
$[\beta ]=r[\alpha ] \in {\operatorname{Ext}}^1_R(\Omega ^{d-1}_Rk,\omega _R)$
for some unit
$r\in R$
. So, we have the following pushout diagram:
Since
$r\in R$
is a unit, we have
$\omega _R \xrightarrow {\cdot r} \omega _R$
is an isomorphism. Hence, by five lemma,
$X_{\alpha }\cong X_{\beta }$
. So, there exists a unique module (up to isomorphism), call it
$E^R$
, such that the middle term of every non-split exact sequence
$0\to \omega _R \to X \to \Omega ^{d-1} k \to 0$
is isomorphic to
$E^R$
. Since
$\Omega ^{d-1}_R k$
has co-depth
$1$
, by [Reference Leuschke and Wiegand27, Prop. 11.21], we get that
$0\to \omega _R \to E^R \to \Omega ^{d-1}_Rk \to 0$
is the minimal maximal Cohen--Macaulay (MCM) approximation of
$\Omega ^{d-1}_R k$
.
Moving forward, we denote
$\operatorname{Hom}_R(-,\omega _R)$
by
$(-)^{\dagger }$
.
We need to collect some properties of
$E^R$
to prove the other direction of Theorem 5.1.1.
Lemma 5.1.6. Let
$(R,\mathfrak {m},k)$
be a local Cohen–Macaulay ring of dimension
$1$
admitting a canonical module
$\omega _R$
. Then,
$E^R\cong \mathfrak {m}^{\dagger }$
.
Proof. Dualizing
$0\to \mathfrak {m} \to R \to k \to 0$
by
$\omega _R$
, we get
$0\to \omega _R \to \mathfrak {m}^{\dagger } \to {\operatorname{Ext}}^1_R(k,\omega _R)\cong k \to 0$
. So, we have the exact sequence
$0\to \omega _R \to \mathfrak {m}^{\dagger } \to k \to 0$
, which is clearly non-split, since
$\mathfrak {m}^{\dagger }$
has positive depth. But
$k\oplus \omega _R$
has depth
$0$
. Thus,
$E^R\cong \mathfrak {m}^{\dagger }$
.
We record the following lemma, which will be used to deduce further properties of the module
$E^R$
.
Lemma 5.1.7. Let
$(R,\mathfrak {m},k)$
be a local ring, and let I be an ideal of R, which is not principal. Then, for every
$x\in I\setminus \mathfrak {m} I$
, we have
$(x\mathfrak {m}:_{\mathfrak {m}} I)=(x\mathfrak {m}:_R I)=((x) :_R I)$
.
Proof. Since trivially
$(x\mathfrak {m}:_{\mathfrak {m}}I)\subseteq (x\mathfrak {m}:_R I)\subseteq ((x) :_R I)$
always holds, it is enough to prove the inclusion
$((x) :_R I)\subseteq (x\mathfrak {m}:_{\mathfrak {m}} I)$
. So, let
$y\in ((x) :_R I)$
, which means
$yI \subseteq (x)$
. If y were a unit, then we would have
$I \subseteq y^{-1}(x)=(x)\subseteq I$
, implying
$I=(x)$
, contradicting our assumption that I is not principal. Thus, we must have
$y\in \mathfrak {m}$
. Now, pick an arbitrary element
$r\in I$
. Then
$yr\in y I\subseteq (x)$
, so
$yr=xs$
for some
$s\in R$
. If
$s\notin \mathfrak {m}$
, then
$x=s^{-1}yr\in \mathfrak {m} I$
, which is a contradiction. Hence,
$s\in \mathfrak {m}$
. So,
$yr=xs\in x\mathfrak {m}$
. Since
$r\in I$
was arbitrary, we get
$yI \subseteq x\mathfrak {m}$
. Hence,
$y\in (x\mathfrak {m}:_{\mathfrak {m}} I)$
.
In the following,
${\operatorname{r}}(-)$
will denote the type of a module, that is,
${\operatorname{r}}(M)= \dim _k {{\operatorname{Ext}}}^{\operatorname{depth}_R M}_R(k,M)$
.
Lemma 5.1.8. Let
$(R,\mathfrak {m},k)$
be a non-regular complete local Cohen–Macaulay ring of dimension
$d\ge 1$
, with canonical module
$\omega _R$
. Then, we have
$\mu (E^R)=\operatorname{r}(R)+\mu (\Omega ^{d-1}_Rk)$
.
Proof. We prove this by induction on d. First, let
$d=1$
. By Lemma 5.1.6, we have
$E^R\cong \mathfrak {m}^{\dagger }$
. Pick
$x\in \mathfrak {m}\setminus \mathfrak {m}^2$
to be R-regular. Then
$\mu _R(E^R)=\mu _{R/xR}(E^R/xE^R)$
. Now,
$E^R/x E^R\cong \mathfrak {m}^{\dagger }/x\mathfrak {m}^{\dagger }\cong \operatorname{Hom}_{R/xR}(\mathfrak {m}/x\mathfrak {m},\omega _R/x\omega _R)\cong (\mathfrak {m}/x\mathfrak {m})^{\dagger }$
(where the last two isomorphisms follow from [Reference Bruns and Herzog3, Prop. 3.3.3(a) and Th. 3.3.5(a)]). Now,
$ \operatorname{Hom}_{R/xR}((\mathfrak {m}/x\mathfrak {m})^{\dagger },k)\cong \operatorname{Hom}_{R/xR}(k^{\dagger }, \mathfrak {m}/x\mathfrak {m})\cong \operatorname{Hom}_{R/xR}(k, \mathfrak {m}/x\mathfrak {m})\cong (x\mathfrak {m}:_{\mathfrak {m}}\mathfrak {m})/x\mathfrak {m}=((x):_R\mathfrak {m})/x\mathfrak {m}$
(where the first isomorphism holds because
$R/xR$
is Artinian, so one can invoke [Reference Dao and Huneke10, Lem. 3.14 and Rem. 3.15], and the last equality follows from Lemma 5.1.7 since R is not regular). Hence,
$\mu _R(E^R)=\mu _{R/xR}(E^R/xE^R)=\dim _k \operatorname{Hom}_{R/xR}((\mathfrak {m}/x\mathfrak {m})^{\dagger },k)=\dim _k ((x):_R\mathfrak {m})/x\mathfrak {m}=\dim _k \left ((xR:_R \mathfrak {m})/xR \right )+ \dim _k (xR/x\mathfrak {m})=\dim _k \operatorname{Soc}(R/xR)+\dim _k (xR/x\mathfrak {m})=\text {r}(R)+\dim _k (xR/x\mathfrak {m})$
(since
$xR/x\mathfrak {m}$
is annihilated by
$\mathfrak {m}$
, hence it is a k-vector space). Since
$xR/x\mathfrak {m}$
is generated by x and it is a k-vector space, we have
$xR/x\mathfrak {m}\cong k$
. Hence,
$\dim _k (xR/x\mathfrak {m})=1$
, so
$\mu (E^R)=\text {r}(R)+1=\text {r}(R)+\mu (k)$
. This concludes the
$d=1$
case.
Now, let
$\dim R=d>1$
and suppose the claim has been proved for all non-regular local Cohen–Macaulay rings of dimension
$1,...,d-1$
admitting a canonical module. Since
$\Omega ^{d-1}_R k$
has co-depth
$1$
, by [Reference Leuschke and Wiegand27, Prop. 11.21] (see 5.1.5), we get that
$0\to \omega _R \to E^R \to \Omega ^{d-1}_Rk \to 0$
is the minimal MCM approximation of
$\Omega ^{d-1}_R k$
. Now, pick an R-regular element
$x\in \mathfrak {m}\setminus \mathfrak {m}^2$
(which is also
$\Omega ^{d-1}_R k$
-regular, since
$d-1>0$
). By [Reference Yoshida39, Cor. 2.5], we have
$0\to \omega _{R/xR}\to E^R/xE^R\to \Omega ^{d-1}_R k/x(\Omega ^{d-1}_Rk)\to 0$
is the minimal MCM approximation of
$\Omega ^{d-1}_R k/x(\Omega ^{d-1}_Rk) \cong \Omega ^{d-1}_{R/xR}k \oplus \Omega ^{d-2}_{R/xR} k$
over
$R/xR$
(the isomorphism follows from [Reference Takahashi37, Cor. 5.3]). Since
$0\to 0 \to \Omega ^{d-1}_{R/xR}k\to \Omega ^{d-1}_{R/xR}k \to 0$
is the minimal MCM approximation of
$\Omega ^{d-1}_{R/xR}k$
over
$R/xR$
and
$0\to \omega _{R/xR}\to E^{R/xR} \to \Omega ^{d-2}_{R/xR} k \to 0$
is the minimal MCM approximation of
$\Omega ^{d-2}_{R/xR} k$
over
$R/xR$
(by 5.1.5 and [Reference Leuschke and Wiegand27, Prop. 11.21]), their direct sum
$0\to \omega _{R/xR}\to E^{R/xR} \oplus \Omega ^{d-1}_{R/xR}k \to \Omega ^{d-2}_{R/xR} k \oplus \Omega ^{d-1}_{R/xR}k \to 0$
is the minimal MCM approximation of
$\Omega ^{d-2}_{R/xR} k \oplus \Omega ^{d-1}_{R/xR}k\cong \Omega ^{d-1}_R k/x(\Omega ^{d-1}_Rk)$
over
$R/xR$
(direct sum preserves minimal MCM approximation by [Reference Yoshida39, Th. 1.4]). By uniqueness of minimal MCM approximation, we have
$E^R/xE^R\cong E^{R/xR} \oplus \Omega ^{d-1}_{R/xR}k$
. So,
$\mu _R(E^R)=\mu _{R/xR}(E^R/xE^R)=\mu _{R/xR}(E^{R/xR})+\mu _{R/xR}(\Omega ^{d-1}_{R/xR}k)$
. Since R is not regular,
$R/xR$
is not regular. Hence, by induction hypothesis, we have
$\mu _{R/xR}(E^{R/xR})=\text {r}(R/xR)+\mu _{R/xR}(\Omega ^{d-2}_{R/xR} k)=\text {r}(R)+\mu _{R/xR}(\Omega ^{d-2}_{R/xR} k)$
. So,
$\mu _R(E^R)=\text {r}(R)+\mu _{R/xR}(\Omega ^{d-2}_{R/xR} k)+\mu _{R/xR}(\Omega ^{d-1}_{R/xR} k)=\text {r}(R)+\mu _{R/xR}(\Omega ^{d-2}_{R/xR} k \oplus \Omega ^{d-1}_{R/xR}k)=\text {r}(R)+\mu _{R/xR}\left ( \Omega ^{d-1}_R k/x(\Omega ^{d-1}_Rk) \right )=\text {r}(R)+\mu _R(\Omega ^{d-1}_R k)$
. This finishes the inductive step, and hence the proof.
Since
$\mu (\omega _R)=\text {r}(R)$
, Lemma 5.1.8 says that the sequence
$0\to \omega _R\to E^R\to \Omega _R^{d-1}k\to 0$
is
$\mu $
-additive. As a consequence of this, we get the following.
Proposition 5.1.9. Let
$(R,\mathfrak {m},k)$
be a non-regular local Cohen–Macaulay ring of dimension
$d\ge 1$
, and let N be a finitely generated nonzero R-module with finite injective dimension. Then
${\operatorname{Ext}}^1_R(\Omega _R^{d-1}k,N)^{\mu }={\operatorname{Ext}}_{R}^1(\Omega _R^{d-1}k,N)\ne 0$
.
Proof. That
${\operatorname{Ext}}_{R}^1(\Omega _R^{d-1}k,N)\cong {\operatorname{Ext}}^d_R(k,N)\ne 0$
follows from [Reference Bruns and Herzog3, Exer. 3.1.24]. So, it is enough to prove that
${\operatorname{Ext}}^1_R(\Omega _R^{d-1}k,N)^{\mu }={\operatorname{Ext}}_{R}^1(\Omega _R^{d-1}k,N)$
.
We first consider the case where R is complete, hence admitting a canonical module
$\omega _R$
. Consider the
$\mu $
-additive (by Lemma 5.1.8) exact sequence
$\sigma :0\to \omega _R\to E^R\to \Omega _R^{d-1}k\to 0$
. Since
$\sigma \in {\cal{S}}|_{{\operatorname{mod}}\ R}^{\mu }$
, by applying
$\operatorname{Hom}_R(-,N)$
to
$\sigma $
, we get the following part of a commutative diagram of exact sequences by Corollary 3.9:
Since
$E^R$
is maximal Cohen–Macaulay, we have
${\operatorname{Ext}}_{R}^1(E^R,N)=0$
by [Reference Bruns and Herzog3, Exer. 3.1.24]. So,
${\operatorname{Ext}}_{R}^1(E^R,N)^{\mu }=0$
as well. Hence, we get the following commutative diagram:
Thus,
$h\circ g=f$
is surjective, so h is surjective. Since h is the natural inclusion map, we have
${\operatorname{Ext}}^1_R(\Omega _R^{d-1}k,N)^{\mu }={\operatorname{Ext}}^1_R(\Omega _R^{d-1}k,N)$
.
Now, we consider the general case. Since
${\operatorname{Ext}}^1_R(\Omega _R^{d-1}k,N)^{\mu }\subseteq {\operatorname{Ext}}_{R}^1(\Omega _R^{d-1}k,N)$
, it is enough to prove the other inclusion. So, let
$\sigma : 0\to N\to X\to \Omega ^{d-1}_R k\to 0 $
be a short exact sequence. We need to show that
$\sigma $
is
$\mu $
-additive. Now, consider the completion
$\widehat \sigma : 0\to \widehat N\to \widehat X\to \widehat {\Omega ^{d-1}_R k}\cong \Omega ^{d-1}_{\widehat R} k \to 0$
. Since
$\widehat R$
is non-regular, Cohen–Macaulay of dimension d admits a canonical module, and
$\widehat N \in {\operatorname{mod}} \widehat R$
has finite injective dimension over
$\widehat R$
, by the first part of the proof, we get
$ {\operatorname{Ext}}^1_{\widehat R}(\Omega ^{d-1}_{\widehat R} k,\widehat N)={\operatorname{Ext}}^1_{\widehat R}(\Omega ^{d-1}_{\widehat R} k,\widehat N)^{\mu }$
. Thus,
$[\widehat \sigma ]\in {\operatorname{Ext}}^1_{\widehat R}(\Omega ^{d-1}_{\widehat R} k,\widehat N)={\operatorname{Ext}}^1_{\widehat R}(\Omega ^{d-1}_{\widehat R} k,\widehat N)^{\mu }$
. Hence,
$\widehat \sigma $
is
$\mu $
-additive. So,
$\mu _{\widehat R}(\widehat N)+\mu _{\widehat R}( \Omega ^{d-1}_{\widehat R} k)=\mu _{\widehat R}(\widehat X)$
. Since the number of generators does not change under completion, we get
$\mu _R(N)+\mu _R(\Omega ^{d-1}_R k)=\mu _R(X)$
. Thus,
$\sigma $
is
$\mu $
-additive, which is what we wanted to prove.
For an arbitrary local ring of positive depth, we give a characterization of the ring being regular in terms of vanishing of
${\operatorname{Ext}}^1_R(k,M)^{\mu }$
for some
$M\in {\operatorname{mod}}\ R$
of finite projective dimension. For this, we first recall the definitions of weakly
$\mathfrak {m}$
-full and Burch submodules of a module from [Reference Dey and Kobayashi12, Defs. 3.1 and 4.1] and subsequently, we relate that property to the vanishing of certain
${\operatorname{Ext}}^1_R(k,-)^{\mu }$
.
Definition 5.1.10. Let
$(R,\mathfrak {m},k)$
be a local ring, and let N be an R-submodule of a finitely generated R-module M. Then N is called a weakly
$\mathfrak {m}$
-full submodule of M if
$(\mathfrak {m} N:_M\mathfrak {m})=N$
. Also, N is called a Burch submodule of M if
$\mathfrak {m} (N:_M \mathfrak {m})\ne \mathfrak {m} N$
.
Proposition 5.1.11. Let
$(R,\mathfrak {m},k)$
be a local ring, and let N be an R-submodule of a finitely generated R-module M such that
${\operatorname{Ext}}^1_R(k,N)^{\mu }=0$
. Then
$(\mathfrak {m} N:_M\mathfrak {m})=N+\operatorname{Soc}(M)$
, that is,
$N+\operatorname{Soc}(M)$
is a weakly
$\mathfrak {m}$
-full submodule of M. So, in particular, if we moreover have
$\operatorname{depth}(M)>0$
, then N is a weakly
$\mathfrak {m}$
-full submodule of M.
Proof. Clearly,
$N\subseteq (\mathfrak {m} N:_M \mathfrak {m})$
and
$\operatorname{Soc}(M)=(0:_M \mathfrak {m})\subseteq (\mathfrak {m} N:_M\mathfrak {m})$
, so
$N+\operatorname{Soc}(M)\subseteq (\mathfrak {m} N:_M\mathfrak {m})$
. Now, choose an element
$f\in (\mathfrak {m} N:_M\mathfrak {m})$
, and we aim to show
$f\in N+\operatorname{Soc}(M)$
. If
$f\in N$
, then we are done. Otherwise, say
$f\notin N$
. As
$f\in (\mathfrak {m} N:_M\mathfrak {m})$
, so
$\mathfrak {m} f\subseteq \mathfrak {m} N\subseteq N$
, and hence
$\mathfrak {m}(N+Rf)=\mathfrak {m} N$
. Thus,
$\dfrac {Rf+N}{N}$
is a nonzero (as
$f\not \in N$
) k-vector space (as
$\mathfrak {m}(N+Rf)\subseteq N$
). Moreover,
$\dfrac {Rf+N}{N}$
is cyclically generated by the image of f in
$M/N$
. Hence,
$\dfrac {Rf+N}{N}\cong k$
, so we have a short exact sequence
$\sigma : 0\to N\to Rf+N \to k\to 0$
. This sequence is
$\mu $
-additive, since
$\mu (N+Rf)=\lambda \left (\frac {N+Rf}{\mathfrak {m}(N+Rf)}\right )=\lambda \left (\frac {N+Rf}{\mathfrak {m} N}\right )=\lambda \left (\frac {N+Rf}{N}\right )+\lambda \left (\frac {N}{\mathfrak {m} N}\right )=\lambda (k)+\mu (N)=\mu (k)+\mu (N)$
. So,
$\sigma \in {\operatorname{Ext}}^1_R(k,N)^{\mu }=0$
. Hence,
$\sigma $
splits. Thus,
$N+Rf=N\oplus g(k)$
, where
$g:k\to N+Rf$
is the splitting map, so
$k\cong g(k)$
. Now,
$f\in N+Rf=N\oplus g(k)$
, so
$f=x+y$
for some
$x\in N$
and
$y\in g(k)\subseteq M$
. Since
$k\cong g(k)$
, we have
$\mathfrak {m} g(k)=0$
. Now,
$\mathfrak {m} y\in \mathfrak {m} g(k)=0$
, so
$y\in (0:_M\mathfrak {m})=\operatorname{Soc}(M)$
. Hence,
$f=x+y\in N+\operatorname{Soc}(M)$
. This finally shows that
$(\mathfrak {m} N:_M\mathfrak {m})\subseteq N+\operatorname{Soc}(M)$
, so
$(\mathfrak {m} N:_M\mathfrak {m})=N+\operatorname{Soc}(M)$
. Since
$\mathfrak {m}(N+\operatorname{Soc}(M))=\mathfrak {m} N$
, we get
$(\mathfrak {m} (N+\operatorname{Soc}(M)):_M\mathfrak {m})=N+\operatorname{Soc}(M)$
, which implies that
$N+\operatorname{Soc}(M)$
is a weakly
$\mathfrak {m}$
-full submodule M. Now note that if
$\operatorname{depth}(M)>0$
, then
$\operatorname{Soc}(M)=0$
. Hence,
$N+\operatorname{Soc}(M)=N$
is a weakly
$\mathfrak {m}$
-full submodule of M.
Now, we give another characterization of regular local rings in terms of vanishing of certain
${\operatorname{Ext}}^1_R(k,-)^{\mu }$
.
Theorem 5.1.12. Let
$(R,\mathfrak {m},k)$
be a local ring of depth
$t>0$
. Then, the following are equivalent:
-
(1) R is regular.
-
(2)
${\operatorname{Ext}}^1_R(k,R/(x_1,...,x_{t-1})R)^{\mu }=0$
for some R-regular sequence
$x_1,...,x_{t-1}$
. -
(3)
${\operatorname{Ext}}^1_R(k,M)^{\mu }=0$
for some finitely generated R-module M of projective dimension
$t-1$
.
Proof.
$(1) \implies (2)$
Since R is regular, we have
$t=\dim R$
. Since R is regular, we get that
$\mathfrak {m}$
is generated by a regular sequence
$x_1,...,x_t$
. Now,
$R/(x_1,...,x_{t-1})R$
has projective dimension
$t-1$
over R and the
$(t-1)$
th Betti number of this module is
$1$
(by looking at the Koszul complex). Hence,
${\operatorname{Ext}}^1_R(k,R/(x_1,...,x_{t-1})R)\cong \operatorname{Tor}^R_{t-1}(k,R/(x_1,...,x_{t-1})R)\cong k$
by [Reference Bruns and Herzog3, Exer. 3.3.26]. Now, we have an exact sequence
$0\to R/(x_1,...,x_{t-1})R\xrightarrow {\cdot x_t} R/(x_1,...,x_{t-1})R\to R/(x_1,...,x_{t-1},x_t)R\cong k\to 0$
, which is clearly not
$\mu $
-additive. Hence,
${\operatorname{Ext}}^1_R(k,R/(x_1,...,x_{t-1})R)\ne {\operatorname{Ext}}^1_R(k,R/(x_1,...,x_{t-1})R)^{\mu }$
. Since
${\operatorname{Ext}}^1_R(k,R/(x_1,...,x_{t-1})R)\cong k$
is cyclic, by Lemma 5.1.2, we get that
${\operatorname{Ext}}^1_R(k,R/(x_1,..., x_{t-1})R)^{\mu }=\mathfrak {m} {\operatorname{Ext}}^1_R(k,R/(x_1,...,x_{t-1})R)=0$
.
$(2) \implies (3)$
Obvious.
$(3) \implies (1)$
By the Auslander–Buchsbaum formula,
$\operatorname{depth}\ M=1$
, so
$\operatorname{Soc}(M)=0$
. Hence, by prime avoidance, we can choose
$x\in \mathfrak {m}$
, which is both R and M-regular. Then
$xM\cong M$
, so
${\operatorname{Ext}}^1_R(k, xM)^{\mu }\cong {\operatorname{Ext}}^1_R(k,M)^{\mu }=0$
. By Proposition 5.1.11, we get that
$xM$
is a weakly
$\mathfrak {m}$
-full submodule of M. Since
$\operatorname{depth} (M/xM)=0$
,
$xM$
is a Burch submodule of M by [Reference Dey and Kobayashi12, Lem. 4.3]. Moreover,
$\operatorname{pd}_{R/xR}(M/xM)=\operatorname{pd}_R M<\infty $
, and
$\operatorname{pd}_R R/xR <\infty $
, so
$\operatorname{pd}_R M/xM<\infty $
. Hence,
$\operatorname{Tor}^R_{\gg 0}(k,M/xM)=0$
. Thus,
$\operatorname{pd}_R k<\infty $
by [Reference Dey and Kobayashi12, Th. 1.2], so R is regular.
Next, we try to relate
${\operatorname{Ext}}^1_R(M,N)^{\mu }$
to
${\operatorname{Ext}}^1_R(M,N)$
for all
$M,N\in {\operatorname{mod}}\ R$
, when
$(R,\mathfrak {m})$
is a regular local ring of dimension
$1$
(i.e., a local PID). For this, we first record a preliminary lemma.
Lemma 5.1.13. Let
$(R,\mathfrak {m})$
be a local ring, and let
$x\in \mathfrak {m}$
be a nonzero divisor. Then,
${\operatorname{Ext}}^1_R(R/xR, R/I)^{\mu }=\mathfrak {m}{\operatorname{Ext}}^1_R(R/xR, R/I)\cong \dfrac {\mathfrak {m}}{I+xR}$
for every proper ideal I of R.
Proof. Calculating
${\operatorname{Ext}}^1_R(R/xR, R/I)$
from the minimal free-resolution
$0\to R \xrightarrow {\cdot x} R\to R/xR\to 0$
of
$R/xR$
, we see that
${\operatorname{Ext}}^1_R(R/xR,R/I)\cong R/(xR+I)$
is a cyclic R-module. By Lemma 5.2.1, we have
$\mathfrak {m}{\operatorname{Ext}}^1_R(R/xR,R/I)\subseteq {\operatorname{Ext}}^1_R(R/xR,R/I)^{\mu }$
. We finally claim that
${\operatorname{Ext}}^1_R(R/xR,R/I)^{\mu }\neq {\operatorname{Ext}}^1_R(R/xR,R/I)$
. Indeed, we have an exact sequence
$\sigma : 0\to xR/xI \to R/xI\to R/xR\to 0$
. Now, we have a natural surjection
$R \xrightarrow {r\mapsto rx+xI} xR/xI$
, whose kernel is
$\{r\in R: rx\in xI\}$
. Since x is a nonzero divisor,
$rx\in xI$
if and only if
$r\in I$
. Hence, the kernel is I. Hence,
$R/I\cong xR/xI$
. So, we get the exact sequence
$\sigma : 0\to R/I\to R/xI \to R/xR\to 0$
. Moreover,
$\sigma $
is not
$\mu $
-additive, since
$\mu (R/xI)=1\ne 1+1= \mu (R/I)+\mu (R/xR)$
. Thus,
$[\sigma ]\in {\operatorname{Ext}}^1_R(R/xR, R/I)\setminus {\operatorname{Ext}}^1_R(R/xR, R/I)^{\mu }$
. Hence,
${\operatorname{Ext}}^1_R(R/xR,R/I)^{\mu }=\mathfrak {m} {\operatorname{Ext}}^1_R(R/xR,R/I)$
by Lemma 5.1.2.
Now, using this lemma, we can compare the structure of
${\operatorname{Ext}}^1_R(M,N)^{\mu }$
to
${\operatorname{Ext}}^1_R(M,N)$
for every pair of finitely generated modules
$M,N$
over a DVR.
Proposition 5.1.14. Let
$(R,\mathfrak {m})$
be a regular local ring of dimension
$1$
. Then,
${\operatorname{Ext}}^1_R(M,N)^{\mu }=\mathfrak {m}{\operatorname{Ext}}^1_R(M,N)$
for all finitely generated R-modules M and N. So, in particular,
${\operatorname{Ext}}^1_R(k,N)^{\mu }=0$
for all finitely generated R-modules N.
Proof. Let
$\mathfrak {m}=xR$
. Then, for every finitely generated R-module X, we have
$X\cong R^{\oplus a}\oplus \left (\oplus _{i=1}^n R/x^{a_i}R\right )$
for some nonnegative integers (depending on X)
$a,a_i$
. Now, fix arbitrary
$N\in {\operatorname{mod}}\ R$
. Applying Lemma 2.4 to the subfunctor
${\operatorname{Ext}}^1_R(-,N)^{\mu }$
of
${\operatorname{Ext}}^1_R(-,N)$
and taking
$I=\mathfrak {m}$
, it is enough to prove that
${\operatorname{Ext}}^1_R(R,N)^{\mu }=\mathfrak {m}{\operatorname{Ext}}^1_R(R,N)$
and
${\operatorname{Ext}}^1_R(R/x^lR,N)^{\mu }=\mathfrak {m}{\operatorname{Ext}}^1_R(R/x^lR,N)$
for every integer
$l\ge 1$
. Now,
${\operatorname{Ext}}^1_R(R,N)^{\mu }=\mathfrak {m}{\operatorname{Ext}}^1_R(R,N)$
is obvious, as both sides are zero. Now, to prove
${\operatorname{Ext}}^1_R(R/x^lR,N)^{\mu }=\mathfrak {m}{\operatorname{Ext}}^1_R(R/x^lR,N)$
for every integer
$l\ge 1$
, first fix an
$l\ge 1$
, and look at the subfunctor
${\operatorname{Ext}}^1_R(R/x^lR,-)^{\mu }$
of
${\operatorname{Ext}}^1_R(R/x^lR,-)$
. Again, by the structure of finitely generated R-modules and Lemma 2.4, it is enough to prove that
${\operatorname{Ext}}^1_R(R/x^lR,R)^{\mu }=\mathfrak {m} {\operatorname{Ext}}^1_R(R/x^lR,R)$
and
${\operatorname{Ext}}^1_R(R/x^lR,R/x^bR)^{\mu }=\mathfrak {m} {\operatorname{Ext}}^1_R(R/x^lR,R/x^bR)$
for every integer
$b\ge 1$
. Now, these equalities follow from Lemma 5.1.13, since x is a nonzero divisor.
When restricting to short exact sequences in
${\operatorname{Ext}}^1$
, on which certain subadditive function, other than
$\mu (-)$
, is additive, one obtains vanishing of the corresponding submodule of
${\operatorname{Ext}}^1_R(M,F)$
for any free R-module F. We make this precise in Proposition 5.1.16, whose proof uses the following lemma.
Lemma 5.1.15. Let
$(R,\mathfrak {m})$
be a local Cohen–Macaulay ring of dimension
$1$
. Let
$x\in \mathfrak {m}$
be R-regular. Let
$\sigma :0\to F_1\to F_2\oplus \frac {R}{x^aR}\to \frac {R}{x^bR}\to 0$
be a short exact sequence, where
$a,b$
are nonnegative integers and
$F_1,F_2$
are free R-modules. Let c be an integer such that
$c\ge b$
and
$\sigma \otimes \frac {R}{x^cR}$
is short exact. Then
$a=b$
and
$\operatorname{rank}(F_2)= \operatorname{rank}(F_1)$
. So, in particular,
$\sigma $
is split exact.
Proof. Since
$x\in R$
is a nonzero divisor, so
$\frac {R}{x^aR}, \frac {R}{x^bR}$
are torsion modules, that is, have constant rank
$0$
. Hence, calculating rank along
$\sigma $
, we get
$\operatorname{rank}(F_2)= \operatorname{rank}(F_1)$
. Next, we will show that
$a\le b$
. Dualizing
$\sigma $
by R, we get the following part of a long exact sequence:
${\operatorname{Ext}}^1_R\left (\frac {R}{x^bR},R\right )\to {\operatorname{Ext}}^1_R\left (\frac {R}{x^aR},R\right )\to 0$
. Now, since
$x^a\in R$
is a nonzero divisor, by taking the resolution
$0\to R\xrightarrow {\cdot x^a}R\to \frac {R}{x^aR}\to 0$
of
$\frac {R}{x^aR}$
, and dualizing by R, and calculating the cohomology, we get that
${\operatorname{Ext}}^1_R\left (\frac {R}{x^aR},R\right )\cong \frac {R}{x^aR}$
. Similarly,
${\operatorname{Ext}}^1_R\left (\frac {R}{x^bR},R\right )\cong \frac {R}{x^bR}$
. Hence, we have the exact sequence
$\frac {R}{x^bR}\to \frac {R}{x^aR}\to 0$
. Since
$x^b$
annihilates
$\frac {R}{x^bR}$
, we have
$x^b$
annihilates
$\frac {R}{x^aR}$
as well, which implies
$x^bR\subseteq x^aR$
. Since
$x\in \mathfrak {m}$
is R-regular, so
$x^bR\subseteq x^aR$
implies that
$a\leq b$
. Since
$\sigma \otimes \frac {R}{x^cR}$
is short exact, we have the following short exact sequence:
$$ \begin{align*}\sigma\otimes \frac{R}{x^cR}:0\to \left(\frac{R}{x^cR}\right)^{\oplus l_1}\to\left(\frac{R}{x^cR}\right)^{\oplus l_2}\oplus\frac{R}{x^aR}\to \frac{R}{x^bR}\to 0,\end{align*} $$
where we have used that
$\frac {R}{x^cR}\otimes \frac {R}{x^aR}\cong \frac {R}{x^cR+x^aR}\cong \frac {R}{x^aR}$
and
$\frac {R}{x^cR}\otimes \frac {R}{x^bR}\cong \frac {R}{x^cR+x^bR}\cong \frac {R}{x^bR}$
, as
$a\leq b\leq c$
. Since R is local Cohen–Macaulay of dimension
$1$
and
$x\in \mathfrak {m}$
is R-regular, we have
$x^aR, x^b R, x^c R$
are
$\mathfrak {m}$
-primary ideals. Hence, by calculating the length along the short exact sequence
$\sigma \otimes \frac {R}{x^cR}$
, we get that
$\lambda \left (\frac {R}{x^bR}\right )-\lambda \left (\frac {R}{x^aR}\right )=(l_2-l_1)\lambda \left (\frac {R}{x^cR}\right )=0$
. Since
$x^bR\subseteq x^aR$
, so
$\lambda \left (\frac {R}{x^bR}\right )=\lambda \left (\frac {R}{x^aR}\right )$
now implies
$x^aR=x^bR$
by calculating length along the short exact sequence
$0\to \frac {x^aR}{x^bR}\to \frac {R}{x^bR}\to \frac {R}{x^aR}\to 0$
. Since
$x\in \mathfrak {m}$
is R-regular, we have
$a=b$
. Finally, since
$F_1\cong F_2$
and
$a=b$
, we get
$\sigma $
is split exact.
In the following,
$\operatorname{H}^0_{\mathfrak {m}}(-)$
denotes the zeroth local cohomology module. For a finite length R-module M,
$\ell \ell (M)$
will stand for the smallest integer
$n\ge 0$
such that
$\mathfrak {m}^n M=0$
.
Proposition 5.1.16. Let
$(R,\mathfrak {m})$
be a regular local ring of dimension
$1$
. Let
$L\in {\operatorname{mod}}(R)$
. Let c be an integer such that
$c\ge \ell \ell (\operatorname{H}^0_{\mathfrak {m}}(L))$
. Consider the function
$\phi _L(-):=\lambda (\frac {R}{\mathfrak {m}^{c}}\otimes -):{\operatorname{mod}}(R)\to \mathbb {N}\cup \{0\}$
. Then, for any free R-module F, we have
${\operatorname{Ext}}_R^1(L,F)^{\phi _L}=0$
.
Proof. Note that R is not a field, since
$\dim R=1$
. Since R is a regular local ring, we have
$L\cong G\oplus L'$
, where G is a finite free R-module and
$L'$
is an R-module of finite length. We have
$\operatorname{H}^0_{\mathfrak {m}}(L)\cong L'$
. Since
${\operatorname{Ext}}^1_R(L,F)^{\phi _L}\cong {\operatorname{Ext}}^1_R(G,F)^{\phi _L}\oplus {\operatorname{Ext}}^1_R(L',F)^{\phi _L}\cong {\operatorname{Ext}}^1_R(L',F)^{\phi _L}$
, we may replace L by
$L'$
, and assume without loss of generality that L has finite length, and
$c\ge \ell \ell (L)$
. For simplicity, we denote
$\phi _L=\phi $
. Now, it is enough to show that
${\operatorname{Ext}}_R^1(L,R)^{\phi }=0$
. Since
$(R,\mathfrak {m})$
is a regular local ring of dimension
$1$
, R is a PID. Hence,
$\mathfrak {m}=xR$
for some
$x\neq 0$
. Since L is a finite length module, by the structure theorem of modules over a PID, we have
$L\cong \oplus _{i=1}^{n} \frac {R}{x^{b_i}R}$
, where
$b_i>0$
. So,
$\ell \ell (L)=\max _{1\le i\le n}{b_i}$
. Hence, it is enough to show that
${\operatorname{Ext}}_R^1\left (\frac {R}{x^{b_i}R},R\right )^{\phi }=0$
for all
$i=1,\ldots ,n$
. Fix an
$i\in \{1,\ldots ,n\}$
and consider a short exact sequence
$\sigma :0\to R\to X\to \frac {R}{x^{b_i}R}\to 0$
in
${\operatorname{Ext}}_R^1\left (\frac {R}{x^{b_i}R},R\right )^{\phi }$
. By the structure theorem of modules over a PID, we also have
$X\cong R^{\oplus s}\oplus \left (\oplus _{j=1}^t\frac {R}{x^{a_j}R}\right )$
, where
$a_j>0$
. As R is an integral domain and
$\frac {R}{x^{b_i}R}$
and
$\frac {R}{x^{a_j}R}$
are all torsion R-modules, so calculating rank along the short exact sequence
$\sigma $
, we obtain
$s=1$
. Moreover,
$1+t=s+t=\mu (X)\leq 2$
, since
$\mu $
is subadditive. Thus,
$t\leq 1$
. If
$t=0$
, then we obtain
$\sigma :0\to R\to R \oplus \frac {R}{x^0R}\to \frac {R}{x^{b_i}R}\to 0$
. Since
$\sigma $
is
$\phi $
-additive, by using Lemma 4.12 with the functor
$\lambda _R\left (\frac {R}{\mathfrak {m}^{c}} \otimes _R -\right )$
, we obtain that
$\sigma \otimes \frac {R}{\mathfrak {m}^{c}}=\sigma \otimes \frac {R}{x^{c}R}$
is short exact. As
$c\geq b_i>0$
, we obtain a contradiction from Lemma 5.1.15. Thus,
$t=1$
. So, we get
$X\cong R\oplus \frac {R}{x^{a}R}$
for some
$a=a_j$
. Thus, we have the short exact sequence
$\sigma :0\to R\to R\oplus \frac {R}{x^{a}R}\to \frac {R}{x^{b_i}R}\to 0$
.
Since
$c\ge \ell \ell (L)=\max _{1\le i\le n}{b_i}$
, we have
$c\ge b_i$
for all
$i=1,\ldots ,n$
. Hence, by applying Lemma 5.1.15 on
$\sigma $
, we get that
$\sigma $
is split exact. Since
$\sigma $
is an arbitrary element of
${\operatorname{Ext}}_R^1\left (\frac {R}{x^{b_i}R},R\right )^{\phi }$
, we get
${\operatorname{Ext}}_R^1\left (\frac {R}{x^{b_i}R},R\right )^{\phi }=0$
for all
$i=1,\ldots ,n$
. This implies that
${\operatorname{Ext}}_R^1\left (L,R\right )^{\phi }=0$
, so
${\operatorname{Ext}}_R^1\left (L,F\right )^{\phi }=0$
.
Taking
$L=k$
and
$c=1=\ell \ell (k)$
, we see that
$\phi _L(-)=\mu (-)$
in Proposition 5.1.16. So, we also get another proof of Corollary 5.1.4 in dimension
$1$
.
Next, we compare
${\operatorname{Ext}}^1_R(M,R)^{\mu }$
with
${\operatorname{Ext}}^1_R(M,R)$
. In arbitrary dimension, we only consider this for local Cohen–Macaulay rings of minimal multiplicity. We first record some general preliminary lemmas.
Lemma 5.1.17. Let
$(R,\mathfrak {m},k)$
be a local ring such that
$\mathfrak {m}^2=0$
and
$\mathfrak {m}\neq 0$
. Let
$e:=\mu (\mathfrak {m})$
. Then
$\mu (\omega _R)=e$
,
$\mu (\Omega _R\omega _R)=e^2-1$
, and
$\Omega _R\omega _R\cong k^{\oplus (e^2-1)}$
.
Proof. Since
$\mathfrak {m}^2=0$
and
$\mathfrak {m}\neq 0$
, we have
$\mathfrak {m}\subseteq (0:_R\mathfrak {m})\subsetneq R$
. Hence,
$\mathfrak {m}=(0:_R\mathfrak {m})$
. Then we have
${\mu }({\omega }_R)={\operatorname{r}}(R)=\dim _k(0:_R\mathfrak {m})=\dim _k\mathfrak {m}=\dim _k\left (\frac {\mathfrak {m}}{\mathfrak {m}^2}\right )=\mu (\mathfrak {m})=e$
. Also, note that
$\lambda (R)=\lambda \left (\frac {R}{\mathfrak {m}^2}\right )=\lambda \left (\frac {R}{\mathfrak {m}}\right )+\lambda \left (\frac {\mathfrak {m}}{\mathfrak {m}^2}\right )=1+\mu (\mathfrak {m})$
. Next, consider the short exact sequence
$0\to \Omega _R\omega _R\to R^{\oplus \mu (\omega _R)}\to \omega _R\to 0$
, so
$\Omega _R\omega _R\subseteq \mathfrak {m} R^{\oplus \mu (\omega _R)}$
. Hence,
$\mathfrak {m}\Omega _R\omega _R=0$
. This implies that
$\mu (\Omega _R\omega _R)=\lambda \left (\frac {\Omega _R\omega _R}{\mathfrak {m}\Omega _R\omega _R}\right )=\lambda (\Omega _R\omega _R)=\lambda (R^{\oplus \mu (\omega _R)})-\lambda (\omega _R)=\mu (\omega _R)\lambda (R)-\lambda (R)=\lambda (R)(\mu (\omega _R)-1)=\lambda (R)(\mu (\mathfrak {m})-1)=(1+\mu (\mathfrak {m}))(\mu (\mathfrak {m})-1)=e^2-1$
, where
$\lambda (\omega _R)=\lambda (R)$
follows from Matlis duality.
Lemma 5.1.18. Let
$(R,\mathfrak {m},k)$
be a local ring, and let M be a finitely generated R-module. If x is M-regular, then
$\frac {\Omega _R M}{x\Omega _R M}\cong \Omega _{R/xR}\left (\frac {M}{xM}\right )$
.
Proof. Since x is M-regular, we have
$\operatorname{Tor}_1^R(M,\frac {R}{xR})=0$
. Hence, tensoring the short exact sequence
$0\to \Omega _R M\to R^{\oplus \mu _R(M)}\to M\to 0$
with
$R/xR$
, we get the exact sequence
$0\to \frac {\Omega _R M}{x\Omega _R M}\to (\frac {R}{xR})^{\oplus \mu _R(M)}\to \frac {M}{xM}\to 0$
. Since
$\mu _R(M)=\mu _{\frac {R}{xR}}(\frac {M}{xM})$
, we have
$\frac {\Omega _R M}{x\Omega _R M}\cong \Omega _{R/xR}\left (\frac {M}{xM}\right )$
.
5.1.19. Since
$\mu _R(-)=\lambda _R\left (\dfrac {(-)}{\mathfrak {m}(-)}\right )$
, it follows by 4.3 that
$\mu _R((-))=\mu _S(S\otimes _R(-))$
, where
$S=R[X]_{\mathfrak {m}[X]}$
. Since tensoring with S preserves exactness, so tensoring a minimal free resolution
$(F_{\bullet },\partial _{\bullet })$
of an R-module M with S and remembering
$S\otimes \partial $
now have entries in
$\mathfrak {m} S$
, the maximal ideal of S, we see that
$S\otimes _R \Omega _R M\cong \Omega _S(S\otimes _S M)$
. Also, if
$\omega _R$
exists, then owing to the fact that
$S/\mathfrak {m} S$
is a field, we see that
$\omega _S$
also exists and
$\omega _S\cong S\otimes _R \omega _R$
by [Reference Bruns and Herzog3, Th. 3.3.14(a)]. Hence,
$S\otimes _R \operatorname{Hom}_R(-,\omega _R)\cong \operatorname{Hom}_S\left (S\otimes _R(-),\omega _S\right )$
. Finally, we also have
${\operatorname{r}}(R)={\operatorname{r}}(S)$
by [Reference Bruns and Herzog3, Prop. 1.2.16(b)].
Proposition 5.1.20. Let
$(R,\mathfrak {m},k)$
be a local Cohen–Macaulay ring of minimal multiplicity admitting a canonical module
$\omega _R$
, and let
$\mathfrak {m}\neq 0$
. Then
$\mu ((\Omega _R\omega _R)^{\dagger })={\operatorname{r}}(R)^2-1$
.
Proof. Due to 5.1.19, we may pass to the faithfully flat extension
$S:=R[X]_{\mathfrak {m}[X]}$
and assume that the residue field is infinite.
We will prove the claim by induction on
$\dim R=d$
. First, let
$d=0$
. Since R has minimal multiplicity, we have
$\mathfrak {m}^2=0$
. Hence, by Lemma 5.1.17, we get that
$\Omega _R\omega _R\cong k^{\oplus ({\operatorname{r}}(R)^2-1)}$
. This implies
$(\Omega _R\omega _R)^{\dagger }\cong k^{\oplus ({\operatorname{r}}(R)^2-1)}$
, so
$\mu ((\Omega _R\omega _R)^{\dagger })={\operatorname{r}}(R)^2-1$
. Now, let
$\dim R=d\ge 1$
, and let the claim be true for rings with dimension
$d-1$
. Let
$x\in \mathfrak {m}$
be such that
$\mathfrak {m}^2=(x,x_1,...,x_{d-1})\mathfrak {m}$
(see [Reference Bruns and Herzog3, Exer. 4.6.14(c)]). So,
$\frac {R}{xR}$
has minimal multiplicity. Now, we have
$$ \begin{align*} \frac{(\Omega_R\omega_R)^{\dagger}}{x(\Omega_R\omega_R)^{\dagger}} &\cong\left(\frac{\Omega_R\omega_R}{x\Omega_R\omega_R}\right)^{\dagger}\cong \left(\Omega_{\frac{R}{xR}}\left(\frac{\omega_R}{x\omega_R}\right)\right)^{\dagger}\cong \left(\Omega_{\frac{R}{xR}}\omega_{\frac{R}{xR}}\right)^{\dagger}, \end{align*} $$
where the first isomorphism follows from [Reference Bruns and Herzog3, Prop. 3.3.3(a)], and the second isomorphism is by Lemma 5.1.18. So,
$$ \begin{align*} \mu_R((\Omega_R\omega_R)^{\dagger}) =\mu_{\frac{R}{xR}}\left(\frac{(\Omega_R\omega_R)^{\dagger}}{x(\Omega_R\omega_R)^{\dagger}}\right)&=\mu_{\frac{R}{xR}}\left(\left(\Omega_{\frac{R}{xR}}\omega_{\frac{R}{xR}}\right)^{\dagger}\right)\\ &=r\left(\frac{R}{xR}\right)^2-1\text{ [By induction hypothesis]}\\ &={\operatorname{r}}(R)^2-1.\\[-45pt] \end{align*} $$
Proposition 5.1.21. Let
$(R,\mathfrak {m},k)$
be a local Cohen–Macaulay ring of minimal multiplicity admitting a canonical module
$\omega _R$
. Then the exact sequence
$0\to R\to \omega _R^{\oplus \mu (\omega _R)}\to (\Omega _R\omega _R)^{\dagger }\to 0$
is
$\mu $
-additive.
Proof. Consider the exact sequence
$0\to \Omega _R\omega _R\to R^{\oplus \mu (\omega _R)}\to \omega _R\to 0$
. Since
${\operatorname{Ext}}^1_R(\omega _R,\omega _R)=0$
, we have the exact sequence
$0\to R\to \omega _R^{\oplus \mu (\omega _R)}\to (\Omega _R\omega _R)^{\dagger }\to 0$
. From Proposition 5.1.20, we have
$\mu ((\Omega _R\omega _R)^{\dagger })={\operatorname{r}}(R)^2-1=\mu (\omega _R)^2-\mu (R)=\mu (\omega _R^{\oplus \mu (\omega _R)})-\mu (R)$
. Hence,
$0\to R\to \omega _R^{\oplus \mu (\omega _R)}\to (\Omega _R\omega _R)^{\dagger }\to 0$
is
$\mu $
-additive.
Proposition 5.1.22. Let
$(R,\mathfrak {m},k)$
be a local Cohen–Macaulay ring of minimal multiplicity. Then
${\operatorname{Ext}}^1_R(M,F)={\operatorname{Ext}}^1_R(M,F)^{\mu }$
for any maximal Cohen–Macaulay R-module M and any finitely generated free R-module F.
Proof. By Lemma 2.4, it is enough to prove the claim for
$F=R$
.
We first consider the case when R has a canonical module
$\omega _R$
. Consider the
$\mu $
-additive (by Proposition 5.1.21) exact sequence
$\sigma :0\to R\to \omega _R^{\oplus \mu (\omega _R)}\to (\Omega _R\omega _R)^{\dagger }\to 0$
. Applying
$\operatorname{Hom}_R(M,-)$
to
$\sigma $
, we get the following part of a commutative diagram of exact sequences by Corollary 3.9:
Since
$\omega _R$
has finite injective dimension, we have
${\operatorname{Ext}}_{R}^1(M,\omega _R^{\oplus \mu (\omega _R)})=0$
by [Reference Bruns and Herzog3, Exer. 3.1.24]. So,
${\operatorname{Ext}}_{R}^1(M,\omega _R^{\oplus \mu (\omega _R)})^{\mu }=0$
as well. Hence, we get the following commutative diagram:
Thus,
$h\circ g=f$
is surjective, so h is surjective. Since h is the natural inclusion map, we have
${\operatorname{Ext}}^1_R(M,R)={\operatorname{Ext}}^1_R(M,R)^{\mu }$
.
Now, we consider the general case. Since
${\operatorname{Ext}}^1_R(M,R)^{\mu }\subseteq {\operatorname{Ext}}_{R}^1(M,R)$
, it is enough to prove the other inclusion. So, let
$\sigma : 0\to R\to X\to M\to 0 $
be an exact sequence. We need to show
$\sigma $
is
$\mu $
-additive. Now, consider the completion
$\widehat \sigma : 0\to \widehat R\to \widehat X\to \widehat M \to 0$
. Since
$\widehat R$
is Cohen–Macaulay, having minimal multiplicity, admitting a canonical module, and
$\widehat M$
is maximal Cohen–Macaulay over
$\widehat R$
, by the first part of the proof, we get
$ {\operatorname{Ext}}^1_{\widehat R}(\widehat M,\widehat R)={\operatorname{Ext}}^1_{\widehat R}(\widehat M,\widehat R)^{\mu }$
. Thus,
$[\widehat \sigma ]\in {\operatorname{Ext}}^1_{\widehat R}(\widehat M,\widehat R)={\operatorname{Ext}}^1_{\widehat R}(\widehat M,\widehat R)^{\mu }$
. Hence,
$\widehat \sigma $
is
$\mu $
-additive. Since the number of generators does not change under completion, we get
$\sigma $
is
$\mu $
-additive, which is what we wanted to prove.
Corollary 5.1.23. Let
$(R,\mathfrak {m},k)$
be a local Cohen–Macaulay ring of dimension d and minimal multiplicity. If
${\operatorname{Ext}}^1_R(\Omega ^{i+d}_R k,R)^{\mu }=0$
for some
$i\geq 0$
, then R is a hypersurface.
Proof. By Proposition 5.1.22, we have
${\operatorname{Ext}}^{d+i+1}_R(k,R)\cong {\operatorname{Ext}}^1_R(\Omega ^{d+i}_R k, R)={\operatorname{Ext}}^1_R(\Omega ^{d+i}_R k, R)^{\mu }=0$
. Hence, R has finite injective dimension by [Reference Roberts32, II, Th. 2], so R is Gorenstein. Since R has minimal multiplicity, R is a hypersurface.
One can also detect when a local ring has depth
$0$
by comparing
${\operatorname{Ext}}^1_R(M,R)$
to
${\operatorname{Ext}}^1_R(M,R)^{\mu }$
as shown in the following proposition. In the following,
$\operatorname{Tr}(-)$
stands for Auslander transpose (see [Reference Leuschke and Wiegand27, Def. 12.3]).
Proposition 5.1.24. Let I be an ideal of a local ring
$(R,\mathfrak {m},k)$
. Then,
${\operatorname{Ext}}^1_R(\operatorname{Tr}(R/I),R/\operatorname{ann}_R(I))^{\mu }=\mathfrak {m} {\operatorname{Ext}}^1_R(\operatorname{Tr}(R/I),R/\operatorname{ann}_R(I))$
. Moreover, the following are equivalent:
-
(1)
$\operatorname{depth} R=0$
. -
(2)
${\operatorname{Ext}}^1_R(M,F)^{\mu }={\operatorname{Ext}}^1_R(M,F)$
for all finitely generated R-modules M and F, where F is free. -
(3)
${\operatorname{Ext}}^1_R(\operatorname{Tr} k,R)^{\mu }= {\operatorname{Ext}}^1_R(\operatorname{Tr} k,R)$
.
Proof. Since everywhere in this proposition the Auslander transpose is in the first component of
${\operatorname{Ext}}^1$
, our claim does not depend on the choice of
$\operatorname{Tr}$
. First, we prove
${\operatorname{Ext}}^1_R(\operatorname{Tr}(R/I),R/\operatorname{ann}_R(I))^{\mu }=\mathfrak {m} {\operatorname{Ext}}^1_R(\operatorname{Tr}(R/I),R/\operatorname{ann}_R(I))$
. We may assume
$\operatorname{Tr}(R/I)$
is non-free. Dualizing the exact sequence
$R^{\oplus \mu (I)} \to R \xrightarrow {\pi } R/I \to 0$
by R, we get
$0\to \operatorname{Hom}_R(R/I,R)\xrightarrow {\pi ^*} R\to R^{\oplus \mu (I)}\to \operatorname{Tr}(R/I)\to 0$
. Under the natural identification
$\operatorname{Hom}_R(R/I,R)\cong \operatorname{ann}_R(I)$
, the map
$\operatorname{Hom}_R(R/I,R)\xrightarrow {\pi ^*} R$
can be identified with the inclusion map
$\operatorname{ann}_R(I)\to R$
, giving us
$0\to \operatorname{ann}_R(I) \to R \to R^{\oplus \mu (I)}\to \operatorname{Tr}(R/I)\to 0$
. Hence, we get an exact sequence
$0\to R/\operatorname{ann}_R(I) \to R^{\oplus \mu (I)}\to \operatorname{Tr}(R/I)\to 0$
. Since
$\operatorname{Tr}(R/I)$
is non-free, this sequence gives us
$\Omega _R \operatorname{Tr}(R/I)\cong R/\operatorname{ann}_R(I)$
and the first Betti number of
$\operatorname{Tr}(R/I)$
is
$1$
. Hence,
${\operatorname{Ext}}^1_R(\operatorname{Tr}(R/I),R/\operatorname{ann}_R(I))^{\mu }=\mathfrak {m} {\operatorname{Ext}}^1_R(\operatorname{Tr}(R/I),R/\operatorname{ann}_R(I))$
by Lemma 5.1.3.
Now, we prove the equivalence of the three conditions as follows:
$(1) \implies (2)$
: Let
$\sigma : 0\to F\to X\to M\to 0$
be an exact sequence, where F is a finitely generated free R-module. Since
$\operatorname{depth} R=0$
, we get that k embeds inside R, that is, k is torsionless. Hence,
${\operatorname{Ext}}^1_R(\operatorname{Tr} k, R)=0$
by [Reference Leuschke and Wiegand27, Prop. 12.5]. Hence,
${\operatorname{Ext}}^1_R(\operatorname{Tr} k, F)=0$
. So, we get an exact sequence
$0\to \operatorname{Hom}_R(\operatorname{Tr} k,F)\to \operatorname{Hom}_R(\operatorname{Tr} k, X) \to \operatorname{Hom}_R(\operatorname{Tr} k, M)\to 0$
. Then, by [Reference Leuschke and Wiegand27, Exer. 13.36], we get that the sequence
$0\to F \otimes _R k \to X \otimes _R k \to M \otimes _R k \to 0$
is also exact, which means
$\sigma $
is
$\mu $
-additive. Thus,
${\operatorname{Ext}}^1_R(M,F)^{\mu }={\operatorname{Ext}}^1_R(M,F)$
.
$(2) \implies (3)$
Obvious.
$(3) \implies (1)$
: Assume
${\operatorname{Ext}}^1_R(\operatorname{Tr} k,R)^{\mu }={\operatorname{Ext}}^1_R(\operatorname{Tr} k,R)$
. Now, if possible let
$\operatorname{depth} R>0$
. Then
$\mathfrak {m}$
contains a nonzero divisor, so
$\operatorname{ann}_R(\mathfrak {m})=0$
. Then, by the first part of this proposition, we get
${\operatorname{Ext}}^1_R(\operatorname{Tr} k,R)^{\mu }=\mathfrak {m} {\operatorname{Ext}}^1_R(\operatorname{Tr} k,R)$
. Hence,
${\operatorname{Ext}}^1_R(\operatorname{Tr} k,R)=\mathfrak {m} {\operatorname{Ext}}^1_R(\operatorname{Tr} k,R)$
. Then, by Nakayama’s lemma,
${\operatorname{Ext}}^1_R(\operatorname{Tr} k,R)=0$
. Hence, by [Reference Leuschke and Wiegand27, Prop. 12.5], we have an embedding
$k \to k^{**}$
. Consequently,
$k^*\ne 0$
, that is,
$\operatorname{depth} R=0$
.
For general local Cohen–Macaulay rings of dimension
$1$
, we now show that if
${\operatorname{Ext}}^1_R(M,R)^{\mu }=0$
for some I-Ulrich module [Reference Dao, Maitra and Sridhar11, Def. 4.1]
$M\subseteq Q(R)$
containing a nonzero divisor of R, then I is principal. For this, we first need the following lemma. For the remainder of this section, given R-submodules
$M,N$
of
$Q(R)$
, by
$(M:N)$
, we will mean
$\{x\in Q(R): xN\subseteq M\}$
.
Lemma 5.1.25. Let
$(R,\mathfrak {m})$
be a local Cohen–Macaulay ring of dimension
$1$
. Let I be an
$\mathfrak {m}$
-primary ideal of R admitting a principal reduction
$a\in I$
. Assume
$(I:I)=R$
. If
$M\subseteq Q(R)$
is an I-Ulrich module, and contains a nonzero divisor of R, then the natural inclusion map
$\operatorname{Hom}_R\left (M,(a)\right )\to \operatorname{Hom}_R(M,I)$
, induced by the inclusion
$(a) \to I$
, is an isomorphism.
Proof. The natural inclusion
$0\to (a) \xrightarrow {i} I$
induces the following commutative diagram:
where the rows are natural inclusion maps, and the vertical arrows are isomorphisms due to [Reference Kobayashi and Takahashi26, Prop. 2.4(1)]. So, it is enough to show that
$(I:M)\subseteq ((a):M)$
. Indeed, if
$x\in (I:M)$
, then
$xM\subseteq I$
. Since M is I-Ulrich, M is a
$B(I)=R\left [\dfrac Ia\right ]$
-module (see [Reference Dao, Maitra and Sridhar11, Rem. 4.4 and Th. 4.6]). Hence,
$\dfrac I a M\subseteq M$
, so
$\dfrac I a xM\subseteq xM\subseteq I$
. Thus,
$\dfrac 1 a xM\subseteq (I:I)=R$
, so
$xM\subseteq (a)$
. Therefore,
$x\in \left ((a):M\right )$
.
Proposition 5.1.26. Let
$(R,\mathfrak {m})$
be a local Cohen–Macaulay ring of dimension
$1$
. Let I be an
$\mathfrak {m}$
-primary ideal of R admitting a principal reduction
$a\in I$
. Assume
$(I:I)=R$
. If there exists an I-Ulrich module
$M\subseteq Q(R)$
, containing a nonzero divisor of R such that
${\operatorname{Ext}}^1_R(M,R)^{\mu }=0$
, then
$I\cong R$
.
Proof. Consider the short exact sequence
$0\to (a) \xrightarrow {i} I \to I/(a)\to 0$
. Since
$I^{n+1}=aI^n$
for all
$n\gg 0$
and I contains a nonzero divisor, we have that a is a nonzero divisor. So,
$(a)\cong R$
. If
$a\in \mathfrak {m} I$
, then
$I^{n+1}\subseteq \mathfrak {m} II^n=\mathfrak {m} I^{n+1}$
, which implies that
$I^{n+1}=0$
by Nakayama’s lemma. This contradicts the fact that I is
$\mathfrak {m}$
-primary. So,
$a\notin \mathfrak {m} I$
, and hence
$\mu (I/(a))=\mu (I)-1$
. Thus, the above short exact sequence is
$\mu $
-additive. So, by Corollary 3.9, we get the following induced exact sequence:
$$ \begin{align*}0\to \operatorname{Hom}_R(M,(a))\to \operatorname{Hom}_R(M,I)\to \operatorname{Hom}_R(M,I/(a))\to {\operatorname{Ext}}^1_R(M,(a))^{\mu}\cong {\operatorname{Ext}}^1_R(M,R)^{\mu}=0,\end{align*} $$
where the induced map
$ \operatorname{Hom}_R(M,(a))\to \operatorname{Hom}_R(M,I)$
is an isomorphism by Lemma 5.1.25. Hence, we get
$\operatorname{Hom}_R(M,I/(a))=0$
. Now,
$I/(a)$
has finite length. So, if
$I/(a)\ne 0$
, then
$\operatorname{Ass}(I/(a))=\{\mathfrak {m}\}$
. Hence,
$\operatorname{Ass}\left (\operatorname{Hom}_R(M,I/(a))\right )=\operatorname{Supp}(M)\cap \operatorname{Ass}(I/(a))=\operatorname{Supp}(M)\cap \{\mathfrak {m}\}=\{\mathfrak {m}\}$
, contradicting
$\operatorname{Hom}_R(M,I/(a))=0$
. Thus, we must have
$I/(a)=0$
, that is,
$I=(a)\cong R$
.
5.2 Some applications of the subfunctor
${\operatorname{Ext}}^1_{\operatorname{Ul}_I(R)}(-,-)$
Let I be an
$\mathfrak {m}$
-primary ideal. In this subsection, we give various applications of the subfunctor
${\operatorname{Ext}}^1_{\operatorname{Ul}_I^s(R)}(-,-):\operatorname{Ul}_I^s (R)^{op}\times \operatorname{Ul}_I^s(R) \to {\operatorname{mod}}\ R$
(see 4.9 for notation), where we recall from Corollary 4.7 that
$\operatorname{Ul}_I^s(R)$
along with its all short exact sequences form an exact subcategory of
${\operatorname{mod}}\ R$
. Hence,
${\operatorname{Ext}}^1_{\operatorname{Ul}_I^s(R)}(-,-):\operatorname{Ul}_I^s (R)^{op}\times \operatorname{Ul}_I^s(R) \to {\operatorname{mod}}\ R$
is a subfunctor of
${\operatorname{Ext}}^1_R(-,-):\operatorname{Ul}_I^s (R)^{op}\times \operatorname{Ul}_I^s(R) \to {\operatorname{mod}}\ R$
by Proposition 3.8.
We begin by observing a general connection between
${\operatorname{Ext}}^1_R(M,N)$
and
${\operatorname{Ext}}^1_R(M,N)^{\nu _I}$
, where
$M,N\in {\operatorname{mod}}\ R$
and
$\nu _I(-):=\lambda ((-)\otimes _R R/I):{\operatorname{mod}}\ R\to \mathbb Z$
and (see Definition 4.10 for notation of
${\operatorname{Ext}}^1_R(-,-)^{\phi }$
).
Lemma 5.2.1. Let
$(R,\mathfrak {m})$
be a Noetherian local ring, and let I be an
$\mathfrak {m}$
-primary ideal. Then,
$I\ {\operatorname{Ext}}^1_R(M,N)\subseteq {\operatorname{Ext}}^1_R(M,N)^{\nu _I}$
for all
$M,N\in {\operatorname{mod}}\ R$
.
Proof. Let
$\sigma :0\to N \to X \to M \to 0$
be a short exact sequence in
$I\ {\operatorname{Ext}}^1_R(M,N)$
. Then
$\sigma \otimes _R R/I$
splits by [Reference Striuli36, Th. 1.1], so
$M/IM \oplus N/IN \cong X/IX$
. Hence, taking length, we get
$\lambda (M/IM)+\lambda (N/IN)=\lambda (X/IX)$
, so
$\sigma $
is
$\nu _I$
-additive.
This allows us to prove a general connection between
${\operatorname{Ext}}^1_{\operatorname{Ul}^s_I(R)}(M,N)$
and
${\operatorname{Ext}}^1_R(M,N)^{\nu _I}$
, when
$M,N\in \operatorname{Ul}^s_I(R)$
(recall the definition of
$\operatorname{Ul}^s_I(R)$
from Definition 4.5).
Lemma 5.2.2. Let
$(R,\mathfrak {m})$
be a local ring, let
$s\ge 0$
be an integer, and let I be an
$\mathfrak {m}$
-primary ideal. Then,
${\operatorname{Ext}}^1_R(M,N)^{\nu _I}={\operatorname{Ext}}^1_{\operatorname{Ul}^s_I(R)}(M,N)$
for all
$M,N\in \operatorname{Ul}^s_I(R)$
. So, in particular,
$I{\operatorname{Ext}}^1_R(M,N)\subseteq {\operatorname{Ext}}^1_{\operatorname{Ul}_I^s(R)}(M,N)$
for all
$M,N\in \operatorname{Ul}^s_I(R)$
.
Proof. Let
$M,N\in \operatorname{Ul}^s_I(R)$
, and let
$\sigma : 0\to N \to X \to M \to 0$
be a short exact sequence of R-modules. Then
$X\in \operatorname{CM}^s(R)$
(see 4.2), and
$\lambda (M/IM)+\lambda (N/IN)=e_R(I,M)+e_R(I,N)=e_R(I,X)$
. Now,
$\sigma \in {\operatorname{Ext}}^1_R(M,N)^{\nu _I}$
, if and only if
$\lambda (X/IX)=\lambda (M/IM)+\lambda (N/IN)$
, if and only if
$\lambda (X/IX)=e_R(I,X)$
, if and only if
$X\in \operatorname{Ul}^s_I(R)$
if and only if
$\sigma \in {\operatorname{Ext}}^1_{\operatorname{Ul}_I^s(R)}(M,N)$
. This proves the desired equality
${\operatorname{Ext}}^1_R(M,N)^{\nu _I}={\operatorname{Ext}}^1_{\operatorname{Ul}^s_I(R)}(M,N)$
. The last inclusion in the statement now follows from this and Lemma 5.2.1.
When R is Cohen–Macaulay of dimension
$1$
and
$M,N\in \operatorname{Ul}_I(R)$
, then one can improve the inclusion
$I{\operatorname{Ext}}^1_R(M,N)\subseteq {\operatorname{Ext}}^1_{\operatorname{Ul}_I(R)}(M,N)$
of Lemma 5.2.2 quite a bit. For this, we first record a general lemma about trace ideals. In the following, we say that
$\dfrac {a}{b}\in Q(R)$
is a nonzero divisor if a is a nonzero divisor in R. Note that this does not depend on the choice of representative, since
$\frac ab=\frac {a'} {b'}$
in
$Q(R)$
implies
$ab'=a'b$
, and since
$b,b'$
are nonzero divisors in R, so a is a nonzero divisor in R if and only if
$a'$
is a nonzero divisor in R.
Lemma 5.2.3. Let I be an ideal of R containing a nonzero divisor. Then the following holds:
-
(1) There exist nonzero divisors
$x_1,...,x_n\in R$
such that
$I=(x_1,...,x_n)$
. -
(2) There exist nonzero divisors
$y_1,...,y_n \in (R:I)$
such that
$\operatorname{tr}_R(I)=\sum _{i=1}^n y_iI$
.
Proof. (1) Let S be the collection of all nonzero divisors of R that are in I. Let
$\langle S\rangle $
be the ideal of R generated by S. Then
$I\subseteq (\cup _{\mathfrak {p}\in \operatorname{Ass}(R)} \mathfrak {p}) \cup \langle S\rangle $
. By prime avoidance, either
$I \subseteq \mathfrak {p}$
for some
$\mathfrak {p} \in \operatorname{Ass}(R)$
, or
$I\subseteq \langle S\rangle $
. But I contains a nonzero divisor, so
$I\nsubseteq \mathfrak {p}$
for every
$\mathfrak {p}\in \operatorname{Ass}(R)$
. Hence,
$I\subseteq \langle S\rangle $
. Since S is a subset of I, we conclude
$I= \langle S\rangle $
. Since R is Noetherian, there exist finitely many elements
$x_1,...,x_n\in S$
such that
$I=(x_1,...,x_n)$
.
(2) We know that
$\operatorname{tr}_R(I)=(R:I)I$
in
$Q(R)$
(see [Reference Kobayashi and Takahashi26, Prop. 2.4(2)]). Pick a nonzero divisor
$a\in I$
, so
$J=a(R:I)\subseteq R$
is an ideal of R. Then
$\operatorname{tr}_R(I)=J(\frac 1 aI)$
. Now, J contains a nonzero divisor, so by part (1), we have
$J=(x_1,...,x_n)$
for some nonzero divisors
$x_1,...,x_n$
. Then
$\operatorname{tr}_R(I)=\sum _{i=1}^n \frac 1 a x_iI$
. Denoting
$y_i:=\frac 1 a x_i$
, we see that each
$y_i\in Q(R)$
is a nonzero divisor, and
$y_i\in (R:I)$
as
$x_i\in a(R:I)$
. Hence the claim.
In the proof of the next result, for an
$\mathfrak {m}$
-primary ideal I,
$B(I)$
denotes blow-up of I in the sense of [Reference Dao, Maitra and Sridhar11, Def. 4.3 and Rem. 4.4], namely
$B(I) :=\cup _{n>0}(I^n:_{Q(R)} I^n)$
.
Proposition 5.2.4. Let
$(R,\mathfrak {m})$
be a local Cohen–Macaulay ring of dimension
$1$
, and let I be an
$\mathfrak {m}$
-primary ideal of R. Then
$\operatorname{tr}_R(I){\operatorname{Ext}}^1_R(M,N)\subseteq {\operatorname{Ext}}^1_{\operatorname{Ul}_I(R)}(M,N)$
for all
$M,N\in \operatorname{Ul}_I(R)$
.
Proof. Let
$M,N\in \operatorname{Ul}_I(R)$
, and let
$a\in (R:I)$
be a nonzero divisor. Then
$aI$
is an
$\mathfrak {m}$
-primary ideal of R and
$B(I)=B(aI)$
. So,
$\operatorname{Ul}_I(R)=\operatorname{Ul}_{aI}(R)$
by [Reference Dao, Maitra and Sridhar11, Prop. 4.24]. Hence, for every nonzero divisor
$a\in (R:I)$
, we have
$aI{\operatorname{Ext}}^1_R(M,N)\subseteq {\operatorname{Ext}}^1_{\operatorname{Ul}_I(R)}(M,N)$
by Lemma 5.2.2. Now, by Lemma 5.2.3, there exist nonzero divisors
$y_1,...,y_n\in (R:I)$
such that
$\operatorname{tr}_R(I)=\sum _{i=1}^n y_iI$
. Thus,
$\operatorname{tr}_R(I){\operatorname{Ext}}^1_R(M,N)= \sum _{i=1}^n y_iI{\operatorname{Ext}}^1_R(M,N)\subseteq {\operatorname{Ext}}^1_{\operatorname{Ul}_I(R)}(M,N)$
.
When
$s=1$
,
$\operatorname{depth} R>0$
, and
$I=\mathfrak {m}$
(so
$\nu _I(-)=\mu (-)$
), the inclusion
$I{\operatorname{Ext}}^1_R(M,N)\subseteq {\operatorname{Ext}}^1_{\operatorname{Ul}^s_I(R)}(M,N)$
of Lemma 5.2.2 is actually an equality as we prove next. For this, we first record an easy lemma about flat extensions.
Lemma 5.2.5. Let
$R\to S$
be a flat extension of rings. Let M be an R-module, and let I an ideal of R. Then, the following holds:
-
(1)
$S \otimes _R (IM)=(IS)(S \otimes _R M)$
when identified as submodules of
$S\otimes _R M$
. -
(2) If
$N\subseteq M$
is an R-submodule, S is a faithfully flat extension of R, and
$S \otimes _R M=S \otimes _R N$
, then
$M=N$
.
Proof. (1) Consider the exact sequence
$0\to IM \to M \to M/IM\to 0$
, which after tensoring with S gives
$0\to S \otimes _R (IM) \to S\otimes _R M \to S\otimes _R M/IM \to 0$
. Now,
$S \otimes _R M/IM \cong S \otimes _R (M\otimes _R R/I)\cong (S\otimes _R M)\otimes _R R/I\cong (S\otimes _R M)/I(S \otimes _R M)$
. Now, by the natural S-module structure on
$S\otimes _R M$
, we see that
$I(S \otimes _R M)=(IS)(S \otimes _R M)$
. Thus, we get the exact sequence
$0\to S \otimes _R (IM) \to S\otimes _R M \to (S \otimes _R M)/(IS)(S \otimes _R M)\to 0$
, and by naturality of the isomorphisms, we see that the map
$S\otimes _R M \to (S \otimes _R M)/(IS)(S \otimes _R M)$
in the exact sequence has kernel
$(IS)(S \otimes _R M)$
. Hence,
$S \otimes _R (IM)=(IS)(S \otimes _R M).$
(2) Tensoring the exact sequence
$0\to N \to M \to M/N \to 0$
with S and using
$S \otimes _R M=S \otimes _R N$
, we get
$(M/N)\otimes _R S=0$
. Since S is faithfully flat, we have
$M/N=0$
. Hence,
$M=N$
.
Now, we prove the desired equality between
$\mathfrak {m}{\operatorname{Ext}}^1_R(M,N)$
and
${\operatorname{Ext}}^1_{\operatorname{Ul}^1(R)}(M,N)$
, when R has positive depth.
Proposition 5.2.6. Let
$(R,\mathfrak {m},k)$
be a local ring of positive depth. Let
$M,N\in \operatorname{Ul}^1(R)$
. Then, the following holds:
-
(1) We always have
$\mathfrak {m}{\operatorname{Ext}}^1_R(M,N)={\operatorname{Ext}}^1_{\operatorname{Ul}^1(R)}(M,N)$
. -
(2) If R is moreover Cohen–Macaulay of dimension
$1$
(so
$\operatorname{Ul}^1(R)=\operatorname{Ul}(R)$
) and if
$x\in \mathfrak {m}$
is a minimal reduction of
$\mathfrak {m}$
, then
$x{\operatorname{Ext}}^1_R(M,N)={\operatorname{Ext}}^1_{\operatorname{Ul}(R)}(M,N)$
.
Proof. (1) Due to Lemma 5.2.2, we only need to prove
${\operatorname{Ext}}^1_{\operatorname{Ul}^1(R)}(M,N) \subseteq \mathfrak {m}{\operatorname{Ext}}^1_R(M,N)$
. We may assume
$M,N\ne 0$
.
First, we assume that the residue field is infinite. Let
$\sigma : 0\to N\to X\to M\to 0$
be a short exact sequence such that
$X\in \operatorname{Ul}^1(R)$
. Choose
$x\in \mathfrak {m}$
to be
$R\oplus M \oplus N\oplus X$
-superficial (which exists by [Reference Huneke and Swanson24, Prop. 8.5.7], since we are assuming R has infinite residue field). Now,
$\operatorname{depth}_R(R\oplus M\oplus N\oplus X)=\inf \{\operatorname{depth} R,\operatorname{depth}_R M,\operatorname{depth}_R N,\operatorname{depth}_R X\}>0$
, so x is
$R\oplus M\oplus N\oplus X$
-regular by [Reference Huneke and Swanson24, Lem. 8.5.4]. Hence,
$x\in \mathfrak {m}$
is regular on
$R,M$
, and N, and superficial on
$M,N$
, and X. Thus,
$M/xM,N/xN,X/xX$
are zero-dimensional Ulrich modules by [Reference Goto, Takahashi and Taniguchi21, Prop. 2.2(4)], so these are k-vector spaces by [Reference Goto, Takahashi and Taniguchi21, Prop. 2.2(1)]. Hence,
$\sigma \otimes _R R/xR$
, being a short exact sequence of k-vector spaces, is split exact. Since x is
$R,M,N$
-regular, by [Reference Striuli36, Prop. 2.8], we have
$\sigma \in x{\operatorname{Ext}}^1_R(M,N)\subseteq \mathfrak {m}{\operatorname{Ext}}^1_R(M,N)$
.
Now, we prove the general case. This part will use extensively that
${\operatorname{Ext}}^1_{\operatorname{Ul}^1(R)}(M,N)$
is a submodule of
${\operatorname{Ext}}^1_R(M,N)$
, where
$M,N\in \operatorname{Ul}^1(R)$
. Consider the faithfully flat extension of R as follows:
$S :=R[X]_{\mathfrak {m}[X]}$
, with maximal ideal
$\mathfrak {m} S$
, and infinite residue field. Since we already know,
$\mathfrak {m}{\operatorname{Ext}}^1_R(M,N)\subseteq {\operatorname{Ext}}^1_{\operatorname{Ul}^1(R)}(M,N)$
by Lemma 5.2.2 and since S is faithfully flat, it is enough to prove
$S\otimes _R \mathfrak {m}{\operatorname{Ext}}^1_R(M,N)=S \otimes _R {\operatorname{Ext}}^1_{\operatorname{Ul}^1(R)}(M,N)$
(by Lemma 5.2.5(2)). Now,
$\mathfrak {m}{\operatorname{Ext}}^1_R(M,N)\subseteq {\operatorname{Ext}}^1_{\operatorname{Ul}^1(R)}(M,N)$
, with flatness of S, already implies
$S \otimes _R \mathfrak {m}{\operatorname{Ext}}^1_R(M,N)\subseteq S \otimes _R {\operatorname{Ext}}^1_{\operatorname{Ul}^1(R)}(M,N)$
. So, to prove equality, it is enough to prove that
$S \otimes _R {\operatorname{Ext}}^1_{\operatorname{Ul}^1(R)}(M,N) \subseteq S\otimes _R \mathfrak {m}{\operatorname{Ext}}^1_R(M,N)$
. Now, due to Lemma 5.2.5(1), it is enough to prove
$S \otimes _R {\operatorname{Ext}}^1_{\operatorname{Ul}^1(R)}(M,N) \subseteq (\mathfrak {m} S)(S \otimes _R {\operatorname{Ext}}^1_R(M,N))$
, and the latter object here is naturally identified with
$(\mathfrak {m} S){\operatorname{Ext}}^1_S(S \otimes _R M, S \otimes _R N)$
. Since the S-module
$S \otimes _R {\operatorname{Ext}}^1_{\operatorname{Ul}^1(R)}(M,N)$
is generated by
$1\otimes _R \sigma $
as
$\sigma $
runs over all elements of
${\operatorname{Ext}}^1_{\operatorname{Ul}^1(R)}(M,N)$
, we need to prove that
$1 \otimes _R \sigma \in (\mathfrak {m} S){\operatorname{Ext}}^1_S(S \otimes _R M, S \otimes _R N)$
for all
$\sigma \in {\operatorname{Ext}}^1_{\operatorname{Ul}^1(R)}(M,N)$
. So, let
$\sigma : 0\to N\to X\to M\to 0$
be a short exact sequence such that
$X\in \operatorname{Ul}^1(R)$
. Then,
$1\otimes _R \sigma \in S \otimes _R {\operatorname{Ext}}^1_{\operatorname{Ul}^1(R)}(M,N)\subseteq S \otimes _R {\operatorname{Ext}}^1_R(M,N)\cong {\operatorname{Ext}}^1_S(S \otimes _R M, S \otimes _R N)$
is naturally identified with the exact sequence
$S \otimes _R \sigma : 0\to S \otimes _R N \to S \otimes _R X \to S \otimes _R M \to 0$
in
${\operatorname{Ext}}^1_S(S \otimes _R M, S \otimes _R N)$
. Now,
$S \otimes _R \sigma $
is a short exact sequence of modules in
$\operatorname{Ul}^1(S)$
by [Reference Goto, Takahashi and Taniguchi21, Prop. 2.2(3)]. Hence,
$1\otimes _R \sigma \in {\operatorname{Ext}}^1_{\operatorname{Ul}^1(S)}(S \otimes _R M, S \otimes _R N)$
. Since S also has positive depth and infinite residue field, by the proof of the infinite residue field case, we get
$1\otimes _R \sigma \in {\operatorname{Ext}}^1_{\operatorname{Ul}^1(S)}(S \otimes _R M, S \otimes _R N) \subseteq (\mathfrak {m} S){\operatorname{Ext}}^1_S(S \otimes _R M, S \otimes _R N)$
, which is what we wanted to prove.
(2) Due to Lemma 5.2.2, we only need to prove
${\operatorname{Ext}}^1_{\operatorname{Ul}(R)}(M,N) \subseteq x{\operatorname{Ext}}^1_R(M,N)$
. So, let
$\sigma : 0\to N\to X\to M\to 0$
be a short exact sequence such that
$X\in \operatorname{Ul}(R)$
. Since
$xR$
is a reduction of
$\mathfrak {m}$
, we have
$xR$
is
$\mathfrak {m}$
-primary. Hence, x is R-regular, so x is
$M,N,X$
-regular. Also,
$\mathfrak {m} M=xM,\mathfrak {m} N=xN,\mathfrak {m} X=xX$
. So,
$\sigma \otimes _R R/xR: 0\to N/xN \to X/xX \to M/xM\to 0$
is an exact sequence of k-vector spaces. Hence,
$\sigma \otimes _R R/xR$
is split exact. Since x is
$R,M,N,X$
-regular, by [Reference Striuli36, Prop. 2.8], we have
$\sigma \in x{\operatorname{Ext}}^1_R(M,N)$
. Thus, we get
${\operatorname{Ext}}^1_{\operatorname{Ul}(R)}(M,N) \subseteq x{\operatorname{Ext}}^1_R(M,N)$
.
Following is an interesting consequence of Proposition 5.2.6. Here, for R-modules
$X,Y$
, we denote by
$X*Y$
the collection of all R-modules Z that fit into a short exact sequence
$0\to X \to Z \to Y \to 0 $
.
Corollary 5.2.7. Let
$(R,\mathfrak {m})$
be a local one-dimensional Cohen–Macaulay ring. Then, the following are equivalent:
-
(1) R is regular.
-
(2)
$\operatorname{Ul}(R)$
is closed under taking extensions. -
(3) There exists
$M,N\in \operatorname{Ul}(R)$
such that N is faithful,
$M\ne 0$
, and
$N*M\subseteq \operatorname{Ul}(R)$
.
Proof.
$(1)\implies (2)$
: If R is regular, then
$\operatorname{Ul}(R)=\operatorname{CM}(R)$
, which is closed under taking extensions.
$(2)\implies (3)$
: One can take
$M=N=\mathfrak {m}^n$
for some
$n\gg 0$
.
$(3) \implies (1)$
: By hypothesis, we get
${\operatorname{Ext}}^1_{R}(M,N)={\operatorname{Ext}}^1_{\operatorname{Ul}(R)}(M,N)$
. Then from Proposition 5.2.6 (remembering
$\operatorname{Ul}^1(R)=\operatorname{Ul}(R)$
in our case), we get
${\operatorname{Ext}}^1_{R}(M,N)=\mathfrak {m}{\operatorname{Ext}}^1_{R}(M,N)$
. So, by Nakayama’s lemma, we get
${\operatorname{Ext}}^1_{R}(M,N)=0$
. By [Reference Dao, Maitra and Sridhar11, Th. 4.6] (see also [Reference Dey and Kobayashi12, Lem. 5.2]), we get
$M\cong \mathfrak {m} M$
and
$N\cong \mathfrak {m} N$
. So,
${\operatorname{Ext}}^1_R(\mathfrak {m} M,\mathfrak {m} N)=0$
. Hence,
$\operatorname{pd}_R (\mathfrak {m} M)<\infty $
by [Reference Dey and Kobayashi12, Prop. 2.6]. Thus,
$\operatorname{Tor}^R_{\gg 0}(\mathfrak {m}, \mathfrak {m} M)=0$
. Since
$0\ne M$
, we have
$\mathfrak {m} M\cong M\ne 0$
. Thus,
$\operatorname{pd}_R(\mathfrak {m})<\infty $
by [Reference Dey and Kobayashi12, Cor. 3.17]. Hence, R is regular.
We mention in passing that when R is local Cohen–Macaulay of minimal multiplicity,
$(2)\implies (1)$
of Corollary 5.2.7 also follows from [Reference Dey and Takahashi13, Prop. 3.2(5)].
Before proceeding further with applications to
$\operatorname{Ul}_I(R)$
, we outline an alternative proof that
$\operatorname{Ul}_I(R)$
is an exact subcategory of
${\operatorname{mod}}\ R$
when R is a local one-dimensional Cohen–Macaulay ring, via birational extensions. For this, we first record some preliminary results.
The following is easy to see from the definition of an exact category.
Lemma 5.2.8. Let
$R,S$
be two commutative rings such that
$R\to S$
is a ring extension. Let
$({\cal{X}},{\cal{E}})$
be a strictly full exact subcategory of
${\operatorname{Mod}}\ S$
. Assume that
${\cal{X}}$
is closed under R-linear isomorphism of R-modules. Also, assume that every R-linear map between any two modules in
${\cal{X}}$
is S-linear, that is,
$\operatorname{Hom}_R(M,N)=\operatorname{Hom}_S(M,N)$
for all
$M,N\in {\cal{X}}$
. Then,
$({\cal{X}},{\cal{E}})$
is a strictly full exact subcategory of
${\operatorname{Mod}}\ R$
.
For the consequences of Lemma 5.2.8, we need the following lemma, which is essentially [Reference Leuschke and Wiegand27, Prop. 4.14(i)]. Before proceeding, we note that for a birational extension
$R\subseteq S\subseteq Q(R)$
and an S-module N, being torsion-free as an S-module and as an R-module are the same.
Lemma 5.2.9. Let R be a commutative Noetherian ring, and let
$R\subseteq S \subseteq Q(R)$
be a ring extension. Then,
$\operatorname{Hom}_R(M,N)=\operatorname{Hom}_S(M,N)$
for all
$M,N\in {\operatorname{Mod}} (S)$
where N is torsion-free S-module.
Proof.
$\operatorname{Hom}_S(M,N)\subseteq \operatorname{Hom}_R(M,N)$
is clear. Hence, it is enough to show that
$\operatorname{Hom}_R(M,N)\subseteq \operatorname{Hom}_S(M,N)$
. Let
$\dfrac a b\in S \subseteq Q(R)$
, where
$a,b\in R$
, so
$b\in R$
is a nonzero divisor. Let
$m\in M$
and
$f\in \operatorname{Hom}_R(M,N)$
. Then
$b\left (f\left (\dfrac a b m\right )-\dfrac a b f(m)\right )=bf\left (\dfrac a b m\right )-af(m)=f\left (b\dfrac a b m\right )-af(m)=f(am)-af(m)=0$
. Since N is torsion-free and b is a nonzero divisor on R, it is a nonzero divisor on N. So,
$f\left (\dfrac a b m\right )=\dfrac a b f(m)$
. As
$\dfrac a b\in S, m\in M$
, and
$f\in \operatorname{Hom}_R(M,N)$
were arbitrary, we conclude
$\operatorname{Hom}_R(M,N)\subseteq \operatorname{Hom}_S(M,N)$
.
Now, as a consequence, we can deduce the following.
Proposition 5.2.10. Let
$R\subseteq S$
be a birational extension (
$S\subseteq Q(R)$
) of commutative rings. Let
${\cal{S}}_R$
be the standard exact structure on
${\operatorname{mod}}\ R$
. Let
${\cal{X}}$
be the strictly full subcategory of
${\operatorname{Mod}} S$
consisting of all torsion-free S-modules. Then it holds that
${\cal{S}}_R|_{\cal{X}}={\cal{S}}_S|_{\cal{X}}$
, and
$({\cal{X}}, {\cal{S}}_R|_{\cal{X}})$
is a strictly full exact subcategory of
${\operatorname{mod}}\ R$
, and
$({\cal{X}}\cap {\operatorname{mod}}\ R, {\cal{S}}_R|_{{\cal{X}}\cap {\operatorname{mod}}\ R})$
is an exact subcategory of
${\operatorname{Mod}}\ R$
(hence, also of
${\operatorname{mod}}\ R$
).
Proof. By definition,
${\cal{S}}_R|_{{\operatorname{mod}}\ R\cap {\cal{X}}}={\cal{S}}_R|_{{{\operatorname{mod}}\ R}}\cap {\cal{S}}_R|_{\cal{X}}$
. Since
$({\operatorname{mod}}\ R,{\cal{S}}_R|_{{{\operatorname{mod}}\ R}})$
is an exact subcategory of
${\operatorname{Mod}}\ R$
, the second part of the statement would readily follow from the first part of the statement and Lemma 3.2. First, we show that
$\cal{X}$
is strictly full in
${\operatorname{mod}}\ R$
. If
$M,N$
are two R-modules and
$f:M\to N$
is an R-linear isomorphism and N is moreover a torsion-free S-module extending its R-module structure, then M has an S-module structure, extending its R-module structure, given by
$s\cdot m:=f^{-1}(sf(m))$
for all
$s\in S, m \in M$
. Note that, with this structure, M is moreover a torsion-free S-module. Indeed, let
$s\in S$
be a nonzero divisor such that
$s\cdot m=0$
. Then
$sf(m)=f(s\cdot m)=0$
, so
$f(m)=0$
as N is S-torsion-free. Thus,
$m=0$
as f is an isomorphism. Moreover,
$\operatorname{Hom}_R(M,N)=\operatorname{Hom}_S(M,N)$
for all torsion-free S-modules
$M,N$
by Lemma 5.2.9. Consequently, we notice that
${\cal{S}}_R|_{{\cal{X}}}={\cal{S}}_S|_{{\cal{X}}}$
. So, to show that
$({\cal{X}}, {\cal{S}}_R|_{\cal{X}})$
is an exact subcategory of
${\operatorname{Mod}}\ R$
, it is enough to show that
$({\cal{X}}, {\cal{S}}_S|_{\cal{X}})$
is an exact subcategory of
${\operatorname{Mod}}\ S$
(by Lemma 5.2.8). Now, it is well known that for any ring S, the subcategory of all S-torsion-free modules is closed under extensions in
${\operatorname{Mod}}\ S$
. Hence,
$({\cal{X}}, {\cal{S}}_S|_{\cal{X}})$
is an exact subcategory of
${\operatorname{Mod}}\ S$
by Proposition 3.5.
Corollary 5.2.11. Let R be a local Cohen–Macaulay ring of dimension
$1$
. Let
$R\subseteq S$
be a finite birational extension. Then,
$(\operatorname{CM}(S),{\cal{S}}_R|_{{\operatorname{CM}(S)}})$
is a strictly full exact subcategory of
${\operatorname{mod}}\ R$
.
Proof. We note that
${\operatorname{mod}} S={\operatorname{Mod}}\ S\cap {\operatorname{mod}}\ R$
. Since S is also one-dimensional Cohen–Macaulay, we have
$\operatorname{CM}(S)=$
collection of all finitely generated torsion-free S-modules
$ = $
collection of all torsion-free S-modules
$\cap {\operatorname{mod}}\ R$
. Hence, the conclusion follows from Proposition 5.2.10.
When
$(R,\mathfrak {m})$
is a local Cohen–Macaulay ring of dimension
$1$
and I an
$\mathfrak {m}$
-primary ideal, then considering the finite birational extension
$R\subseteq B(I)$
, where
$B(I)$
denotes blow-up of I [Reference Dao, Maitra and Sridhar11, Def. 4.3 and Rem. 4.4], it is shown in [Reference Dao, Maitra and Sridhar11, Th. 4.6] that
$\operatorname{CM}(B(I))=\operatorname{Ul}_I(R)$
. Thus, in this particular case, we get another proof of the fact that
$\operatorname{Ul}_I(R)$
is an exact subcategory of
${\operatorname{mod}}\ R$
, which is very different from Corollary 4.7.
For further applications, we first record a lemma connecting
${\operatorname{Ext}}^1_{\operatorname{Ul}_I(R)}(-,-)$
to
${\operatorname{Ext}}^1_{B(I)}(-,-)$
. Here, we will keep in mind that when
$\dim R=1$
, we write
$\operatorname{Ul}^1_I(R)=\operatorname{Ul}_I(R)$
in our notation (see Definition 4.5).
Lemma 5.2.12. Let
$(R,\mathfrak {m})$
be a local Cohen–Macaulay ring of dimension
$1$
. Let I be an
$\mathfrak {m}$
-primary ideal of R. Then, for all I-Ulrich modules
$M,N$
, we have a natural map
${\operatorname{Ext}}^1_{\operatorname{Ul}_I(R)}(M,N)\xrightarrow {[\sigma ]_R \to [\sigma ]_{B(I)}} {\operatorname{Ext}}^1_{B(I)}(M,N)$
, which is an isomorphism of R-modules.
Proof. We recall that
${\operatorname{Ext}}^1_{\operatorname{Ul}_I(R)}(M,N)$
is the equivalence class of all short exact sequences
$0\to N \to X \to M\to 0$
of I-Ulrich modules and R-linear maps. Since
$\operatorname{Ul}_I(R)=\operatorname{CM}(B(I))$
[Reference Dao, Maitra and Sridhar11, Th. 4.6], any R-linear map between two Ulrich modules is
$B(I)$
-linear (see [Reference Leuschke and Wiegand27, Prop. 4.14(i)]). Hence, any short exact sequence
$0\to N \to X \to M\to 0$
of I-Ulrich modules and R-linear maps is a short exact sequence of maximal Cohen–Macaulay
$B(I)$
-modules and
$B(I)$
-linear maps, and any R-linear morphism between two short exact sequences of I-Ulrich R-modules is actually a
$B(I)$
-morphism. This proves the well-definedness of the map. Injectivity is similarly obvious. To prove surjectivity, we note that if Y is a
$B(I)$
-module, and
$0\to N \to Y \to M \to 0$
is a short exact sequence in
${\operatorname{Mod}} (B(I))$
, then
$M,N\in \operatorname{CM}(B(I))$
implies
$Y\in \operatorname{CM}(B(I))$
. Hence,
$Y\in \operatorname{Ul}_I(R)$
. Since
$B(I)$
-linear maps are R-linear, this gives a pre-image in
${\operatorname{Ext}}^1_{\operatorname{Ul}_I(R)}(M,N)$
. Finally, to show the map is R-linear, we recall that the Baer-sum structure and multiplication by R on
${\operatorname{Ext}}^1$
are given by certain pullback and pushout diagrams. So, it is enough to prove that given an exact sequence
$\sigma : 0\to N \to X \to M \to 0$
in
$\operatorname{Ul}_I(R)$
and R-linear maps
$f: N\to N', g:M'\to M$
with
$M',N'\in \operatorname{Ul}_I(R)$
, the pushout and pullback of
$\sigma $
by f and g in
${\operatorname{mod}}\ R$
, respectively, are actually pushout and pullback in
${\operatorname{mod}} (B(I))$
. Let us look at the pushout case, as the pullback case is similar. If we have the following pushout diagram:
then in the bottom row, we again have
$Y\in \operatorname{Ul}_I(R)$
, since
$\operatorname{Ul}_I(R)$
along with all its short exact sequences is an exact subcategory of
${\operatorname{mod}}\ R$
by Corollary 4.7. Hence, in the above diagram, all the modules are in
$\operatorname{Ul}_I(R)=\operatorname{CM}(B(I))$
, so all the R-linear maps are moreover
$B(I)$
-linear. Since
$\operatorname{CM}(B(I))$
is an extension closed subcategory of
${\operatorname{mod}}(B(I))$
, it is an exact subcategory of
${\operatorname{mod}} (B(I))$
by Proposition 3.5. Thus, this is a pushout diagram in
${\operatorname{mod}}(B(I))$
as well by [Reference Bühler6, Prop. 2.12]. This is what we wanted to show.
Corollary 5.2.13. Let
$(R,\mathfrak {m})$
be a local Cohen–Macaulay ring of dimension
$1$
, and let I be an
$\mathfrak {m}$
-primary ideal of R. If
$M,N\in \operatorname{Ul}_I(R)$
are such that
${\operatorname{Ext}}^1_{B(I)}(M,N)=0$
, then
$\operatorname{tr}_R(I){\operatorname{Ext}}^1_R(M,N)=0$
.
In the following, for a module X over a ring R, by
$\operatorname{add}_R(X)$
, we denote the collection of all R-modules Y such that there exist an R-module Z and an isomorphism of R-modules
$Y\oplus Z\cong X^{\oplus n}$
for some integer
$n\ge 0$
.
Theorem 5.2.14. Let
$(R,\mathfrak {m})$
be a local Cohen–Macaulay ring of dimension
$1$
. Let I be an
$\mathfrak {m}$
-primary ideal of R. Then the following hold:
-
(1) If
$M\in \operatorname{add}_R(B(I))$
, then
$\operatorname{tr}_R(I){\operatorname{Ext}}^1_R(M, \operatorname{Ul}_I(R))=0$
. -
(2) Let
$M\in \operatorname{Ul}(R)$
. Then
$M\in \operatorname{add}_R(B(\mathfrak {m}))$
if and only if
$\mathfrak {m} {\operatorname{Ext}}^1_R(M, \operatorname{Ul}(R))=0$
. -
(3) If
$B(I)$
is a Gorenstein ring, then
$\operatorname{tr}_R(I){\operatorname{Ext}}^1_R(\operatorname{Ul}_I(R),B(I))=0$
. The converse holds when
$I=\mathfrak {m}$
.
Proof. (1) Denote
$S:=B(I)$
. If
$M\in \operatorname{add}_R(S)(\subseteq \operatorname{Ul}_I(R))$
, then there exist
$X\in {\operatorname{mod}}(R)$
and an isomorphism of R-modules
$M\oplus X\cong B(I)^{\oplus n}$
for some
$n\ge 0$
. Then
$M,X$
are I-Ulrich, so
$M,X\in \operatorname{CM}(S)$
. Hence, the isomorphism is also S-linear by [Reference Leuschke and Wiegand27, Prop. 4.14(i)], so M is a projective S-module. Thus,
${\operatorname{Ext}}^1_S(M,N)=0$
for all
$N\in \operatorname{CM}(S)=\operatorname{Ul}_I(R)$
. Now, the conclusion follows from Corollary 5.2.13.
(2) If
$M\in \operatorname{add}_R(B(\mathfrak {m}))$
, then the conclusion follows by (1) as
$\mathfrak {m}\subseteq \operatorname{tr}_R(\mathfrak {m})$
. So, assume
$M\in \operatorname{Ul}(R)$
and
$\mathfrak {m} {\operatorname{Ext}}^1_R(M,N)=0$
for all
$N\in \operatorname{Ul}(R)$
. By Proposition 5.2.6, we then have
${\operatorname{Ext}}^1_{\operatorname{Ul}(R)}(M,N)=0$
for all
$N\in \operatorname{Ul}(R)$
. Now, denote
$S=B(\mathfrak {m})$
. Then, by Lemma 5.2.12, we have
${\operatorname{Ext}}^1_{S}(M,N)=0$
for all
$N\in \operatorname{Ul}(R)=\operatorname{CM}(S)$
. We know that, for any Cohen–Macaulay ring S, and
$M\in \operatorname{CM}(S)$
, we have
${\operatorname{Ext}}^1_S(M,\operatorname{CM}(S))=0$
if and only if
$M\in \operatorname{add}_S(S)$
. Indeed, one direction is clear since
$M\in \operatorname{add}_S(S)$
implies M is projective. For the converse, we notice that as M is finitely generated over S, so we have a short exact sequence
$\sigma : 0\to M'\to S^{\oplus n}\to M \to 0$
for some
$n>0$
and S-module
$M'$
. Moreover, S being Cohen–Macaulay implies
$M'\in \operatorname{CM}(S)$
, and now the vanishing of
${\operatorname{Ext}}^1_S(M,M')$
gives that the sequence
$\sigma $
splits. Hence,
$M\in \operatorname{add}_S(S)$
. Now, applying this observation to our scenario with
$S=B(\mathfrak {m})$
, we get
$M\in \operatorname{add}_S(B(\mathfrak {m}))=\operatorname{add}_R(B(\mathfrak {m}))$
, where this last equality follows from the fact that direct summand of
$B(\mathfrak {m})$
is in
$\operatorname{CM}(B(\mathfrak {m}))$
and consequently, R-linear maps are S-linear (see [Reference Leuschke and Wiegand27, Prop. 4.14(i)]).
(3) Nothing to prove if R is regular, so we assume R is singular. If
$B(I)$
is a Gorenstein ring, then
${\operatorname{Ext}}^1_{B(I)}(M,B(I))=0$
for all
$M \in \operatorname{CM}(B(I))=\operatorname{Ul}_I(R)$
. Since
$B(I)$
is I-Ulrich, by Lemma 5.2.12, we get
${\operatorname{Ext}}^1_{\operatorname{Ul}_I(R)}(M,B(I))=0$
for all
$M \in \operatorname{Ul}_I(R)$
. Then, by Corollary 5.2.13, we get
$\operatorname{tr}_R(I){\operatorname{Ext}}^1_R(M,B(I))=0$
for all
$M \in \operatorname{Ul}_I(R)$
. For the converse part, assume
$I=\mathfrak {m}$
and
$\mathfrak {m}{\operatorname{Ext}}^1_R(M,B(\mathfrak {m}))=0$
for all
$M \in \operatorname{Ul}(R)$
. Then, by Proposition 5.2.6, we have
${\operatorname{Ext}}^1_{\operatorname{Ul}(R)}(M,B(\mathfrak {m}))=0$
for all
$M\in \operatorname{Ul}(R)$
. Hence, by Lemma 5.2.12, we have
${\operatorname{Ext}}^1_{B(\mathfrak {m})}(M,B(\mathfrak {m}))=0$
for all
$M\in \operatorname{Ul}(R)=\operatorname{CM}(B(\mathfrak {m}))$
. Now, we note that for a finite-dimensional Cohen–Macaulay ring S,
${\operatorname{Ext}}^1_S(M,S)=0$
for all
$M \in \operatorname{CM}(S)$
if and only if S is Gorenstein. Indeed, if S is Gorenstein, then the vanishing is clear by [Reference Bruns and Herzog3, Def. 3.1.18 and Props. 3.1.9 and 3.1.24]. Conversely, let
$d=\dim S$
. Then, for every
$\mathfrak {p} \in \operatorname{Spec}(S)$
, we have
$\Omega ^d_S(S/\mathfrak {p}) \in \operatorname{CM}(S)$
(by localizing and applying depth lemma). So, now the vanishing of
${\operatorname{Ext}}^{d+1}_S(S/\mathfrak {p}, S)\cong {\operatorname{Ext}}^1_S(\Omega ^d_S(S/\mathfrak {p}),S)$
for every
$\mathfrak {p}\in \operatorname{Spec}(S)$
implies
$\operatorname{inj dim}_S S <\infty $
. Thus, S is Gorenstein by [Reference Bruns and Herzog3, Def. 3.1.18 and Prop. 3.1.9]. Now, applying this observation to our scenario with
$S=B(\mathfrak {m})$
, we get that
$B(\mathfrak {m})$
is Gorenstein.
Since for one-dimensional local Cohen–Macaulay rings
$(R,\mathfrak {m})$
of minimal multiplicity it holds that
$B(\mathfrak {m})=(\mathfrak {m}:\mathfrak {m})\cong \mathfrak {m}$
, the following corollary is an immediate consequence of Theorem 5.2.14(3) and [Reference Goto, Matsuoka and Phuong20, Th. 5.1].
Corollary 5.2.15. Let
$(R,\mathfrak {m})$
be a local Cohen–Macaulay ring of dimension
$1$
and minimal multiplicity. Then, R is almost Gorenstein if and only if
$\mathfrak {m}{\operatorname{Ext}}^1_R(\operatorname{Ul}(R),\mathfrak {m})=0$
.
Finally, we give one more application of Theorem 5.2.14(3), for which we first record an easy observation.
Lemma 5.2.16. Let
$(R,\mathfrak {m})$
be a local Cohen–Macaulay ring of dimension
$1$
. Let I be an
$\mathfrak {m}$
-primary ideal of R admitting a principal reduction
$x\in I$
. Then,
$\mathfrak {m}$
is I-Ulrich if and only if
$\mathfrak {m}\subseteq ((x):I)$
.
Proof. If I is principal, then all MCM modules are I-Ulrich. So, it is enough to assume I is not principal. If
$\mathfrak {m}$
is I-Ulrich, then
$I\mathfrak {m} =x\mathfrak {m}$
(see [Reference Dao, Maitra and Sridhar11, Prop. 4.5]). So,
$\mathfrak {m} \subseteq ((x):I)$
. Conversely, let
$\mathfrak {m}\subseteq ((x):I)$
. Then,
$\mathfrak {m}\subseteq ((x):_RI)$
. We claim that
$x\notin \mathfrak {m} I$
. Since
$x\in I$
is a principal reduction of I, we have
$I^{n+1}=xI^n$
for some
$n\ge 1$
. Hence, if
$x\in \mathfrak {m} I$
, then
$I^{n+1}\subseteq \mathfrak {m} I^{n+1}$
. So,
$I^{n+1}=\mathfrak {m} I^{n+1}$
. Hence, by Nakayama’s lemma,
$I^{n+1}=0$
, contradicting I is
$\mathfrak {m}$
-primary. Thus,
$x\in I\setminus \mathfrak {m} I$
. Since
$\mathfrak {m}\subseteq ((x):_R I)$
and I is not principal, we have
$\mathfrak {m} \subseteq (x\mathfrak {m} :_R I)$
by Lemma 5.1.7. So,
$\mathfrak {m} I \subseteq x \mathfrak {m}$
. Hence,
$\mathfrak {m} I=x\mathfrak {m}$
, so
$\mathfrak {m}$
is I-Ulrich.
Corollary 5.2.17. Let
$(R,\mathfrak {m})$
be a local Cohen–Macaulay ring of dimension
$1$
, with infinite residue field, and minimal multiplicity, admitting a canonical module. If
$\mathfrak {m}{\operatorname{Ext}}^1_R(\operatorname{Ul}(R),\mathfrak {m})=0$
, then R admits a canonical ideal
$\omega $
and
$B(\omega )$
is Gorenstein. Consequently,
$\mathfrak {m}{\operatorname{Ext}}^1_{R}(\operatorname{Ul}_{\omega }(R),B(\omega ))=0$
.
Proof. Let
$\omega $
be a canonical module of R. We show that
$\omega $
can be identified with an ideal of R,
$\mathfrak {m}\subseteq \operatorname{tr}_R(\omega )$
, and
$B(\omega )$
is Gorenstein. Since R has minimal multiplicity and
$\mathfrak {m}{\operatorname{Ext}}^1_R(\operatorname{Ul}(R),\mathfrak {m})=0$
, R is almost Gorenstein by Corollary 5.2.15. Hence,
$\mathfrak {m}\subseteq \operatorname{tr}_R(\omega )$
from [Reference Herzog, Hibi and Stamate22, Def. 2.2 and Prop. 6.1], so R is generically Gorenstein (see [Reference Herzog, Hibi and Stamate22, Lem. 2.1]). Hence,
$\omega $
can be identified with an ideal of R by [Reference Bruns and Herzog3, Prop. 3.3.18]. Since R is almost Gorenstein, we have
$\mathfrak {m} \subseteq ((x):\omega )$
for some principal reduction x of
$\omega $
(see [Reference Goto, Matsuoka and Phuong20, Setting 3.4 and Th. 3.11]). Thus,
$\mathfrak {m}$
is
$\omega $
-Ulrich by Lemma 5.2.16. So,
$\mathfrak {m} \in \operatorname{CM}(B(\omega ))$
by [Reference Dao, Maitra and Sridhar11, Th. 4.6]. Hence,
$\mathfrak {m}\subseteq \operatorname{tr}_R(\mathfrak {m}) \subseteq (R:_{Q(R)}B(\omega ))$
by [Reference Dao, Maitra and Sridhar11, Th. 2.9]. Thus, the conductor
$c_R(B(\omega )):=(R:_{Q(R)}B(\omega ))$
of
$B(\omega )$
is either
$\mathfrak {m}$
or R. As R has minimal multiplicity, by using [Reference Bruns and Herzog3, Exercise 4.6.14(c)], we see that
$c_R(B(\omega ))$
satisfies the condition (2) of [Reference Goto, Matsuoka and Phuong20, Cor. 3.8]. Thus,
$B(\omega )$
is Gorenstein by [Reference Goto, Matsuoka and Phuong20, Cor. 3.8]. Now, the claim follows from Theorem 5.2.14(3).
Acknowledgments
We thank David Eisenbud for helpful conversations and for pointing out reference [Reference Eisenbud and Schreyer15]. We also thank the anonymous referee for numerous helpful suggestions that improved the exposition of the paper.
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