1 Introduction
Let
$(R,\mathfrak{m})$
be a Noetherian local ring of prime characteristic
$p>0$
. We have the Frobenius endomorphism
$F:R\rightarrow R$
,
$x\mapsto x^{p}$
. The
$F$
-singularities are certain singularities defined via this Frobenius map. They appear in the theory of tight closure (cf. [Reference Huneke13] for its introduction), which was systematically introduced by Hochster and Huneke [Reference Hochster and Huneke9] and developed by many researchers, including Hara, Schwede, Smith, Takagi, Watanabe, Yoshida and others. A recent active research of
$F$
-singularities is centered around the correspondence with the singularities of the minimal model program. We recommend [Reference Takagi and Watanabe25] as an excellent survey for recent developments.
In this paper we study the deformation of
$F$
-singularities. That is, we consider the problem: if
$R/(x)$
has certain property P for a regular element
$x\in R$
, then does
$R$
has the property P? The classical objects of
$F$
-singularities are
$F$
-regularity,
$F$
-rationality,
$F$
-purity and
$F$
-injectivity (cf. [Reference Huneke13, Reference Takagi and Watanabe25]). It is well known that
$F$
-rationality always deforms while
$F$
-regularity and
$F$
-purity do not deform in general [Reference Singh22, Reference Singh23]. Whether
$F$
-injectivity deforms is a long- standing open problem [Reference Fedder6] (for recent developments, we refer to [Reference Horiuchi, Miller and Shimomoto11, Reference Ma, Schwede and Shimomoto18]). Recall that the Frobenius endomorphism induces a natural Frobenius action on every local cohomology module,
$F$
:
$H_{\mathfrak{m}}^{i}(R)\rightarrow H_{\mathfrak{m}}^{i}(R)$
. The ring
$R$
is called
$F$
-injective if this Frobenius action
$F$
is injective for every
$i\geqslant 0$
. The class of
$F$
-injective singularities contains other classes of
$F$
-singularities. For an ideal-theoretic characterization of
$F$
-injectivity, see [Reference Quy and Shimomoto20, Main Theorem D]. We consider this paper as a step toward a solution of the deformation of
$F$
-injectivity.
We introduce two conditions:
$F$
-full and
$F$
-anti-nilpotent singularities, in terms of the Frobenius actions on local cohomology modules of
$R$
(we refer to Section 2 for detailed definitions). The first condition is motivated by recent results on Du Bois singularities [Reference Ma, Schwede and Shimomoto18]. The second condition has been studied in [Reference Enescu and Hochster5, Reference Ma16], and is known to be equivalent to stably FH-finite, which means all local cohomology modules of
$R$
and
$R[[x_{1},\ldots ,x_{n}]]$
supported at the maximal ideals have only finitely many Frobenius stable submodules. We prove that
$F$
-fullness and
$F$
-anti-nilpotency both deform, and we obtain more evidence on deformation of
$F$
-injectivity. Our results largely generalize earlier results of [Reference Horiuchi, Miller and Shimomoto11] in this direction. We list some of our main results here:
Theorem 1.1. (Theorem 4.2, Corollary 5.16)
$(R,\mathfrak{m})$
be a Noetherian local ring of characteristic
$p>0$
and
$x$
a regular element of
$R$
. Then we have:
(1) if
$R/(x)$
is
$F$
-anti-nilpotent, then so is
$R$
;(2) if
$R/(x)$
is
$F$
-full, then so is
$R$
;(3) if
$R/(x)$
is
$F$
-full and
$F$
-injective, then so is
$R$
.
Theorem 1.2. (Theorem 5.11)
Let
$(R,\mathfrak{m})$
be a Noetherian local ring of characteristic
$p>0$
. Suppose the residue field
$k=R/\mathfrak{m}$
is perfect. Let
$x$
be a regular element of
$R$
such that
$\operatorname{Coker}(H_{\mathfrak{m}}^{i}(R)\overset{x}{\rightarrow }H_{\mathfrak{m}}^{i}(R))$
has finite length for every
$i$
. If
$R/(x)$
is
$F$
-injective, then the map
$x^{p-1}F$
:
$H_{\mathfrak{m}}^{i}(R)\rightarrow H_{\mathfrak{m}}^{i}(R)$
is injective for every
$i$
, in particular
$R$
is
$F$
-injective.
2 Definitions and basic properties
2.1 Modules with Frobenius structure
Let
$(R,\mathfrak{m})$
be a local ring of characteristic
$p>0$
. A Frobenius action on an
$R$
-module
$M$
,
$F$
:
$M\rightarrow M$
, is an additive map such that for all
$u\in M$
and
$r\in R$
,
$F(ru)=r^{p}u$
. Such an action induces a natural
$R$
-linear map
$\mathscr{F}_{R}(M)\rightarrow M$
,Footnote
1
where
$\mathscr{F}_{R}(-)$
denotes the Peskine–Szpiro’s Frobenius functor. We say
$N$
is an
$F$
-stable submodule of
$M$
if
$F(N)\subseteq N$
. We say the Frobenius action on
$M$
is nilpotent if
$F^{e}(M)=0$
for some
$e$
.
We note that having a Frobenius action on
$M$
is the same as saying that
$M$
is a left module over the ring
$R\{F\}$
, which may be viewed as a noncommutative ring generated over
$R$
by the symbols
$1,F,F^{2},\ldots \,$
by requiring that
$Fr=r^{p}F$
for
$r\in R$
. Moreover,
$N$
is an
$F$
-stable submodule of
$M$
equivalent to requiring that
$N$
is an
$R\{F\}$
-submodule of
$M$
. We will not use this viewpoint in this article though.
Let
$M$
be an (typically Artinian)
$R$
-module with a Frobenius action
$F$
. We say the Frobenius action on
$M$
is full (or simply
$M$
is full), if the map
$\mathscr{F}_{R}^{e}(M)\rightarrow M$
is surjective for some (equivalently, every)
$e\geqslant 1$
. This is the same as saying that the
$R$
-span of all the elements of the form
$F^{e}(u)$
is the whole
$M$
for some (equivalently, every)
$e\geqslant 1$
. We say the Frobenius action on
$M$
is anti-nilpotent (or simply
$M$
is anti-nilpotent), if for any
$F$
-stable submodule
$N\subseteq M$
, the induced Frobenius action
$F$
on
$M/N$
is injective (note that this in particular implies that
$F$
acts injectively on
$M$
).
Lemma 2.1. The Frobenius action on
$M$
is anti-nilpotent if and only if every
$F$
-stable submodule
$N\subseteq M$
is full. In particular, if
$M$
anti-nilpotent, then
$M$
is full.
Proof. Suppose
$M$
is anti-nilpotent. Let
$N\subseteq M$
be an
$F$
-stable submodule. Consider the
$R$
-span of
$F(N)$
, call it
$N^{\prime }$
. Clearly,
$N^{\prime }\subseteq N$
is another
$F$
-stable submodule of
$M$
and
$F(N)\subseteq N^{\prime }$
. But since
$M$
is anti-nilpotent,
$F$
acts injectively on
$M/N^{\prime }$
. Thus we have
$N=N^{\prime }$
and hence
$N$
is full.
Conversely, suppose every
$F$
-stable submodule of
$M$
is full. Suppose there exists an
$F$
-stable submodule
$N\subseteq M$
such that the Frobenius action on
$M/N$
is not injective. Pick
$y\notin N$
such that
$F(y)\in N$
. Let
$N^{\prime \prime }=N+Ry$
. It is clear that
$N^{\prime \prime }$
is an
$F$
-stable submodule of
$M$
and the
$R$
-span of
$F(N^{\prime \prime })$
is contained in
$N\subsetneq N^{\prime \prime }$
. This shows
$N^{\prime \prime }$
is not full, a contradiction.◻
We also mention that whenever
$M$
is endowed with a Frobenius action
$F$
, then
$\widetilde{F}=rF$
defines another Frobenius action on
$M$
for every
$r\in R$
. It is easy to check that if the action
$\widetilde{F}$
is full or anti-nilpotent, then so is
$F$
.
2.2
$F$
-singularities
We collect some definitions about singularities in positive characteristic. Let
$(R,\mathfrak{m})$
be a Noetherian local ring of characteristic
$p>0$
with the Frobenius endomorphism
$F:R\rightarrow R;x\mapsto x^{p}$
.
$R$
is called
$F$
-finite if
$R$
is a finitely generated as an
$R$
-module via the homomorphism
$F$
.
$R$
is called
$F$
-pure if the Frobenius endomorphism is pure.Footnote
2
It is worth to note that if
$R$
is either
$F$
-finite or complete, then
$R$
being
$F$
-pure is equivalent to the condition that the Frobenius endomorphism
$F:R\rightarrow R$
is split [Reference Hochster and Roberts12]. Let
$I=(x_{1},\ldots ,x_{t})$
be an ideal of
$R$
. Then we denote by
$H_{I}^{i}(R)$
the
$i$
th local cohomology module with support at
$I$
(we refer to [Reference Brodmann and Sharp3] for the general theory of local cohomology modules). Recall that local cohomology may be computed as the cohomology of the Čech complex
$$\begin{eqnarray}0\rightarrow R\rightarrow \bigoplus _{i=1}^{t}R_{x_{i}}\rightarrow \cdots \rightarrow \bigoplus _{i=1}^{t}R_{x_{1}\cdots \widehat{x}_{i}\cdots x_{t}}\rightarrow R_{x_{1}\cdots x_{t}}\rightarrow 0.\end{eqnarray}$$
The Frobenius endomorphism
$F:R\rightarrow R$
induces a natural Frobenius action
$F:H_{I}^{i}(R)\rightarrow H_{I^{[p]}}^{i}(R)\cong H_{I}^{i}(R)$
. A local ring
$(R,\mathfrak{m})$
is called
$F$
-injective if the Frobenius action on
$H_{\mathfrak{m}}^{i}(R)$
is injective for all
$i\geqslant 0$
. This is the case if
$R$
is
$F$
-pure [Reference Hochster and Roberts12, Lemma 2.2]. One can also characterize
$F$
-injectivity using certain ideal closure operations (see [Reference Ma17, Reference Quy and Shimomoto20] for more details).
Example 2.2. Let
$I=(x_{1},\ldots ,x_{t})\subseteq R$
be an ideal generated by
$t$
elements. By the above discussion we have
$$\begin{eqnarray}H_{I}^{t}(R)\cong R_{x_{1}\cdots x_{t}}\bigg/\text{Im}\biggl(\bigoplus _{i=1}^{t}R_{x_{1}\cdots \widehat{x}_{i}\cdots x_{t}}\rightarrow R_{x_{1}\cdots x_{t}}\biggr)\end{eqnarray}$$
and the natural Frobenius action on
$H_{I}^{t}(R)$
sends
$1/(x_{1}\cdots x_{t})$
to
$1/(x_{1}^{p}\cdots x_{t}^{p})$
. Therefore, it is easy to see the Frobenius action on
$H_{I}^{t}(R)$
is full (in fact,
$\mathscr{F}_{R}(H_{I}^{t}(R))\rightarrow H_{I}^{t}(R)$
is an isomorphism). On the other hand, one cannot expect
$H_{I}^{t}(R)$
is always anti-nilpotent even when
$R$
is regular. For example, let
$R=k[[x,y]]$
be a formal power series ring in two variables and
$I=(x)$
. We have
$$\begin{eqnarray}H_{(x)}^{1}(R)\cong k[[y]]x^{-1}\oplus \cdots \oplus k[[y]]x^{-n}\oplus \cdots \,.\end{eqnarray}$$
Let
$N$
be the submodule of
$H_{(x)}^{1}(R)$
generated by
$\{y^{2}x^{-n}\}_{n=1}^{\infty }$
, then it is easy to see
$N$
is an
$F$
-stable submodule of
$H_{(x)}^{1}(R)$
. However,
$F(yx^{-1})=y^{p}x^{-p}\in N$
while
$yx^{-1}\notin N$
. So the Frobenius action on
$H_{(x)}^{1}(R)/N$
is not injective and hence
$H_{(x)}^{1}(R)$
is not anti-nilpotent.
We are mostly interested in the Frobenius actions on local cohomology modules of
$R$
supported at the maximal ideal. We introduce two notions of
$F$
-singularities.
Definition 2.3.
(1) We say that
$(R,\mathfrak{m})$
is
$F$
-full, if the Frobenius action on
$H_{\mathfrak{m}}^{i}(R)$
is full for every
$i\geqslant 0$
. This means
$\mathscr{F}_{R}(H_{\mathfrak{m}}^{i}(R))\rightarrow H_{\mathfrak{m}}^{i}(R)$
is surjective for every
$i\geqslant 0$
.(2) We say that
$(R,\mathfrak{m})$
is
$F$
-anti-nilpotent, if the Frobenius action on
$H_{\mathfrak{m}}^{i}(R)$
is anti-nilpotent for every
$i\geqslant 0$
.
The concept of
$F$
-anti-nilpotency is not new, it was introduced and studied in [Reference Enescu and Hochster5] and [Reference Ma16] under the name stably FH-finite: that is, all local cohomology modules of
$R$
and
$R[[x_{1},\ldots ,x_{n}]]$
supported at their maximal ideals have only finitely many
$F$
-stable submodules. It is a nontrivial result [Reference Enescu and Hochster5, Theorem 4.15] that this is equivalent to
$R$
being
$F$
-anti-nilpotent.
Remark 2.4.
(1) It is clear that
$F$
-anti-nilpotent implies
$F$
-injective and
$F$
-full (see Lemma 2.1). Moreover,
$F$
-pure local rings are
$F$
-anti-nilpotent [Reference Ma16, Theorem 1.1]. In particular,
$F$
-pure local rings are
$F$
-full.(2) We can construct many
$F$
-anti-nilpotent (equivalently, stably FH-finite) rings that are not
$F$
-pure [Reference Quy and Shimomoto20, Sections 5 and 6].(3) Cohen–Macaulay rings are automatically
$F$
-full, since
$\mathscr{F}_{R}(H_{\mathfrak{m}}^{d}(R))\rightarrow H_{\mathfrak{m}}^{d}(R)$
is an isomorphism. But even
$F$
-injective Cohen–Macaulay rings are not necessarily
$F$
-anti-nilpotent [Reference Enescu and Hochster5, Example 2.16].
We give some simple examples of rings that are not
$F$
-full, we see a family of such rings in Example 3.6.
Example 2.5.
(1) Let
$R=k[s^{4},s^{3}t,st^{3},t^{4}]$
where
$k$
is a field of characteristic
$p>0$
. Then
$R$
is a graded ring with
$s^{4},t^{4}$
a homogeneous system of parameters. A simple computation shows that the class spans the local cohomology module$$\begin{eqnarray}\left[\frac{(s^{3}t)^{2}}{s^{4}},-\frac{(st^{3})^{2}}{t^{4}}\right]\in R_{s^{4}}\oplus R_{t^{4}}\end{eqnarray}$$
$H_{\mathfrak{m}}^{1}(R)$
. In particular,
$[H_{\mathfrak{m}}^{1}(R)]$
sits only in degree 2 and thus the natural Frobenius map kills
$H_{\mathfrak{m}}^{1}(R)$
.
$R$
is not
$F$
-full.(2) Let
$R=(k[x,y,z]/(x^{3}+y^{3}+z^{3}))\#k[s,t]$
be the Segre product of
$A=(k[x,y,z]/(x^{3}+y^{3}+z^{3}))$
and
$B=k[s,t]$
, where
$k$
is a field of characteristic
$p>0$
with
$p\equiv 2$
mod
$3$
. Then
$R$
is a normal domain, since it is a direct summand of
$A\otimes _{k}B=A[s,t]$
. Moreover, a direct computation (for example see [Reference Ma, Schwede and Shimomoto18, Examples 4.11 and 4.16]) shows that Since$$\begin{eqnarray}H_{\mathfrak{m}_{R}}^{2}(R)=[H_{\mathfrak{ m}_{R}}^{2}(R)]_{0}\cong [H_{\mathfrak{m}_{A}}^{2}(A)]_{0}=k.\end{eqnarray}$$
$p\equiv 2$
mod 3, we know the natural Frobenius map kills
$[H_{\mathfrak{m}_{A}}^{2}(A)]_{0}$
. Hence
$R$
is not
$F$
-full. On the other hand, if
$p\equiv 1$
mod
$3$
, then it is well known that
$R$
is
$F$
-pure (since
$A$
is) and hence
$F$
-anti-nilpotent [Reference Ma16, Theorem 1.1].
Remark 2.6.
(1) When
$R$
is a homomorphic image of a regular ring
$A$
, say
$R=A/I$
,
$R$
is
$F$
-full if and only if
$H_{\mathfrak{m}}^{i}(A/J)\rightarrow H_{\mathfrak{m}}^{i}(A/I)$
is surjective for every
$J\subseteq I\subseteq \sqrt{J}$
. This is because by [Reference Lyubeznik15, Lemma 2.2], the
$R$
-span of
$F^{e}(H_{\mathfrak{m}}^{i}(R))$
is the same as the image
$H_{\mathfrak{m}}^{i}(A/I^{[p^{e}]})\rightarrow H_{\mathfrak{m}}^{i}(A/I)$
, and for every
$J\subseteq I\subseteq \sqrt{J}$
,
$I^{[p^{e}]}\subseteq J$
for
$e\gg 0$
. As an application, when
$R=A/I$
is
$F$
-full, we have
$H_{\mathfrak{m}}^{i}(A/I)=0$
provided
$H_{\mathfrak{m}}^{i}(A/J)=0$
. Hence
$\operatorname{depth}A/I\geqslant \operatorname{depth}A/J$
for every
$J\subseteq I\subseteq \sqrt{J}$
.(2) Suppose
$R$
is a local ring essentially of finite type over
$\mathbb{C}$
and
$R$
is Du Bois (we refer to [Reference Schwede21] or [Reference Ma, Schwede and Shimomoto18] for the definition and basic properties of Du Bois singularities). In this case we do have
$H_{\mathfrak{m}}^{i}(A/J)\rightarrow H_{\mathfrak{m}}^{i}(A/I)$
is surjective for every
$J\subseteq I=\sqrt{J}$
[Reference Ma, Schwede and Shimomoto18, Lemma 3.3]. This is the main ingredient in proving singularities of dense
$F$
-injective type deform [Reference Ma, Schwede and Shimomoto18, Theorem C].(3) Since
$F$
-injective singularity is the conjectured characteristic
$p>0$
analog of Du Bois singularity [Reference Bhatt, Schwede and Takagi1, Reference Schwede21], it is thus quite natural to ask whether
$F$
-injective local rings are always
$F$
-full. It turns out that this is false in general [Reference Ma, Schwede and Shimomoto18, Example 3.5]. However, constructing such examples seems hard. In fact, [Reference Enescu and Hochster5, Example 2.16] (or its variants like [Reference Ma, Schwede and Shimomoto18, Example 3.5]) is the only example we know that is
$F$
-injective but not
$F$
-anti-nilpotent.
The above remarks motivate us to introduce and study
$F$
-fullness and a stronger notion of
$F$
-injectivity (see Section 5).
We end this subsection by proving that
$F$
-full rings localize. Note that it is proved in [Reference Ma16, Theorem 5.10] that
$F$
-anti-nilpotent rings localize.
For convenience, we use
$R^{(1)}$
to denote the target ring of the Frobenius map
$R\overset{F}{\rightarrow }R^{(1)}$
. If
$M$
is an
$R$
-module, then
$\operatorname{Hom}_{R}(R^{(1)},M)$
has a structure of an
$R^{(1)}$
-module. We can then identify
$R^{(1)}$
with
$R$
, and
$\operatorname{Hom}_{R}(R^{(1)},M)$
corresponds to an
$R$
-module which we call
$F^{\flat }(M)$
(we refer to [Reference Blickle and Böckle2, Section 2.3] for more details on this). When
$R$
is
$F$
-finite, we have
$\operatorname{Hom}_{R}(R^{(1)},E_{R})\cong E_{R^{(1)}}$
and
$F^{\flat }(E)\cong E_{R}$
, where
$E_{R}$
denotes the injective hull of the residue field of
$(R,\mathfrak{m})$
.
Proposition 2.7. Let
$(R,\mathfrak{m})$
be an
$F$
-finite and
$F$
-full local ring. Then
$R_{\mathfrak{p}}$
is also
$F$
-full for every
$\mathfrak{p}\in \operatorname{Spec}R$
.
Proof. By a result of Gabber [Reference Gabber7, Remark 13.6],
$R$
is a homomorphic image of a regular ring
$A$
. Let
$n=\dim A$
. We have
$$\begin{eqnarray}\displaystyle & & \displaystyle \operatorname{Hom}_{R^{(1)}}(\operatorname{Hom}_{R}(R^{(1)},\operatorname{Ext}_{A}^{n-i}(R,A)),E_{R^{(1)}})\nonumber\\ \displaystyle & & \displaystyle \quad \cong \operatorname{Hom}_{R^{(1)}}(\operatorname{Hom}_{R}(R^{(1)},\operatorname{Ext}_{A}^{n-i}(R,A)),\operatorname{Hom}_{R}(R^{(1)},E_{R}))\nonumber\\ \displaystyle & & \displaystyle \quad \cong \operatorname{Hom}_{R}(\operatorname{Hom}_{R}(R^{(1)},\operatorname{Ext}_{A}^{n-i}(R,A)),E_{R})\nonumber\\ \displaystyle & & \displaystyle \quad \cong R^{(1)}\otimes \operatorname{Hom}_{R}(\operatorname{Ext}_{A}^{n-i}(R,A),E_{R})\nonumber\\ \displaystyle & & \displaystyle \quad \cong R^{(1)}\otimes H_{\mathfrak{m}}^{i}(R)\nonumber\end{eqnarray}$$
where the last isomorphism is by local duality. Thus after identifying
$R^{(1)}$
with
$R$
, we have
$\mathscr{F}_{R}(H_{\mathfrak{m}}^{i}(R))$
is the Matlis dual of
$F^{\flat }(\operatorname{Ext}_{A}^{n-i}(R,A))$
. So
$\mathscr{F}_{R}(H_{\mathfrak{m}}^{i}(R))\rightarrow H_{\mathfrak{m}}^{i}(R)$
is surjective for every
$i$
if and only if
$\operatorname{Ext}_{A}^{n-i}(R,A)\rightarrow F^{\flat }(\operatorname{Ext}_{A}^{n-i}(R,A))$
is injective for every
$i$
. The latter condition clearly localizes. So
$R$
is
$F$
-full implies
$R_{\mathfrak{p}}$
is
$F$
-full for every
$\mathfrak{p}\in \operatorname{Spec}R$
.◻
3 On surjective elements
The following definition was introduced in [Reference Horiuchi, Miller and Shimomoto11] and was the key tool in [Reference Horiuchi, Miller and Shimomoto11].
Definition 3.1. Let
$(R,\mathfrak{m})$
be a Noetherian local ring and
$x$
a regular element of
$R$
.
$x$
is called a surjective element if the natural map on the local cohomology module
$H_{\mathfrak{m}}^{i}(R/(x^{n}))\rightarrow H_{\mathfrak{m}}^{i}(R/(x))$
induced by
$R/(x^{n})\rightarrow R/(x)$
is surjective for all
$n>0$
and
$i\geqslant 0$
.
The next proposition is a restatement of [Reference Horiuchi, Miller and Shimomoto11, Lemma 3.2], so we omit the proof.
Proposition 3.2. The following are equivalent:
(i)
$x$
is a surjective element.(ii) For all
$0<h\leqslant k$
the multiplication map induces an injection$$\begin{eqnarray}R/(x^{h})\overset{x^{k-h}}{\rightarrow }R/(x^{k})\end{eqnarray}$$
for each$$\begin{eqnarray}H_{\mathfrak{m}}^{i}(R/(x^{h}))\rightarrow H_{\mathfrak{ m}}^{i}(R/(x^{k}))\end{eqnarray}$$
$i\geqslant 0$
.(iii) For all
$0<h\leqslant k$
the short exact sequence induces a short exact sequence$$\begin{eqnarray}0\rightarrow R/(x^{h})\overset{x^{k-h}}{\rightarrow }R/(x^{k})\rightarrow R/(x^{k-h})\rightarrow 0\end{eqnarray}$$
for each$$\begin{eqnarray}0\rightarrow H_{\mathfrak{m}}^{i}(R/(x^{h}))\rightarrow H_{\mathfrak{ m}}^{i}(R/(x^{k}))\rightarrow H_{\mathfrak{ m}}^{i}(R/(x^{k-h}))\rightarrow 0\end{eqnarray}$$
$i\geqslant 0$
.
Proposition 3.3. The following are equivalent:
(i)
$x$
is a surjective element.(ii) The multiplication map
$H_{\mathfrak{m}}^{i}(R)\overset{x}{\rightarrow }H_{\mathfrak{m}}^{i}(R)$
is surjective for all
$i\geqslant 0$
.
Proof. By Proposition 3.2,
$x$
is a surjective element if and only if all maps in the direct limit system
$\{H_{\mathfrak{m}}^{i}(R/(x^{h}))\}_{h\geqslant 1}$
are injective. This is equivalent to the condition
$$\begin{eqnarray}\unicode[STIX]{x1D719}_{h}:H_{\mathfrak{m}}^{i}(R/(x^{h}))\rightarrow \mathop{\varinjlim }\nolimits_{h}H_{\mathfrak{m}}^{i}(R/(x^{h}))\cong H_{\mathfrak{ m}}^{i}(H_{(x)}^{1}(R))\cong H_{\mathfrak{ m}}^{i+1}(R)\end{eqnarray}$$
is injective for all
$h\geqslant 1$
and all
$i\geqslant 0$
(the last isomorphism comes from an easy computation using local cohomology spectral sequences and noting that
$x$
is a nonzero divisor on
$R$
, see also [Reference Horiuchi, Miller and Shimomoto11, Lemma 2.2]).
Claim 3.4.
$\unicode[STIX]{x1D719}_{h}$
is exactly the connection maps in the long exact sequence of local cohomology induced by
$0\rightarrow R\xrightarrow[{}]{\cdot x^{h}}R\rightarrow R/(x^{h})\rightarrow 0$
:
$$\begin{eqnarray}\cdots \rightarrow H_{\mathfrak{m}}^{i}(R/(x^{h}))\overset{\unicode[STIX]{x1D719}_{h}}{\rightarrow }H_{\mathfrak{m}}^{i+1}(R)\overset{x^{h}}{\rightarrow }H_{\mathfrak{ m}}^{i+1}(R)\rightarrow \cdots \,.\end{eqnarray}$$
Proof of claim.
Observe that by definition,
$\unicode[STIX]{x1D719}_{h}$
is the natural map in the long exact sequence of local cohomology
$$\begin{eqnarray}\cdots \rightarrow H_{\mathfrak{m}}^{i}(R/(x^{h}))\xrightarrow[{}]{\unicode[STIX]{x1D719}_{h}}H_{\mathfrak{m}}^{i}(R_{x}/R)\xrightarrow[{}]{\cdot x}H_{\mathfrak{m}}^{i}(R_{x}/R)\rightarrow \cdots\end{eqnarray}$$
which is induced by
$0\rightarrow R/(x^{h})\rightarrow R_{x}/R\xrightarrow[{}]{\cdot x^{h}}R_{x}/R\rightarrow 0$
(note that
$x^{h}$
is a nonzero divisor on
$R$
and
$H_{x}^{1}(R)\cong R_{x}/R$
). However, it is easy to see that the multiplication by
$x^{h}$
map
$H_{\mathfrak{m}}^{i}(R_{x}/R)\xrightarrow[{}]{\cdot x^{h}}H_{\mathfrak{m}}^{i}(R_{x}/R)$
can be identified with the multiplication by
$x^{h}$
map
$H_{\mathfrak{m}}^{i+1}(R)\xrightarrow[{}]{\cdot x^{h}}H_{\mathfrak{m}}^{i+1}(R)$
because we have a natural identification
$H_{\mathfrak{m}}^{i}(R_{x}/R)\cong H_{\mathfrak{m}}^{i}(H_{x}^{1}(R))\cong H_{\mathfrak{m}}^{i+1}(R)$
(see for example [Reference Horiuchi, Miller and Shimomoto11, Lemma 2.2]). This finishes the proof of the claim.◻
From the claim it is immediate that
$x$
is a surjective element if and only if the long exact sequence splits into short exact sequences:
$$\begin{eqnarray}0\rightarrow H_{\mathfrak{m}}^{i}(R/(x^{h}))\rightarrow H_{\mathfrak{ m}}^{i+1}(R)\overset{x^{h}}{\rightarrow }H_{\mathfrak{ m}}^{i+1}(R)\rightarrow 0.\end{eqnarray}$$
But this is equivalent to saying that the multiplication map
$H_{\mathfrak{m}}^{i}(R)\overset{x^{h}}{\rightarrow }H_{\mathfrak{m}}^{i}(R)$
is surjective for all
$h\geqslant 1$
and
$i\geqslant 0$
, and also equivalent to
$H_{\mathfrak{m}}^{i}(R)\overset{x}{\rightarrow }H_{\mathfrak{m}}^{i}(R)$
is surjective for all
$i\geqslant 0$
.◻
We next link the notion of surjective element with
$F$
-fullness. This is inspired by [Reference Ma, Schwede and Shimomoto18, Reference Singh and Walther24].
Proposition 3.5. Let
$x$
be a regular element of
$(R,\mathfrak{m})$
. If
$R/(x)$
is
$F$
-full, then
$x$
is a surjective element. In particular, if
$R/(x)$
is
$F$
-anti-nilpotent, then
$x$
is a surjective element.
Proof. We have natural maps:
$$\begin{eqnarray}\displaystyle \mathscr{F}_{R}^{e}(H_{\mathfrak{m}}^{i}(R/(x))) & \xrightarrow[{}]{\unicode[STIX]{x1D6FC}_{e}} & \displaystyle R/(x)\otimes _{R}\mathscr{F}_{R}^{e}(H_{\mathfrak{m}}^{i}(R/(x)))\cong \mathscr{F}_{R/(x)}^{e}(H_{\mathfrak{m}}^{i}(R/(x)))\nonumber\\ \displaystyle & \xrightarrow[{}]{\unicode[STIX]{x1D6FD}_{e}} & \displaystyle H_{\mathfrak{m}}^{i}(R/(x)).\nonumber\end{eqnarray}$$
If
$R/(x)$
is
$F$
-full, then
$\unicode[STIX]{x1D6FD}_{e}$
is surjective for every
$e$
. Since
$\unicode[STIX]{x1D6FC}_{e}$
is always surjective, the natural map
$\mathscr{F}_{R}^{e}(H_{\mathfrak{m}}^{i}(R/(x)))\rightarrow H_{\mathfrak{m}}^{i}(R/(x))$
is surjective for every
$e$
. Now simply notice that for every
$e>0$
, the map
$\mathscr{F}_{R}^{e}(H_{\mathfrak{m}}^{i}(R/(x)))\rightarrow H_{\mathfrak{m}}^{i}(R/(x))$
factors through
$H_{\mathfrak{m}}^{i}(R/(x^{p^{e}}))\rightarrow H_{\mathfrak{m}}^{i}(R/(x))$
, so
$H_{\mathfrak{m}}^{i}(R/(x^{p^{e}}))\rightarrow H_{\mathfrak{m}}^{i}(R/(x))$
is surjective for every
$e>0$
. This clearly implies that
$x$
is a surjective element.◻
The above propositions allow us to construct a family of non
$F$
-full local rings:
Example 3.6. Let
$(R,\mathfrak{m})$
be a local ring with finite length cohomology, that is,
$H_{\mathfrak{m}}^{i}(R)$
has finite length for every
$i<\dim R$
(under mild conditions, this is equivalent to saying that
$R$
is Cohen–Macaulay on the punctured spectrum). Let
$x$
be an arbitrary regular element in
$R$
. If
$R$
is not Cohen–Macaulay, then we claim that
$R/(x)$
is not
$F$
-full (and hence not
$F$
-anti-nilpotent). For suppose it is, then
$x$
is a surjective element by Proposition 3.5, hence
$H_{\mathfrak{m}}^{i}(R)\overset{x}{\rightarrow }H_{\mathfrak{m}}^{i}(R)$
is surjective for every
$i$
by Proposition 3.3. But since
$R$
has finite length cohomology, we also know that a power of
$x$
annihilates
$H_{\mathfrak{m}}^{i}(R)$
for every
$i<\dim R$
. This implies
$H_{\mathfrak{m}}^{i}(R)=0$
for every
$i<\dim R$
. So
$R$
is Cohen–Macaulay, a contradiction.
We learned the following argument from [Reference Horiuchi, Miller and Shimomoto11, Lemma A.1]. Since it is a crucial technique of this paper, we provide a detailed proof.
Proposition 3.7. Let
$(R,\mathfrak{m})$
be a local ring of prime characteristic
$p$
and
$x$
a regular element of
$R$
. Let
$s$
be a positive integer such that the map
$H_{\mathfrak{m}}^{s-1}(R)\overset{x}{\rightarrow }H_{\mathfrak{m}}^{s-1}(R)$
is surjective and the Frobenius action on
$H_{\mathfrak{m}}^{s-1}(R/(x))$
is injective, then the map
$$\begin{eqnarray}H_{\mathfrak{m}}^{s}(R)\overset{x^{p-1}F}{\longrightarrow }H_{\mathfrak{ m}}^{s}(R)\end{eqnarray}$$
is injective.
Proof. The natural commutative diagram
induces the following commutative diagram (the left most
$0$
comes from our hypothesis that the map
$H_{\mathfrak{m}}^{s-1}(R)\overset{x}{\rightarrow }H_{\mathfrak{m}}^{s-1}(R)$
is surjective):
Suppose
$y\in \operatorname{Ker}(x^{p-1}F)\cap \text{Soc}(H_{\mathfrak{m}}^{s}(R))$
. Then we have
$x\cdot y=0$
so there exists
$z\in H_{\mathfrak{m}}^{s-1}(R/(x))$
such that
$\unicode[STIX]{x1D6FC}(z)=y$
. Following the above commutative diagram we have
$$\begin{eqnarray}(\unicode[STIX]{x1D6FC}\circ F)(z)=x^{p-1}F(\unicode[STIX]{x1D6FC}(z))=x^{p-1}F(y)=0.\end{eqnarray}$$
However, since both
$F$
and
$\unicode[STIX]{x1D6FC}$
are injective, we have
$z=0$
and hence
$y=0$
. This shows
$x^{p-1}F$
is injective and hence completes the proof.◻
Proposition 3.7 immediately generalizes the main result of [Reference Horiuchi, Miller and Shimomoto11]:
Corollary 3.8. (Compare with [Reference Horiuchi, Miller and Shimomoto11], Main Theorem)
Let
$(R,\mathfrak{m})$
be a local ring of prime characteristic
$p$
and
$x$
a regular element of
$R$
. Suppose
$R/(x)$
is
$F$
-injective. Then we have
(i) The map
$H_{\mathfrak{m}}^{t}(R)\overset{x^{p-1}F}{\longrightarrow }H_{\mathfrak{m}}^{t}(R)$
is injective where
$t=\operatorname{depth}R$
. In particular, the natural Frobenius action on
$H_{\mathfrak{m}}^{t}(R)$
is injective.(ii) Suppose
$x$
is a surjective element. Then the map
$H_{\mathfrak{m}}^{i}(R)\overset{x^{p-1}F}{\longrightarrow }H_{\mathfrak{m}}^{i}(R)$
is injective for all
$i\geqslant 0$
. In particular,
$R$
is
$F$
-injective.(iii) If
$R/(x)$
is
$F$
-full (e.g.,
$R$
is
$F$
-anti-nilpotent or
$R$
is
$F$
-pure), then
$R$
is
$F$
-injective.
Proof. (i) Follows from Proposition 3.7 applied to
$s=t$
, (ii) also follows from Proposition 3.7 (because
$H_{\mathfrak{m}}^{i}(R)\overset{x}{\rightarrow }H_{\mathfrak{m}}^{i}(R)$
is surjective for every
$i\geqslant 0$
by Proposition 3.3), (iii) follows from (ii), because we know
$x$
is a surjective element by Proposition 3.5.◻
In the next two sections, we show that
$F$
-full and
$F$
-anti-nilpotent singularities both deform. We also prove new cases of deformation of
$F$
-injectivity. These results are generalizations of Proposition 3.7 and Corollary 3.8.
4 Deformation of
$F$
-full and
$F$
-anti-nilpotent singularities
In this section we prove that the condition
$F$
-full and
$F$
-anti-nilpotent both deform. Throughout this section we assume that
$(R,\mathfrak{m})$
is a local ring of prime characteristic
$p$
. We begin with a crucial lemma.
Lemma 4.1. Let
$x$
be a surjective element of
$R$
. Let
$N\subseteq H_{\mathfrak{m}}^{i}(R)$
be an
$F$
-stable submodule. Let
$L=\bigcap _{t}x^{t}N$
. Then
$L$
is an
$F$
-stable submodule of
$H_{\mathfrak{m}}^{i}(R)$
and we have the following commutative diagram (for every
$e\geqslant 1$
):
where
$\unicode[STIX]{x1D719}$
is the map
$H_{\mathfrak{m}}^{i-1}(R/(x))\rightarrow H_{\mathfrak{m}}^{i}(R)$
.
Proof. Since
$x$
is a surjective element, by Proposition 3.3 we know that the map
$$\begin{eqnarray}H_{\mathfrak{m}}^{i}(R)\overset{x}{\rightarrow }H_{\mathfrak{ m}}^{i}(R)\text{ is surjective for every }i>0.\quad (\star )\end{eqnarray}$$
Applying the local cohomology functor to the following commutative diagram:
we have the following commutative diagram:
for all
$i\geqslant 1$
and
$e\geqslant 1$
, where the rows are short exact sequences by
$(\star )$
.
Therefore, to prove the lemma, it suffices to show that
$L$
is
$F$
-stable and
$$\begin{eqnarray}0\rightarrow H_{\mathfrak{m}}^{i-1}(R/(x))/\unicode[STIX]{x1D719}^{-1}(L)\overset{\unicode[STIX]{x1D719}}{\rightarrow }H_{\mathfrak{ m}}^{i}(R)/L\overset{x}{\rightarrow }H_{\mathfrak{ m}}^{i}(R)/L\rightarrow 0\end{eqnarray}$$
is exact. It is clear that
$L$
is
$F$
-stable since it is an intersection of
$F$
-stable submodules of
$H_{\mathfrak{m}}^{i}(R)$
. To see the exactness of the above sequence, first note that
$\operatorname{Im}(\unicode[STIX]{x1D719})=0:_{H_{\mathfrak{m}}^{i}(R)}x$
, so
$L+\operatorname{Im}(\unicode[STIX]{x1D719})\subseteq L:_{H_{\mathfrak{m}}^{i}(R)}x$
. Thus it is enough to check that
$L:_{H_{\mathfrak{m}}^{i}(R)}x\subseteq L+\operatorname{Im}(\unicode[STIX]{x1D719})$
. Let
$y$
be an element such that
$xy\in L$
. Since
$L=xL$
by the construction of
$L$
, there exists
$z\in L$
such that
$xy=xz$
. So
$y-z\in \operatorname{Im}(\unicode[STIX]{x1D719})$
and hence
$y\in L+\operatorname{Im}(\unicode[STIX]{x1D719})$
, as desired.◻
We are ready to prove the main result of this section. This answers [Reference Quy and Shimomoto20, Problem 4] for stably FH-finiteness.
Theorem 4.2.
$(R,\mathfrak{m})$
be a local ring of positive characteristic
$p$
and
$x$
a regular element of
$R$
. Then we have:
(i) if
$R/(x)$
is
$F$
-anti-nilpotent, then so is
$R$
;(ii) if
$R/(x)$
is
$F$
-full, then so is
$R$
.
Proof. We first prove (i). Let
$N$
be an
$F$
-stable submodule of
$H_{\mathfrak{m}}^{i}(R)$
. We want to show that the induced Frobenius action on
$H_{\mathfrak{m}}^{i}(R)/N$
is injective. Since
$R/(x)$
is
$F$
-anti-nilpotent,
$x$
is a surjective element by Proposition 3.5. Let
$L=\bigcap _{t}x^{t}N$
. By Lemma 4.1, we have the following commutative diagram:
We first claim that the middle map
$x^{p^{e}-1}F^{e}:H_{\mathfrak{m}}^{i}(R)/L\rightarrow H_{\mathfrak{m}}^{i}(R)/L$
is injective. Let
$y\in \operatorname{Ker}(x^{p^{e}-1}F^{e})\cap \text{Soc}(H_{\mathfrak{m}}^{i}(R)/L)$
. We have
$x\cdot y=0$
, so
$y=\unicode[STIX]{x1D719}(z)$
for some
$z\in H_{\mathfrak{m}}^{i-1}(R/(x))/\unicode[STIX]{x1D719}^{-1}(L)$
. It is easy to see that
$\unicode[STIX]{x1D719}^{-1}(L)$
is an
$F$
-stable submodule of
$H_{\mathfrak{m}}^{i-1}(R/(x))$
and
$F^{e}(z)=0$
. Since
$R/(x)$
is
$F$
-anti-nilpotent, we know the Frobenius action
$F$
, and hence its iterate
$F^{e}$
, on
$H_{\mathfrak{m}}^{i-1}(R/(x))/\unicode[STIX]{x1D719}^{-1}(L)$
is injective. Therefore,
$z=0$
and hence
$y=0$
. This proves that
$x^{p^{e}-1}F^{e}$
and hence
$F$
acts injectively on
$H_{\mathfrak{m}}^{i}(R)/L$
.
Note that we have a descending chain
$N\supseteq xN\supseteq x^{2}N\supseteq \cdots \,.$
Since
$H_{\mathfrak{m}}^{i}(R)$
is Artinian,
$L=\bigcap _{t}x^{t}N=x^{n}N$
for all
$n\gg 0$
. We next claim that
$L=N$
, this will finish the proof because we already showed
$F$
acts injectively on
$H_{\mathfrak{m}}^{i}(R)/L$
. We have
$x^{p^{e}-1}F^{e}(N)\subseteq x^{p^{e}-1}N=L$
for
$e\gg 0$
, but the map
$x^{p^{e}-1}F^{e}:H_{\mathfrak{m}}^{i}(R)/L\rightarrow H_{\mathfrak{m}}^{i}(R)/L$
is injective by the above paragraph. So we must have
$N\subseteq L$
and thus
$L=N$
. This completes the proof of (1).
Next we prove (ii). The method is similar to that of (i). Let
$N$
be the
$R$
-span of
$F(H_{\mathfrak{m}}^{i}(R))$
in
$H_{\mathfrak{m}}^{i}(R)$
, this is the same as the image of
$\mathscr{F}_{R}(H_{\mathfrak{m}}^{i}(R))\rightarrow H_{\mathfrak{m}}^{i}(R)$
. It is clear that
$N$
is an
$F$
-stable submodule. We want to show
$N=H_{\mathfrak{m}}^{i}(R)$
. Since
$R/(x)$
is
$F$
-full,
$x$
is a surjective element by Proposition 3.5. Let
$L=\bigcap _{t}x^{t}N$
. By Lemma 4.1, we have the following commutative diagram:
The descending chain
$N\supseteq xN\supseteq x^{2}N\supseteq \cdots \,$
stabilizes because
$H_{\mathfrak{m}}^{i}(R)$
is Artinian. So
$L=\bigcap _{t}x^{t}N=x^{n}N$
for
$n\gg 0$
. The key point is that in the above diagram, the middle Frobenius action
$x^{p^{e}-1}F^{e}$
is the zero map on
$H_{\mathfrak{m}}^{i}(R)/L$
for
$e\gg 0$
, because for any
$y\in H_{\mathfrak{m}}^{i}(R)$
,
$F^{e}(y)\in N$
and thus
$x^{p^{e}-1}F^{e}(y)\in L$
for
$e\gg 0$
. But then since
$H_{\mathfrak{m}}^{i-1}(R/(x))/\unicode[STIX]{x1D719}^{-1}(L)$
can be viewed as a submodule of
$H_{\mathfrak{m}}^{i}(R)/L$
by the above commutative diagram, the natural Frobenius action
$F^{e}$
on
$H_{\mathfrak{m}}^{i-1}(R/(x))/\unicode[STIX]{x1D719}^{-1}(L)$
is zero, that is,
$F$
is nilpotent on
$H_{\mathfrak{m}}^{i-1}(R/(x))/\unicode[STIX]{x1D719}^{-1}(L)$
.
Since
$F$
is nilpotent on
$H_{\mathfrak{m}}^{i-1}(R/(x))/\unicode[STIX]{x1D719}^{-1}(L)$
, we know that
$\unicode[STIX]{x1D719}^{-1}(L)$
must contain all elements
$F^{e}(H_{\mathfrak{m}}^{i}(R/(x)))$
, hence it contains the
$R$
-span of
$F^{e}(H_{\mathfrak{m}}^{i}(R/(x)))$
. But
$R/(x)$
is
$F$
-full, so we must have
$\unicode[STIX]{x1D719}^{-1}(L)=H_{\mathfrak{m}}^{i-1}(R/(x))$
. But this means the map
$$\begin{eqnarray}H_{\mathfrak{m}}^{i}(R)/L\xrightarrow[{}]{x}H_{\mathfrak{ m}}^{i}(R)/L\end{eqnarray}$$
is an isomorphism, which is impossible unless
$H_{\mathfrak{m}}^{i}(R)=L$
(since otherwise any nonzero socle element of
$H_{\mathfrak{m}}^{i}(R)/L$
maps to zero). Therefore, we have
$H_{\mathfrak{m}}^{i}(R)=N=L$
. This proves
$R$
is
$F$
-full and hence finished the proof of (2).◻
The following is a well-known counter-example of Fedder [Reference Fedder6] and Singh [Reference Singh22] for the deformation of
$F$
-purity.
Example 4.3. (Compare with [Reference Quy and Shimomoto20, Lemma 6.1])
Let
$K$
be a perfect field of characteristic
$p>0$
and let
$$\begin{eqnarray}R:=K[[U,V,Y,Z]]/(UV,UZ,Z(V-Y^{2})).\end{eqnarray}$$
Let
$u,v,y$
and
$z$
denote the image of
$U,V,Y$
and
$Z$
in
$R$
(and its quotients), respectively. Then
$y$
is a regular element of
$R$
and
$R/(y)\cong K[[U,V,Z]]/(UV,UZ,VZ)$
is
$F$
-pure by [Reference Hochster and Roberts12, Proposition 5.38]. So
$R/(y)$
is
$F$
-anti-nilpotent by [Reference Ma16, Theorem 1.1]. By Theorem 4.2 we have
$R$
is also
$F$
-anti-nilpotent, or equivalently,
$R$
is stably
$FH$
-finite.
5
$F$
-injectivity
5.1
$F$
-injectivity and depth
We start with the following definition.
Definition 5.1. (Cf. [Reference Brodmann and Sharp3, Definition 9.1.3]) Let
$M$
be a finitely generated module over a local ring
$(R,\mathfrak{m})$
. The finiteness dimension
$f_{\mathfrak{m}}(M)$
of
$M$
with respect to
$\mathfrak{m}$
is defined as follows:
$$\begin{eqnarray}f_{\mathfrak{m}}(M):=\text{inf}\{i\,|\,H_{\mathfrak{m}}^{i}(M)~\text{is not finitely generated}\}\in \mathbb{Z}_{{\geqslant}0}\cup \{\infty \}.\end{eqnarray}$$
Remark 5.2.
(i) Assume that
$\dim M=0$
or
$M=0$
(recall that a trivial module has dimension
$-1$
). In this case,
$H_{\mathfrak{m}}^{i}(M)$
is finitely generated for all
$i$
and
$f_{\mathfrak{m}}(M)$
is equal to
$\infty$
. It will be essential to know when the finiteness dimension is a positive integer. We mention the following result. Let
$(R,\mathfrak{m})$
be a local ring and let
$M$
be a finitely generated
$R$
-module. If
$d=\dim M>0$
, then the local cohomology module
$H_{\mathfrak{m}}^{d}(M)$
is not finitely generated. For the proof of this result, see [Reference Brodmann and Sharp3, Corollary 7.3.3].(ii) Suppose
$(R,\mathfrak{m})$
is an image of a Cohen–Macaulay local ring. By the Grothendieck finiteness theorem (cf. [Reference Brodmann and Sharp3, Theorem 9.5.2]) we have $$\begin{eqnarray}f_{\mathfrak{m}}(M)=\min \{\operatorname{depth}M_{\mathfrak{p}}+\dim R/\mathfrak{p}\,:\,\mathfrak{p}\in \text{Supp}(M)\setminus \{\mathfrak{m}\}\}.\end{eqnarray}$$
(iii)
$M$
is generalized Cohen–Macaulay if and only if
$\dim M=f_{\mathfrak{m}}(M)$
.
It is clear that
$\operatorname{depth}R\leqslant f_{\mathfrak{m}}(R)\leqslant \dim R$
. The following result says that if
$R/(x)$
is
$F$
-injective, then
$R$
has ‘good’ depth.
Theorem 5.3. If
$R/(x)$
is
$F$
-injective, then
$\operatorname{depth}R=f_{\mathfrak{m}}(R)$
.
Proof. Suppose
$t=\operatorname{depth}R<f_{\mathfrak{m}}(R)$
. The commutative diagram
induces the following commutative diagram
where both
$\unicode[STIX]{x1D6FC}$
and the left vertical map are injective. But
$H_{\mathfrak{m}}^{t}(R)$
has finite length,
$x^{p^{e}-1}F^{e}:H_{\mathfrak{m}}^{t}(R)\rightarrow H_{\mathfrak{m}}^{t}(R)$
vanishes for
$e\gg 0$
, which is a contradiction.◻
Remark 5.4. The assertion of Theorem 5.3 also holds true if
$R/(x)$
is
$F$
-full. Indeed, by Proposition 3.5 we have
$x$
is a surjective element. Hence there is no nonzero
$H_{\mathfrak{m}}^{i}(R)$
of finite length. Thus
$\operatorname{depth}R=f_{\mathfrak{m}}(R)$
.
Remark 5.5. The above result is closely related to the work of Schwede and Singh in [Reference Horiuchi, Miller and Shimomoto11, Appendix]. In the proof of [Reference Horiuchi, Miller and Shimomoto11, Lemma A.2, Theorem A.3], it is claimed that if
$R_{\mathfrak{p}}$
satisfies the Serre condition
$(S_{k})$
for all
$\mathfrak{p}$
in
$\text{Spec}^{\circ }(R)$
, the punctured spectrum of
$R$
, and
$\operatorname{depth}R=t<k$
, then
$H_{\mathfrak{m}}^{t}(R)$
is finitely generated. But this fact may not be true if
$R$
is not equidimensional. For instance, let
$R=K[[a,b,c,d]]/(a)\cap (b,c,d)$
with
$K$
a field. We have
$\operatorname{depth}R=1$
and
$R_{\mathfrak{p}}$
satisfies
$(S_{2})$
for all
$\mathfrak{p}\in \text{Spec}^{\circ }(R)$
. However,
$H_{\mathfrak{m}}^{1}(R)$
is not finitely generated.
The assertion of [Reference Horiuchi, Miller and Shimomoto11, Lemma A.2] (and hence [Reference Horiuchi, Miller and Shimomoto11, Theorem A.3]) is still true. In fact, we can reduce it to the case that
$R$
is equidimensional. We fill this gap below.
Corollary 5.6. [Reference Horiuchi, Miller and Shimomoto11, Lemma A.2]
Let
$(R,\mathfrak{m})$
be an
$F$
-finite local ring. Suppose there exists a regular element
$x$
such that
$R/(x)$
is
$F$
-injective. If
$R_{\mathfrak{p}}$
satisfies the Serre condition
$(S_{k})$
for all
$\mathfrak{p}\in \text{Spec}^{\circ }(R)$
, then
$R$
is
$(S_{k})$
.
Proof. We can assume that
$k\leqslant d=\dim R$
. In fact, we need only to prove that
$t:=\operatorname{depth}R\geqslant k$
. The case
$k=1$
is trivial since
$R$
contains a regular element
$x$
. For
$k\geqslant 2$
, since
$R/(x)$
is
$F$
-injective we have
$R/(x)$
is reduced (cf. [Reference Schwede21, Proposition 4.3]). Hence
$\operatorname{depth}(R/(x))\geqslant 1$
, so
$\operatorname{depth}R\geqslant 2$
. Thus
$R$
satisfies the Serre condition
$(S_{2})$
. On the other hand, since
$R$
is
$F$
-finite,
$R$
is a homomorphic image of a regular ring by a result of Gabber [Reference Gabber7, Remark 13.6]. In particular,
$R$
is universally catenary.Footnote
3
But if a universally catenary ring satisfies
$(S_{2})$
, then it is equidimensional (see [Reference Hochster and Huneke10, Remark 2.2(h)]). By Theorem 5.3 and Remark 5.2(ii), there exists a prime ideal
$\mathfrak{p}\in \text{Spec}^{\circ }(R)$
such that
$\operatorname{depth}R=\operatorname{depth}R_{\mathfrak{p}}+\dim R/\mathfrak{p}$
. It is then easy to see that
$\operatorname{depth}R\geqslant \min \{d,k+1\}\geqslant k$
. The proof is complete.◻
Remark 5.7. In the above argument, we actually proved that if
$k<d$
, then
$\operatorname{depth}R\geqslant k+1$
.
5.2 Deformation of
$F$
-injectivity
We begin with the following generalization of the notion of surjective elements.
Definition 5.8. (Cf. [Reference Cuong, Morales and Nhan4])
A regular element
$x$
is called a strictly filter regular element if
$$\begin{eqnarray}\operatorname{Coker}(H_{\mathfrak{m}}^{i}(R)\overset{x}{\rightarrow }H_{\mathfrak{ m}}^{i}(R))\end{eqnarray}$$
has finite length for all
$i\geqslant 0$
.
Lemma 5.9. Let
$(R,\mathfrak{m})$
be a local ring of characteristic
$p>0$
. Suppose the residue field
$k=R/\mathfrak{m}$
is perfect. Let
$M$
be an
$R$
-module with an injective Frobenius action
$F$
. Suppose
$L$
is an
$F$
-stable submodule of
$M$
of finite length. Then the induced Frobenius action on
$M/L$
is injective.
Proof. First we note that
$L$
is killed by
$\mathfrak{m}$
: suppose
$x\in L$
, then
$F^{e}(\mathfrak{m}\cdot x)=\mathfrak{m}^{[p^{e}]}\cdot x=0$
for
$e\gg 0$
since
$L$
has finite length. But then
$\mathfrak{m}\cdot x=0$
since
$F$
acts injectively. Now we have a Frobenius action
$F$
on a
$k$
-vector space
$L$
. Call the image of
$L^{\prime }\subseteq L$
(which is a
$k^{p}$
-vector subspace of
$L$
). Since
$F$
is injective, the
$k^{p}$
-vector space dimension of
$L^{\prime }$
is equal to the
$k$
-vector space dimension of
$L$
. But since
$k^{p}=k$
, this implies
$L^{\prime }=L$
and thus
$F$
is surjective, hence
$F$
is bijective. Now by the injectivity of
$F$
again we have
$F(x)\notin L$
for all
$x\notin L$
. Thus
$F:M/L\rightarrow M/L$
is injective.◻
Example 5.10. The perfectness of the residue field in Lemma 5.9 is necessary. Let
$A=\mathbb{F}_{p}[t]$
and
$R=k=\mathbb{F}_{p}(t)$
, where
$t$
is an indeterminate. We consider the Frobenius action on the
$A$
-module
$Ae_{1}\oplus Ae_{2}$
defined by
$$\begin{eqnarray}F(f(t),g(t))=(f(t)^{p}+tg(t)^{p},0).\end{eqnarray}$$
It is clear that
$F$
is injective. Moreover,
$Ae_{1}\oplus 0$
is an
$F$
-stable submodule of
$Ae_{1}\oplus Ae_{2}$
. Since
$F(Ae_{1}\oplus Ae_{2})\subseteq Ae_{1}\oplus 0$
, the induced Frobenius action on
$(Ae_{1}\oplus Ae_{2})/(Ae_{1}\oplus 0)$
is the zero map. By localizing, we obtain an injective Frobenius action on
$M=k\cdot e_{1}\oplus k\cdot e_{2}$
with
$L=k\cdot e_{1}\oplus 0$
is an
$F$
-stable submodule of finite length, but the induced Frobenius action on
$M/L$
is not injective.
The following is a generalization of the main result of [Reference Horiuchi, Miller and Shimomoto11] when
$R/\mathfrak{m}$
is perfect.
Theorem 5.11. Let
$(R,\mathfrak{m})$
be a Noetherian local ring of characteristic
$p>0$
. Suppose the residue field
$k=R/\mathfrak{m}$
is perfect. Let
$x$
be a strictly filter regular element. If
$R/(x)$
is
$F$
-injective, then the map
$x^{p-1}F$
:
$H_{\mathfrak{m}}^{i}(R)\rightarrow H_{\mathfrak{m}}^{i}(R)$
is injective for every
$i$
, in particular
$R$
is
$F$
-injective.
Proof. Let
$L_{i}:=\operatorname{Coker}(H_{\mathfrak{m}}^{i}(R)\overset{x}{\rightarrow }H_{\mathfrak{m}}^{i}(R))$
, we have
$L_{i}$
has finite length for all
$i\geqslant 0$
. The commutative diagram
induces the following commutative diagram
Therefore, we have the following commutative diagram
with the Frobenius action
$F:H_{\mathfrak{m}}^{i-1}(R/(x))/L_{i-1}\rightarrow H_{\mathfrak{m}}^{i-1}(R/(x))/L_{i-1}$
is injective by Lemma 5.9. Now by the same method as in the proof of Proposition 3.7 or Theorem 4.2(i), we conclude that the map
$x^{p-1}F:H_{\mathfrak{m}}^{i}(R)\rightarrow H_{\mathfrak{m}}^{i}(R)$
is injective for all
$i\geqslant 0$
.◻
Similarly, we have the following:
Proposition 5.12. Let
$(R,\mathfrak{m})$
be a Noetherian local ring of characteristic
$p>0$
. Suppose the residue field
$k=R/\mathfrak{m}$
is perfect. Let
$x$
be a regular element such that
$R/(x)$
is
$F$
-injective. Let
$s$
be a positive integer such that
$H_{\mathfrak{m}}^{s-1}(R/(x))$
has finite length. Then the map
$x^{p-1}F:H_{\mathfrak{m}}^{s+1}(R)\rightarrow H_{\mathfrak{m}}^{s+1}(R)$
is injective.
Proof. The short exact sequence
$$\begin{eqnarray}0\rightarrow R\overset{x}{\rightarrow }R\rightarrow R/(x)\rightarrow 0\end{eqnarray}$$
induces the exact sequence
$$\begin{eqnarray}\cdots \rightarrow H_{\mathfrak{m}}^{s-1}(R/(x))\rightarrow H_{\mathfrak{ m}}^{s}(R)\overset{x}{\rightarrow }H_{\mathfrak{ m}}^{s}(R)\rightarrow H_{\mathfrak{ m}}^{s}(R/(x))\rightarrow H_{\mathfrak{ m}}^{s+1}(R)\rightarrow \cdots \,.\end{eqnarray}$$
Since
$H_{\mathfrak{m}}^{s-1}(R/(x))$
has finite length, so is
$\operatorname{Ker}(H_{\mathfrak{m}}^{s}(R)\overset{x}{\rightarrow }H_{\mathfrak{m}}^{s}(R))$
. We claim that
$$\begin{eqnarray}L_{s}:=\operatorname{Coker}(H_{\mathfrak{m}}^{s}(R)\overset{x}{\rightarrow }H_{\mathfrak{ m}}^{s}(R))\end{eqnarray}$$
also has finite length: to see this we may assume
$R$
is complete, since
$\operatorname{Ker}(H_{\mathfrak{m}}^{s}(R)\overset{x}{\rightarrow }H_{\mathfrak{m}}^{s}(R))$
has finite length, this means
$H_{\mathfrak{m}}^{s}(R)^{\vee }\overset{x}{\rightarrow }H_{\mathfrak{m}}^{s}(R)^{\vee }$
is surjective when localizing at any
$\mathfrak{p}\neq \mathfrak{m}$
. But by [Reference Matsumura19, Theorem 2.4] this implies
$H_{\mathfrak{m}}^{s}(R)^{\vee }\overset{x}{\rightarrow }H_{\mathfrak{m}}^{s}(R)^{\vee }$
is an isomorphism when localizing at any
$\mathfrak{p}\neq \mathfrak{m}$
. Thus
$\operatorname{Ker}(H_{\mathfrak{m}}^{s}(R)^{\vee }\overset{x}{\rightarrow }H_{\mathfrak{m}}^{s}(R)^{\vee })$
has finite length which, after dualizing, shows that
$\operatorname{Coker}(H_{\mathfrak{m}}^{s}(R)\overset{x}{\rightarrow }H_{\mathfrak{m}}^{s}(R))$
has finite length.
We have proved
$L_{s}=\operatorname{Coker}(H_{\mathfrak{m}}^{s}(R)\overset{x}{\rightarrow }H_{\mathfrak{m}}^{s}(R))$
has finite length. Now the map
$x^{p-1}F:H_{\mathfrak{m}}^{s+1}(R)\rightarrow H_{\mathfrak{m}}^{s+1}(R)$
is injective by the same argument as in Theorem 5.11.◻
The following immediate corollary of the above proposition recovers (and in fact generalizes) results in [Reference Horiuchi, Miller and Shimomoto11].
Corollary 5.13. [Reference Horiuchi, Miller and Shimomoto11, Corollary 4.7]
Let
$(R,\mathfrak{m})$
be a Noetherian local ring of characteristic
$p>0$
. Suppose the residue field
$k=R/\mathfrak{m}$
is perfect. Let
$x$
be a regular element such that
$R/(x)$
is
$F$
-injective. Then the map
$x^{p-1}F:H_{\mathfrak{m}}^{i}(R)\rightarrow H_{\mathfrak{m}}^{i}(R)$
is injective for all
$i\leqslant f_{\mathfrak{m}}(R/(x))+1$
. In particular, if
$R/(x)$
is generalized Cohen–Macaulay, then
$R$
is
$F$
-injective.
Because of the deep connections between
$F$
-injective and Du Bois singularities [Reference Bhatt, Schwede and Takagi1, Reference Schwede21] and Remark 2.6, we believe that it is rarely the case that an
$F$
-injective ring fails to be
$F$
-full (again, the only example we know this happens is [Reference Ma, Schwede and Shimomoto18, Example 3.5], which is based on the construction of [Reference Enescu and Hochster5, Example 2.16]). Therefore, we introduce:
Definition 5.14. We say
$(R,\mathfrak{m})$
is strongly
$F$
-injective if
$R$
is
$F$
-injective and
$F$
-full.
Remark 5.15. In general we have:
$F$
-anti-nilpotent
$\Rightarrow$
strongly
$F$
-injective
$\Rightarrow$
$F$
-injective. Moreover, when
$R$
is Cohen–Macaulay, strongly
$F$
-injective is equivalent to
$F$
-injective.
We can prove that strong
$F$
-injectivity deform.
Corollary 5.16. Let
$x$
be a regular element on
$(R,\mathfrak{m})$
. If
$R/(x)$
is strongly
$F$
-injective, then
$R$
is strongly
$F$
-injective.
