Proof. The nontrivial implication is to show that if $\varphi \otimes _R \operatorname {\mathrm {id}}_S$ is an injective S-linear map for all R-algebras S that are a first order nilpotent thickening of R, then $\varphi $ is a pure R-linear map. Let P be an R-module and let $R \ltimes P$ be the R-algebra obtained by Nagata’s principle of idealization [Reference Nagata19]. That is, the underlying additive group of $R \ltimes P$ is $R \oplus P$ with multiplication given by $(r_1,p_1)\cdot (r_2,p_2) = (r_1r_2, r_1p_2 + r_2p_1)$ . The R-algebra structure is induced by the ring homomorphism

$$ \begin{align*} R &\to R \ltimes P.\\ r & \mapsto (r,0) \end{align*} $$

and coincides with the R-module structure on $R \oplus P$ . Note that $R \ltimes P$ is a first order nilpotent thickening of R because the ideal $I = \{(0,p) \colon p \in P\}$ satisfies $I^2 = 0$ and we have $(R \ltimes P)/I \cong R$ . The canonical injection $P \hookrightarrow R \ltimes P$ given by mapping $p \to (0,p)$ is a split R-linear map (the usual projection $R \ltimes P \twoheadrightarrow P$ is a left-inverse), and hence, it is R-pure. Therefore, in the commutative diagram

the vertical maps are injective. Since the bottom horizontal map is injective by assumption so is the top horizontal map.

Proof. Since purity is preserved under restriction of scalars, the $R/I$ -purity of $\varphi \otimes _R \operatorname {\mathrm {id}}_{R/I}$ implies the R-purity of $\varphi \otimes _R \operatorname {\mathrm {id}}_{R/I}$ . The result now follows by [Reference Raynaud and Gruson23, Part I, Lem. 4.2.1].

Proof. First suppose the canonical map is surjective. Let $j: M \to \widehat M$ be the completion map. Then there is some $h\in \operatorname {\mathrm {Hom}}_R(N, \widehat {M})$ such that $h \circ \varphi = j$ . But since j is faithfully flat, it is pure, and hence $\varphi $ is pure.

Conversely, suppose $\varphi $ is pure. Then $\varphi \otimes 1_{M^\vee } :M \otimes _R M^\vee \to N \otimes _R M^\vee $ is injective, where $(-)^\vee = \operatorname {\mathrm {Hom}}_R(-, E_R(R/{\mathfrak {m}}))$ is the Matlis duality functor. Applying said functor to the resulting map, we get a surjective map $(N \otimes _R M^\vee )^\vee {\xrightarrow {\ \ }\hspace {-0.8em}\rightarrow }(M \otimes _R M^\vee )^\vee $ . But then by Hom-tensor adjointness, along with the fact that the double Matlis dual of a finite module is its completion, we get $\operatorname {\mathrm {Hom}}_R(N,\widehat M) \cong \operatorname {\mathrm {Hom}}_R(N, M^{\vee \vee }) \cong (N \otimes _R M^\vee )^\vee {\xrightarrow {\ \ }\hspace {-0.8em}\rightarrow }(M \otimes _R M^\vee )^\vee \cong \operatorname {\mathrm {Hom}}_R(M, M^{\vee \vee }) = \operatorname {\mathrm {Hom}}_R(M, \widehat {M})$ . Moreover, since everything is natural, the map in question is the canonical map given in (b).

For the final statement, in this case we have $M = \widehat {M}$ , so if $\varphi $ is pure, then by what we have proved, the canonical map $\operatorname {\mathrm {Hom}}_R(N,M) \to \operatorname {\mathrm {Hom}}_R(M,M)$ is surjective. In particular, there exists $h \in \operatorname {\mathrm {Hom}}_R(N,M)$ with $1_M = h \circ \varphi $ . The converse is trivial.

Proof. (1) follows by [Reference Authors27, Tag 0AGW].

(2) Consider the short exact sequence

$$\begin{align*}0 \to N \xrightarrow{\varphi} M \to \operatorname{\mathrm{coker}}(\varphi) \to 0. \end{align*}$$

Since $\varphi $ is R-pure and M is R-flat, both N and $\operatorname {\mathrm {coker}}(\varphi )$ are flat R-modules. This follows easily by the long exact sequence on $\operatorname {\mathrm {Tor}}$ and purity, but see for instance, [Reference Authors27, Tag 058P] for a more elementary diagram-chase proof. We then get a short exact sequence of flat $\widehat {R}^I$ -modules

$$\begin{align*}0 \to \widehat{N}^I \xrightarrow{\widehat{\varphi}^I} \widehat{M}^I \to \widehat{\operatorname{\mathrm{coker}}{\varphi}}^I \to 0 \end{align*}$$

by (1) and [Reference Authors27, Tag 0315 (3)]. Then $\widehat {\varphi }^I$ is a pure map of $\widehat {R}^I$ -modules by looking at the long-exact sequence on Tor because $\widehat {\operatorname {\mathrm {coker}}(\varphi )}^I$ is $\widehat {R}^I$ -flat, or alternatively, by using [Reference Authors27, Tag 058M].

Proof. Let $f \colon P \to Q$ be an injective R-linear map. We get a commutative diagram

The map $\operatorname {\mathrm {id}}_P \otimes _R \iota $ is injective by purity of $\iota $ and $f \otimes _R \operatorname {\mathrm {id}}_M$ is injective by flatness of M. Thus, by the commutativity of the above diagram, $f \otimes _R \operatorname {\mathrm {id}}_N$ is also injective, that is, N is R-flat.

Proof. Since the property of being a radical ideal is preserved under arbitrary localization, we may assume A is local and ${\mathfrak {p}}$ is the maximal ideal ${\mathfrak {m}}$ of A. We let $\widehat {A}$ denote the ${\mathfrak {m}}$ -adic completion of A. By hypothesis, the fibers of $A \to \widehat {A}$ are reduced. Let I be a radical ideal of A. Then $ A/I \to \widehat {A}/I\widehat {A} $ is also a flat map of Noetherian rings with reduced fibers. Since $A/I$ is reduced by assumption, $\widehat {A}/I\widehat {A}$ is reduced by [Reference Authors27, Tag 0C21]. Thus, $I\widehat {A}$ is a radical ideal of $\widehat A$ .

The second statement follows easily from the first and its proof is omitted.

Proof. $(1)$ is [Reference Authors27, Tag 059C].

For $(2)$ , if g factors through f via $\varphi $ , then it follows that for any R-module N,

$$\begin{align*}\mathrm{ker}(g \otimes_R \operatorname{\mathrm{id}}_N) = \mathrm{ker}((\varphi \otimes_R \operatorname{\mathrm{id}}_N) \circ (f \otimes_R \operatorname{\mathrm{id}}_N)) \supseteq \mathrm{ker}(f \otimes_R \operatorname{\mathrm{id}}_N). \end{align*}$$

Hence g dominates f. The converse follows by [Reference Authors27, Tag 059D].

$(3)$ follows by [Reference Authors27, Tag 0AUM].

$(4)$ Since $\varphi $ is pure, for all R-modules N, $\mathrm{ker} (\varphi \otimes _R \operatorname {\mathrm {id}}_N) = 0$ . Thus,

$$\begin{align*}\mathrm{ker}(g \otimes_R \operatorname{\mathrm{id}}_N) = \mathrm{ker}((\varphi \otimes_R \operatorname{\mathrm{id}}_N) \circ (f \otimes_R \operatorname{\mathrm{id}}_N)) = \mathrm{ker}(f \otimes_R \operatorname{\mathrm{id}}_N), \end{align*}$$

that is, g and f dominate each other.

$(5)$ Since h factors through g and g factors through f, for all R-modules N, we have

$$\begin{align*}\mathrm{ker}(h \otimes_R \operatorname{\mathrm{id}}_N) = \mathrm{ker}(f \otimes_R \operatorname{\mathrm{id}}_N) \subseteq \mathrm{ker}(g \otimes_R \operatorname{\mathrm{id}}_N) \subseteq \mathrm{ker}(h \otimes_R \operatorname{\mathrm{id}}_N) = \mathrm{ker}(f \otimes_R \operatorname{\mathrm{id}}_N). \end{align*}$$

Thus, g and h dominate each other and g and f dominate each other.

$(6)$ Note g dominates $h \circ f$ by (2). On the other hand, for all R-modules N

$$\begin{align*}\mathrm{ker}((h \circ f) \otimes_R \operatorname{\mathrm{id}}_N) = \mathrm{ker}((h \otimes_R \operatorname{\mathrm{id}}_N) \circ (f \otimes_R \operatorname{\mathrm{id}}_N)) \supseteq \mathrm{ker}(f \otimes_R \operatorname{\mathrm{id}}_N) = \mathrm{ker}(g \otimes_R \operatorname{\mathrm{id}}_N), \end{align*}$$

where the last equality follows because f and g dominate each other. This shows that $h \circ f$ dominates g.

$(7)$ is [Reference Lazard17, Lem. 1.1]. It is the key lemma that leads to a proof of Lazard’s Theorem on flat modules being precisely the ones that are filtered colimits of free modules of finite rank [Reference Lazard17, Thm. 1.2].

$(8)$ Since tensor product is right exact, for every R-module N, the induced map $ \varphi \otimes _R \operatorname {\mathrm {id}}_N \colon P' \otimes _R N \to P \otimes _R N $ is also surjective. Now note that $\mathrm{ker} \left ((g \circ \varphi ) \otimes _R \operatorname {\mathrm {id}}_N\right )$ (resp. $\mathrm{ker} \left ((f \circ \varphi ) \otimes _R \operatorname {\mathrm {id}}_N\right )$ ) is the pre-image of $\mathrm{ker} (g \otimes _R \operatorname {\mathrm {id}}_N)$ (resp. $\mathrm{ker} (f \otimes _R \operatorname {\mathrm {id}}_N)$ ) along the surjective map $\varphi \otimes _R \operatorname {\mathrm {id}}_N$ . Thus, we must have

$$\begin{align*}\mathrm{ker}(g \otimes_R \operatorname{\mathrm{id}}_N) \supseteq \mathrm{ker}(f \otimes_R \operatorname{\mathrm{id}}_N) \Longleftrightarrow \mathrm{ker}\left((g \circ \varphi) \otimes_R \operatorname{\mathrm{id}}_N\right) \supseteq \mathrm{ker}\left((f \circ \varphi) \otimes_R \operatorname{\mathrm{id}}_N\right). \end{align*}$$

Hence, g dominates f if and only if $g \circ \varphi $ dominates $f \circ \varphi $ .

Proof. By Lemma 3.1.2 (3) the map $f'$ in the pushout square

is R-pure. Hence for all ${\mathfrak {p}} \in \operatorname {\mathrm {Spec}}(R)$ , $f^{\prime }_{\mathfrak {p}}$ is a pure map of $R_{\mathfrak {p}}$ -modules. Thus, if $g_{\mathfrak {p}}$ is $R_{\mathfrak {p}}$ -pure, then $ g^{\prime }_{\mathfrak {p}} \circ f_{\mathfrak {p}} = f^{\prime }_{\mathfrak {p}} \circ g_{\mathfrak {p}} $ is $R_{\mathfrak {p}}$ -pure as well, and consequently, $f_{\mathfrak {p}}$ is a pure map of $R_{\mathfrak {p}}$ -modules.

Proof. $(1)$ If f stabilizes g then by definition Q is a finitely presented R-module. Then g dominates f and $\operatorname {\mathrm {coker}}(f)$ is finitely presented because this cokernel is the quotient of a finitely presented module by a finitely generated submodule; see [Reference Authors27, Tag 0519 (4)]. Thus, g factors through f by Lemma 3.1.2 $(2)$ .

$(2)$ Let $f \colon P \to Q$ be a stabilizer of g. Then g factors through f by $(1)$ . Hence choose a map $\varphi \colon Q \to M$ such that $ g = \varphi \circ f. $ Since P and Q are finitely presented and M is the filtered colimit of finitely presented modules $L_i$ , there exists an index $i \in I$ such that $\varphi \colon Q \to M$ admits a lift to a map $\varphi _i \colon Q \to L_i$ along $u_i$ , that is,

For all $j \geq i$ , define

Then

$$\begin{align*}u_j \circ f_j = ((u_j \circ u_{ij}) \circ \varphi_i ) \circ f = (u_i \circ \varphi_i) \circ f = \varphi \circ f = g \end{align*}$$

by the choice of the index i. Moreover, since $u_{ii} = \operatorname {\mathrm {id}}_{L_i}$ , it follows that $f_i = \varphi _i \circ f$ , and so, $ f_j = u_{ij} \circ f_i $ by definition of $f_j$ for all $j \geq i$ . It remains to show that $f_j$ stabilizes g.

Since g factors through $f_j$ , it follows that g dominates $f_j$ by Lemma 3.1.2 $(2)$ . On the other hand, for any R-module N

$$\begin{align*}\mathrm{ker}(f_j \otimes_R \operatorname{\mathrm{id}}_N) = \mathrm{ker}((u_{ij} \circ \varphi_i \circ f) \otimes_R \operatorname{\mathrm{id}}_N) \supseteq \mathrm{ker}(f \otimes_R \operatorname{\mathrm{id}}_N) = \mathrm{ker}(g \otimes_R \operatorname{\mathrm{id}}_N), \end{align*}$$

where the last equality follows because f stabilizes g. This shows that $f_j$ dominates g, and so, g and $f_j$ dominate each other, that is, $f_j$ stabilizes g, as desired.

Now suppose the stabilizer f also factors through g, that is, there exists $\phi \colon M \to Q$ such that $ f = \phi \circ g. $ For the index $i \in I$ and the maps $f_j \colon P \to L_j$ for $j \geq i$ chosen as above, if we define

$$\begin{align*}\phi_j {:=}q M \xrightarrow{\phi} Q \xrightarrow{\varphi_i} L_i \xrightarrow{u_{ij}} L_j, \end{align*}$$

then $ \phi _j \circ g = (u_{ij} \circ \varphi _i \circ \phi ) \circ g = u_{ij} \circ \varphi _i \circ f = f_j, $ where the last equality follows by the definition of $f_j$ . Thus $f_j$ stabilizes g and factors through g for all $j \geq i$ .

$(3)$ For brevity, we will write the map $\operatorname {\mathrm {Hom}}_R(L_j, N) \xrightarrow {- \circ f_j} \operatorname {\mathrm {Hom}}_R(P, N)$ as $\operatorname {\mathrm {Hom}}_R(f_j, N)$ . In this new notation, we have to show that for all $j \geq i$ ,

$$\begin{align*}\operatorname{\mathrm{im}}\left(\operatorname{\mathrm{Hom}}_R(f_j, N)\right) = \operatorname{\mathrm{im}}\left(\operatorname{\mathrm{Hom}}_R(f_i, N)\right). \end{align*}$$

Since $f_j = u_{ij} \circ f_i$ for all $j \geq i$ , it follows that

$$\begin{align*}\operatorname{\mathrm{im}}\left(\operatorname{\mathrm{Hom}}_R(f_j,N)\right) = \operatorname{\mathrm{im}}\left(\operatorname{\mathrm{Hom}}_R(f_i,N) \circ \operatorname{\mathrm{Hom}}_R(u_{ij},N)\right) \subseteq \operatorname{\mathrm{im}}\left(\operatorname{\mathrm{Hom}}_R(f_i,N)\right). \end{align*}$$

For all $j \geq i$ , we have a commutative diagram

Since $f_i$ and g dominate each other, by Lemma 3.1.5 (5), for all $j \geq i$ , $f_i$ and $f_j$ also dominate each other. Since $\operatorname {\mathrm {coker}}(f_j)$ is finitely presented, it follows that $f_i$ factors through $f_j$ , that is, there exists $v_{ji} \colon L_j \to L_i$ such that

Then $ \operatorname {\mathrm {im}}\left (\operatorname {\mathrm {Hom}}_R(f_i,N)\right ) = \operatorname {\mathrm {im}}\left (\operatorname {\mathrm {Hom}}_R(f_j,N) \circ \operatorname {\mathrm {Hom}}_R(v_{ji},N)\right ) \subseteq \operatorname {\mathrm {im}}\left (\operatorname {\mathrm {Hom}}_R(f_j,N)\right ). $

Proof. If L satisfies $(1)-(3)$ then the map f stabilizes g by Lemma 3.1.2 (4). So now we show the existence of such an L.

By Lazard’s Theorem [Reference Lazard17, Thm. 1.2], M is a colimit of a system $\{(L_{i}, u_{ij}) \colon i, j \in I, i\leq j\}$ of free R-modules of finite rank indexed by a filtered poset $(I,\leq )$ . Here $u_{ij}$ denotes the transition map $L_i \to L_j$ for $i \leq j$ . We also let $u_i \colon L_i \to M$ denote the associated maps for all i. Since g admits a stabilizer, by Lemma 3.1.5 parts (2) and (3), there exists an index $i \in I$ such that for all $j \geq i$ , we have a map $ f_j \colon P \to L_j $ such that $g = u_j \circ f_j, f_j = u_{ij} \circ f_i$ , $f_j$ stabilizes g and $ \operatorname {\mathrm {im}}(\operatorname {\mathrm {Hom}}_R(f_j,R)) = \operatorname {\mathrm {im}}(\operatorname {\mathrm {Hom}}_R(f_i,R)). $ Let us call this stable image as S. Note that S is a finitely generated R-module, since it is a quotient of the free module $\operatorname {\mathrm {Hom}}_R(L_i,R)$ . Let n be the minimal number of generators of S, that is, $ n = \dim _k(S/\mathfrak {m} S). $ We then have a surjective R-linear map $ p \colon R^{\oplus n} \twoheadrightarrow S. $ By the choice of n, the induced map $ p \otimes _R \operatorname {\mathrm {id}}_\kappa \colon \kappa ^{\oplus n} \to S/\mathfrak {m} S $ is an isomorphism. Consider the diagram

The vertical map is the one obtained by restricting the codomain of $\operatorname {\mathrm {Hom}}_R(f_i,R)$ to $S = \operatorname {\mathrm {im}}(\operatorname {\mathrm {Hom}}_R(f_i,R))$ . Since $\operatorname {\mathrm {Hom}}_R(L_i,R)$ is a free, hence projective R-module, there exists an R-linear map $q \colon \operatorname {\mathrm {Hom}}_R(L_i,R) \to R^{\oplus n}$ such that

(3.1.6.1) $$ \begin{align} \operatorname{\mathrm{Hom}}_R(L_i,R) \twoheadrightarrow S = p \circ q. \end{align} $$

After tensoring by $\kappa $ and using that $p \otimes _R \operatorname {\mathrm {id}}_\kappa $ is an isomorphism, it follows that $q \otimes _R \operatorname {\mathrm {id}}_\kappa $ is also surjective. Then by Nakayama’s lemma, we see that q is surjective. Since the composition

$$\begin{align*}\operatorname{\mathrm{Hom}}_R(L_j,R) \xrightarrow{\operatorname{\mathrm{Hom}}_R(u_{ij}, R)} \operatorname{\mathrm{Hom}}_R(L_i,R) \twoheadrightarrow S \end{align*}$$

is surjective for all $j \geq i$ (S is the stable image $\operatorname {\mathrm {im}}(\operatorname {\mathrm {Hom}}_R(f_j,R))$ for all $j \geq i$ ), it follows by the exact same reasoning as above that for all $j \geq i$ ,

(3.1.6.2) $$ \begin{align} q \circ \operatorname{\mathrm{Hom}}_R(u_{ij}, R) \colon \operatorname{\mathrm{Hom}}_R(L_j,R) \to R^{\oplus n} \end{align} $$

is also surjective. Applying $\operatorname {\mathrm {Hom}}_R(\hspace {2mm}, R)$ to the maps and $ \operatorname {\mathrm {Hom}}_R(L_i,R) \xrightarrow {q} R^{\oplus n}, $ we get maps

and

Here $P \to \operatorname {\mathrm {Hom}}_R(\operatorname {\mathrm {Hom}}_R(P,R),R)$ is the canonical map to the double dual, and we can make the identification $\operatorname {\mathrm {Hom}}_R(\operatorname {\mathrm {Hom}}_R(L_i,R),R) \cong L_i$ since $L_i$ is free of finite rank. By (3.1.6.1), $ t\circ s = f_i. $ Then $u_i \circ (t \circ s) = u_i \circ f_i = g$ , and so,

(3.1.6.3) $$ \begin{align} \operatorname{\mathrm{im}}(g) \subseteq \operatorname{\mathrm{im}}(u_i \circ t). \end{align} $$

We also have $ u_i \circ t = \operatorname {\mathrm {colim}}_{j \geq i} u_{ij} \circ t, $ where each $u_{ij} \circ t$ admits a left-inverse because it can be identified with the map obtained by applying $\operatorname {\mathrm {Hom}}_R(\bullet , R)$ to the surjective map of free modules (3.1.6.2) $ q \circ \operatorname {\mathrm {Hom}}_R(u_{ij},R) $ which always has a right-inverse. Since a filtered colimit of split maps is pure, it follows that $u_i \circ t \colon \operatorname {\mathrm {Hom}}_R(R^{\oplus n},R) \to M$ is a pure map. Taking $L {:=}q \operatorname {\mathrm {im}}(u_i \circ t) \cong R^{\oplus n}$ , (3.1.6.3) shows that L satisfies properties $(1)-(3)$ .

Proof. (1) follows from [Reference Raynaud and Gruson23, Part II, Prop. 2.1.4 $(i)\iff (iii)$ ] (alternate reference [Reference Authors27, Tag 059E]). Note that the implication $\implies $ in (1) follows from Lemma 3.1.5. The assertion in (2) follows from [Reference Raynaud and Gruson23, Part II, Prop. 2.3.2 $(i)\iff (iii)$ ].

Proof. (1) A filtered colimit of free modules is flat. Thus, the “if” implication follows by Remark 3.2.4 (e) because free modules of finite rank are finitely presented, and hence, ML. Now suppose M is a flat ML R-module. We know that M is a filtered union of its finitely generated submodules. Thus, it suffices to show that if N is a finitely generated submodule of M, then there exists a submodule L of M such that $N \subseteq L$ , L is free of finite rank and $L \hookrightarrow M$ is pure. But this follows upon choosing a map $g \colon F \to M$ from a free module F of finite rank that surjects onto N and observing that since g admits a stabilizer (M is ML), it must admit one $f \colon F \to L$ where L has the desired properties by Proposition 3.1.6.

(2) Since a SML module is ML, it follows by (a) that a flat and SML R-module is a filtered union of finitely generated free submodules that are pure in M. But such a submodule must also be direct summand of M by Remark 3.2.4 (g). This proves the forward implication. Conversely, suppose M is a filtered union of finitely generated free submodules that are direct summands of M. Then M is flat since it is a filtered colimit of free modules. Moreover, free modules are SML by Remark 3.2.4 (h).

Thus, it suffices to show that if M is a filtered union of SML submodules that are direct summands of M, then M is also SML. Let P be a finitely presented R-module and $g \colon P \to M$ be R-linear. Then there exists a SML direct summand $M'$ of M such that $\operatorname {\mathrm {im}}(g) \subseteq M'$ . Let $u \colon M \to M'$ be a left-inverse of the inclusion $M' \hookrightarrow M$ . Let $g' \colon P \to M'$ be the composition $P \xrightarrow {g} \operatorname {\mathrm {im}}(g) \hookrightarrow M'$ . Since $M'$ is SML, there exists a finitely presented R-module Q and a map $f \colon P \to Q$ such that $g'$ factors through f and f factors through $g'$ . Since g factors through $g'$ , it follows that g factors through f. Furthermore, since $g'$ factors through g via u, it follows that f also factors through g. Thus, M is SML.

Proof. Since we are working over a Noetherian ring, finitely generated is the same as being finitely presented for modules.

Given Remark 3.2.4 (a), for the equivalence of $(1)$ and $(2)$ it suffices to show that if M is ML then M is SML. Let P be a finitely generated R-module and let $g \colon P \to M$ be an R-linear map. Since M is ML, there exists a finitely generated R-module Q and an R-linear map $f \colon P \to Q$ such that g and f dominate one another. Thus, if one forms the pushout square

then both $f'$ and $g'$ are pure maps of R-modules by Lemma 3.1.2 (3). Since $\operatorname {\mathrm {coker}}(f)$ is finitely presented, it follows by Lemma 3.1.2 $(2)$ that g factors through f. Thus, it remains to show by Definition 3.2.3 that f factors through g.

The map $g' \colon Q \to N$ is a pure map of modules over a Noetherian complete local ring such that Q is a finitely generated R-module. Thus, by Lemma 2.2.3, $g'$ admits an R-linear left-inverse $\varphi \colon N \to Q,$ that is, $\varphi \circ g' = \operatorname {\mathrm {id}}_Q$ . Then by the commutativity of the above diagram, $ (\varphi \circ f') \circ g = \varphi \circ (f' \circ g) = \varphi \circ (g' \circ f) = (\varphi \circ g') \circ f = f. $ That is, f factors through g as desired.

Proof. This is [Reference Authors27, Tag 0AS5].

Proof. The proof of [Reference Authors27, Tag 0AUM] readily adapts to yield the assertion. We omit the details.

Proof. Proof is similar to the proof of Lemma 3.1.2 (4) and is omitted.

Proof. One can readily adapt the proof of the analogous result for mutual domination instead of mutual cyclic domination given in Lemma 3.1.2 (5).

Proof. Suppose $f \colon P \to Q$ is a cyclic stabilizer of g that $vg$ dominates. Since Q is a finitely presented R-module, g factors through f (Lemma 3.1.2 (2)), that is, there exists an R-linear map $ \varphi \colon Q \to M $ such that the following diagram commutes

Since Q is a finitely presented R-module and $M = \operatorname {\mathrm {colim}}_i L_i$ is a filtered colimit of finitely presented modules $L_i$ , there exists $i \in I$ such that $\varphi $ admits a lift $\varphi _i \colon Q \to L_i$ along $u_i \colon L_i \to M$ , that is, the following diagram commutes

Now for all $j \geq i$ , define $ g_j {:=}q (P \xrightarrow {f} Q \xrightarrow {\varphi _i} L_i \xrightarrow {u_{ij}} L_j). $ Then by the commutativity of the above diagrams, $ u_j \circ g_j = ((u_j \circ u_{ij}) \circ \varphi _i) \circ f = (u_i \circ \varphi _i) \circ f = \varphi \circ f = g. $ Thus, for all $j \geq i$ , g dominates $g_j$ (Lemma 3.1.2 (2)). Moreover, for all $j \geq i$ , we have a commutative diagram

Since f and g cyclically dominate each other, by Lemma 3.3.5 we have that $g_j$ and g cyclically dominate each other for all $j \geq i$ .

Proof. By Lazard’s theorem we can express M as a filtered colimit of free modules of finite rank. The Corollary now follows by Proposition 3.3.7.

Proof. Let $I {:=}q \sum _{x \in N} c_M(x)$ . Since M is Ohm-Rush, for all $x \in N$ , $x \in c_M(x)M \subseteq IM$ . Thus, $N \subseteq IM$ , and so, $c_M(N) \subseteq I = \sum _{x \in N} c_M(x)$ . For the other inclusion, let J be an ideal of R such that $N \subseteq JM$ . Then for all $x \in N$ , $x \in JM$ , which implies $c_M(x) \subseteq J$ . Thus $\sum _{x \in N} c_M(x) \subseteq J$ , for all ideals J such that $N \subseteq JM$ , that is, $\sum _{x \in N} c_M(x) \subseteq c_M(N)$ .

The fact that $N \subseteq c_M(N)M$ follows because for all $x \in N$ , $x \in c_M(x)M \subseteq c_M(N)M$ . Then by definition of $c_M(N)$ , the latter is the smallest ideal I of R such that $N \subseteq IM$ .

Proof. (2) follows from (1) because $\sum _{\alpha \in A} N_\alpha $ is the submodule of M generated by $\bigcup _{\alpha \in A} N_\alpha $ . For (1), by a double application of Lemma 3.4.4, we get

$$\begin{align*}c_M(\bigcup_{\alpha \in A} N_\alpha) = \sum_{x \in \bigcup_{\alpha \in A}N_\alpha} c_M(x) = \sum_{\alpha \in A} \sum_{x \in N_\alpha} c_M(x) = \sum_{\alpha \in A} c_M(N_\alpha).\\[-47pt] \end{align*}$$

Proof. Let $\{I_\alpha \colon \alpha \in A\}$ be a collection of ideals of R. Then

$$\begin{align*}\bigcap_{\alpha \in A} I_\alpha M = \bigcap_{\alpha \in A} (I_\alpha S)M = \left(\bigcap_{\alpha \in A} I_\alpha S\right)M = \left(\left(\bigcap_{\alpha \in A} I_\alpha \right)S\right)M = \left(\bigcap_{\alpha \in A} I_\alpha \right)M. \end{align*}$$

The first and fourth equalities follow because M is an S-module, the second equality follows because M is an Ohm-Rush S-module and the third equality follows because S is an Ohm-Rush R-module. Thus, M is an Ohm-Rush R-module by Lemma 3.4.8.

Proof. Suppose $r=0$ . Then $rx=0$ , so $c_M(rx) = 0 = rc_M(x)$ .

Suppose r is nonzerodivisor on R. Let I be an ideal with $x\in IM$ . Then $rx \in rIM$ , so $c_M(rx) \subseteq rI$ . Hence, $c_M(rx)$ is contained in the intersection of all such ideals $rI$ , which, since r is a nonzerodivisor, coincides with r times the intersection of the ideals I. That is, $c(rx) \subseteq rc(x)$ .

If $x \in c_M(x)M$ , then $rx \in rc_M(x)M$ , whence $c_M(rx) \subseteq rc_M(x)$ .

On the other hand, suppose M is flat. Let I be an ideal with $rx \in IM$ . Then $x \in (IM :_M r) = (I:_R r)M$ (since M is flat), so that $c_M(x) \subseteq (I : r),$ whence $rc_M(x) \subseteq I$ . Thus, $rc_M(x)$ is contained in the intersection of all such ideals I, which is to say $rc_M(x) \subseteq c_M(rx)$ .

Proof. For all $x \in M$ and for all ideals I of R

$$\begin{align*}f(x) \in IP \Longleftrightarrow x + I \in \mathrm{ker}(f \otimes_R \operatorname{\mathrm{id}}_{R/I}) \subseteq \mathrm{ker}(g \otimes_R \operatorname{\mathrm{id}}_{R/I}) \implies g(x) \in IQ. \end{align*}$$

This immediately implies that $c_Q(g(x)) \subseteq c_P(f(x))$ .

Proof. For all $x \in N$ and ideals I of R, the injectivity of the induced map $\varphi \otimes \operatorname {\mathrm {id}}_{R/I} \colon N/IN \to M/IM$ shows that $x \in IN$ if and only if $\varphi (x)$ is in $IM$ . Thus, $c_N(x) = c_M(\varphi (x)) = c_M \circ \varphi (x)$ , that is, $c_N = c_M \circ \varphi $ .

Proof. (2) follows from (1). Indeed, if v cyclically dominates u, then since $u(1) \in c_F(u(1))F$ (free modules are Ohm-Rush), by Lemma 3.4.11 (1) we have $ v(1) \in c_F(u(1))M. $ Then v factors through u by (1).

We first prove the backward implication of (1). Let $\{e_i \colon i \in I\}$ be a free basis of F and let $u(1) = r_1e_{i_1} + \dots + r_ne_{i_n},$ where $r_1,\dots ,r_n \in R \setminus \{0\}$ . By Example 3.4.7, $c_F(u(1)) = (r_1,\dots ,r_n).$ Since $v(1) \in c_F(u(1))M$ , there exist $m_1,\dots ,m_n \in M$ such that $ v(1) = r_1m_1 + \dots + r_nm_n. $ Now consider any R-linear map

$$ \begin{align*} \varphi \colon F &\to M\\e_{i_j} &\mapsto m_j, \end{align*} $$

for $j = 1, \dots , n$ . By construction, $ \varphi \circ u (1) = \varphi \left (\sum _{j=1}^nr_je_{i_j}\right ) = \sum _{j=1}^n r_jm_j = v(1). $ Thus, $ \varphi \circ u = v. $ Conversely, suppose there exists $\varphi \colon F \to M$ such that $v = \varphi \circ u$ . Since F is Ohm-Rush, $u(1) \in c_F(u(1))F$ . Thus, $ v(1) = \varphi (u(1)) \in c_F(u(1))M $ by linearity of $\varphi $ . This completes the proof of (1).

Proof. Let I be an ideal of R such that $x \in IN$ . Then by linearity, $\varphi (x) \in IM$ . Thus, $c_M(\varphi (x)) \subset I$ for all such ideals I, and so, $c_M(\varphi (x)) \subseteq c_N(x)$ .

We now assume M is Ohm-Rush, that is, for all $y \in M$ , $y \in c_M(y)M$ . Then it is clear that $y \neq 0$ if and only if $c_M(y) \neq (0)$ . Part (1) readily follows by this observation. Part (2) essentially follows by the argument in [Reference Ohm and Rush20, Thm. 1.3] and we omit the details. Note that $\varphi $ is cyclically pure $\implies c_M \circ \varphi = c_N$ by Lemma 3.4.12. Part (3) is part of [Reference Ohm and Rush20, Cor. 1.6]. Note that the forward implication of $(3)$ is shown in Lemma 3.4.10.

The first part of (4) follows by [Reference Ohm and Rush20, Thm. 3.1] and (3). If N is a submodule of M, then

$$ \begin{align*} c_M(N)(S^{-1}R) &= \left(\sum_{x \in N}c_M(x)\right)S^{-1}R = \sum_{x \in N} c_M(x)(S^{-1}R)\\ &= \sum_{s \in S}\sum_{x \in N}c_{S^{-1}M}(x/s) = c_{S^{-1}M}(S^{-1}N). \end{align*} $$

The first equality follows by Lemma 3.4.4 because M is Ohm-Rush, the second equality follows because expansions of ideals commute with arbitrary sums of ideals, the third equality follows by the first assertion in this statement, and the fourth equality follows because $S^{-1}M$ is an Ohm-Rush $S^{-1}R$ -module by the equality $c_M(x)(S^{-1}R) = c_{S^{-1}M}(x/s)$ .

Proof. By Lemma 3.4.12, we have $c_{M} \circ \tilde {f} = c_{N \otimes _R S}$ since $\tilde {f}$ is S-cyclically pure. Thus, $c_M(y) = c_M(\tilde {f}(x \otimes 1)) = c_{N \otimes _R S}(x \otimes 1)$ , and we reduce to the case where $M = N \otimes _R S$ , $f \colon N \to N \otimes _R S$ is the canonical map that sends $z \mapsto z \otimes 1$ and $y = x \otimes 1$ .

Since $x \in c_N(x)N$ , by linearity of f we get $x \otimes 1 = f(x) \in (c_N(x)S)(N\otimes _R S)$ . Thus, $c_{N \otimes _R S}(x \otimes 1) \subseteq c_N(x)S$ . To finish the proof it remains to show $c_N(x)S \subseteq c_{N \otimes _R S}(x \otimes 1)$ . Let J be an ideal of S such that $x \otimes 1 \in J(N \otimes _R S)$ . Let $J^{\mathfrak c}$ denote the contraction of J to R. Since the induced map $R/J^{\mathfrak c} \to S/J$ is injective, the flatness of N implies that $N/J^{\mathfrak c}N \to (S/J) \otimes _R N$ is injective as well. This induced map sends $x + J^{\mathfrak c}N \mapsto 0$ since $x \otimes 1 \in J(S \otimes _R N)$ . Thus, we have shown that if J is an ideal of R such that $x \otimes 1 \in J(N \otimes _R S)$ , then $x \in J^{\mathfrak c}N$ . In other words, $c_N(x) \subseteq J^{\mathfrak c}$ for all such ideals J. Since taking ideal contraction commutes with arbitrary intersections of ideals, and since the intersection of all the ideals J of S such that $x \otimes 1 \in J(N \otimes _R S)$ is precisely $c_{N \otimes _R S}(x \otimes 1)$ , we get $c_N(x) \subseteq (c_{N \otimes _R S}(x \otimes 1))^{\mathfrak c}$ , and so, $c_{N}(x)S \subseteq c_{N \otimes _R S}(x \otimes 1)$ .

Proof. The implications $(2) \implies (3)$ and $(4) \implies (5)$ are clear. For $(5) \implies (1)$ , suppose $u \colon R \to P$ is a cyclic stabilizer of v such that $ u(1) \in c_P(u(1))P. $ Since u and v cyclically dominate each other, by Lemma 3.4.11 (2) we have $ c_P(u(1)) = c_M(v(1)). $ Then $v(1) \in c_P(u(1))M = c_M(v(1))M$ by Lemma 3.4.11 (1).

Now assume that M is a flat R-module.

$(1) \implies (2):$ Since $v(1) \in c_M(v(1))M$ , $c_M(v(1))$ is a finitely generated ideal of R (Remark 3.4.3). Suppose $ c_M(v(1)) = (r_1,\dots ,r_n). $ Now consider the map $u \colon R \to R^{\oplus n}$ such that $u(1) = (r_1,\dots ,r_n)$ . We claim that u is a cyclic stabilizer of v. Then the fact that v dominates u will follow from Lemma 3.4.13 (2). For an ideal I of R, we have

(3.4.17.1) $$ \begin{align} \mathrm{ker}(u \otimes_R \operatorname{\mathrm{id}}_{R/I}) = (I \colon (r_1,\dots,r_n))/I = \left(I \colon c_M(v(1))\right)/I. \end{align} $$

Now suppose $r \in R$ such that $r + I \in \mathrm{ker} (v \otimes _R \operatorname {\mathrm {id}}_{R/I})$ . This is equivalent to saying that $rv(1) \in IM$ , and so, $c_M(rv(1)) \subseteq I$ . Since M is flat, Lemma 3.4.10 shows that $rc_M(v(1)) \subseteq c_M(rv(1)) \subseteq I$ , and so, $r \in (I \colon c_M(v(1)))$ . Thus, $ \mathrm{ker} (v \otimes _R \operatorname {\mathrm {id}}_{R/I}) \subseteq (I \colon c_M(v(1)))/I. $

Conversely, suppose $r \in (I \colon c_M(v(1)))$ . Then $rc_M(v(1)) \subseteq I$ , and so, $rc_M(v(1))M \subseteq IM$ . Since $v(1) \in c_M(v(1))M$ , we then get $ v(r) = rv(1) \in rc_M(v(1))M \subseteq IM. $ Hence $r + I \in \mathrm{ker} (v \otimes _R \operatorname {\mathrm {id}}_{R/I})$ , which establishes the other inclusion $ (I \colon c_M(v(1)))/I \subseteq \mathrm{ker} (v \otimes _R \operatorname {\mathrm {id}}_{R/I}). $ Then for all ideals I of R

Hence v and u cyclically dominate each other.

Finally, if M is flat then the implication $(3) \implies (4)$ follows by Corollary 3.3.8 because free modules are Ohm-Rush by Example 3.4.7.

Proof. This follows from the implications (4) $\implies $ (5) $\implies $ (1) of Theorem 3.4.18.

Proof. Let $v \colon R \to M$ be a linear map. Since M is ML, v admits a stabilizer $u \colon R \to P$ . That is, P is finitely presented and v and u dominate each other, and so, also cyclically dominate each other. Thus v admits a cyclic stabilizer that v dominates. This shows M is Ohm-Rush by Theorem 3.4.18 (3) $\implies $ (1).

Proof. In our setup, S is a torsion-free R-module. Thus, S is flat by Proposition 3.4.21.

Proof. (1) By definition, $ u_{jk}(x_j) = u_{jk}(u_{i_0j}(x_{i_0})) = u_{i_0k}(x_{i_0}) = x_k. $ Thus, (1) follows by applying Proposition 3.4.14 to the map $u_{i_0j}$ (resp. $u_{jk}$ ) and the element $x_{i_0} \in L_{i_0}$ (resp. $x_j \in L_j$ ).

(2) Since $u_{i_0}$ maps $x_{i_0} \mapsto m$ , we see that for all $j \geq i_0$ , $ u_j(x_j) = u_j(u_{i_0j}(x_{i_0})) = u_{i_0}(x_{i_0}) = m. $ Thus, by Proposition 3.4.14 applied to the maps

$$ \begin{align*} u_{j} \colon L_{j} &\to M\\ x_{j} &\mapsto m, \end{align*} $$

we get

(3.4.25.1) $$ \begin{align} c_M(m) \subseteq \bigcap_{j \geq i_0} c_{L_j}(x_j). \end{align} $$

Now suppose that I is an ideal of R such that $m \in IM$ . Choose $a_1,\dots ,a_n \in I$ and $m_1,\dots ,m_n \in M$ such that $ m = a_1m_1 + \dots + a_nm_n. $ Consider the maps

$$ \begin{align*} f \colon R &\to M\\ 1 &\mapsto m, \end{align*} $$

and

$$ \begin{align*} g \colon R &\longrightarrow R^{\oplus n}\\ 1 &\mapsto (a_1,\dots,a_n). \end{align*} $$

Then f factors through g via the map $ h \colon R^{\oplus n} \to M $ that sends the standard basis element $e_\ell \mapsto m_\ell $ , for $1 \leq \ell \leq n$ . One can choose an index $i \in I$ such that h lifts to a map $ h_i \colon R^{\oplus n} \to L_i $ along $u_i \colon L_i \to M$ , that is, $ u_i \circ h_i = h. $ Then $ u_i(h_i(a_1,\dots ,a_n)) = h(a_1,\dots ,a_n) = h(g(1)) = f(1) = m. $ Since $x_{i_0}$ and $h_i(a_1,\dots ,a_n)$ are both lifts of m, by construction of a filtered colimit, there exists $j \in I$ such that $j \geq i_0, i$ and $ u_{ij}(h_i(a_1,\dots ,a_n)) = u_{i_0j}(x_{i_0}) = x_j. $ Then by Proposition 3.4.14 applied to the map $u_{ij} \circ h_i \colon R^{\oplus n} \to L_j$ , we have $ c_{L_j}(x_j) \subseteq c_{R^{\oplus n}}(a_1,\dots ,a_n) = (a_1,\dots ,a_n) \subseteq I. $ Thus, we have shown that for any ideal I of R such that $m \in IM$ , there exists $j \geq i_0$ and $c_{L_j}(x_j) \subseteq I$ . Consequently, $ \bigcap _{j \geq i_0} c_{L_j}(x_j) \subseteq I $ for all ideals I of R such that $m \in IM$ , and so, $\bigcap _{j \geq i_0} c_{L_j}(x_j) \subseteq c_M(m)$ . This, along with (3.4.25.1), completes the proof of (2).

(3) Suppose $m \in c_M(m)M$ . In order to show that $\{c_{L_j}(x_j) \colon j \geq i_0\}$ stabilizes under inclusion, by (1) and (2) it suffices to show that there exists $j_0 \geq i_0$ such that $ c_{L_{j_0}}(x_{j_0}) \subseteq c_M(m). $ Let $c_M(m) = (a_1,\dots ,a_n)$ (Remark 3.4.3). Then by repeating the argument in (2) above, we see that there exists $j_0 \geq i_0$ such that $ c_{L_{j_0}}(x_{j_0}) \subseteq (a_1,\dots ,a_n) = c_M(m). $

Conversely, suppose $\{c_{L_j}(x_j) \colon j \geq i_0\}$ stabilizes under inclusion, say with stable ideal $c_{L_{j_0}}(x_{j_0})$ . Since I is a directed set, by (1) and (2) we have $ c_M(m) = \bigcap _{j\geq i_0} c_{L_j}(x_j) = \bigcap _{k \geq j_0} c_{L_k}(x_k) = c_{L_{j_0}}(x_{j_0}). $ Since $L_{j_0}$ is an Ohm-Rush R-module, it follows that $x_{j_0} \in c_{L_{j_0}}(x_{j_0})L_{j_0}$ . Using the homomorphism $ u_{j_0} \colon L_{j_0} \to M, $ we get $m = u_{j_0}(x_{j_0}) \in c_{L_{j_0}}(x_{j_0})M = c_M(m)M$ , as desired.

(4) We first show $(b) \implies (a)$ . If there exists $k \geq i_0$ such that $v_k$ stabilizes v, then since $L_k$ is Ohm-Rush, by Theorem 3.4.17 $(4) \implies (1)$ we have that $m = v(1) \in c_M(v(1))M = c_M(m)M$ .

We next prove $(a) \implies (b)$ . So assume that $m \in c_M(m)M$ . By (3) and (2), there exists $j_0 \geq i_0$ such that for all $k \geq j_0$ , $ c_{L_k}(x_k) = c_{L_{j_0}}(x_{j_0}) = c_M(m). $ By assumption, $L_{j_0}$ and $L_k$ are free R-modules, and hence Ohm-Rush. Thus, we have $ v_{j_0}(1) = x_{j_0} \in c_{L_{j_0}}(x_{j_0})L_{j_0} = c_{L_k}(x_k)L_{j_0} = c_{L_k}(v_k(1))L_{j_0}, $ and $ v_{k}(1) = x_k \in c_{L_k}(x_k)L_k = c_{L_{j_0}}(x_{j_0})L_{k} = c_{L_{j_0}}(v_{j_0}(1))L_k. $ By Lemma 3.4.13 (1) we then conclude that $v_{j_0} \colon R \to L_{j_0}$ factors through $v_k \colon R \to L_k$ and vice-versa (here we are using that $L_{j_0}$ and $L_k$ are both free and not just Ohm-Rush). Thus, for all $k \geq j_0$ , $v_{j_0}$ and $v_k$ dominate each other. Since $ v = \operatorname {\mathrm {colim}}_{k \geq j_0} v_j, $ it follows that v and $v_{j_0}$ dominate each other because for all R-modules N,

$$ \begin{align*} \mathrm{ker}(v \otimes_R \mathrm{id}_N) &= \mathrm{ker}((\mathrm{colim}_{k \geq j_0} v_k)\otimes_R \mathrm{id}_N) = \mathrm{ker}(\mathrm{colim}_{k \geq j_0}(v_k \otimes_R \mathrm{id}_N))\\ &= \mathrm{colim}_{k \geq j_0} \mathrm{ker}(v_k \otimes_R \mathrm{id}_N) = \mathrm{colim}_{k \geq j_0} \mathrm{ker}(v_{j_0} \otimes_R \mathrm{id}_N) = \mathrm{ker}(v_{j_0} \otimes_R \mathrm{id}_N). \end{align*} $$

Here the second equality follows because tensor products commute with filtered colimits, the third equality follows because filtered colimits are exact in $\operatorname {\mathrm {Mod}}_R$ [Reference Authors27, Tag 00DB] and the fourth equality follows because $v_{j_0}$ and $v_k$ dominate each other. In fact, since $\mathrm{ker} (v_{j_0} \otimes _R \operatorname {\mathrm {id}}_N) = \mathrm{ker} (v_k \otimes _R \operatorname {\mathrm {id}}_N)$ , the above chain of equalities also shows that $\mathrm{ker} (v \otimes _R \operatorname {\mathrm {id}}_N) = \mathrm{ker} (v_k \otimes _R \operatorname {\mathrm {id}}_N)$ for all $k \geq j_0$ . Since $L_k$ is a finitely presented R-module by hypothesis, we get that $v_k$ stabilizes v for all $k \geq j_0$ . This completes the proof of $(a) \implies (b)$ .

It remains to show that (b) $\implies $ (c). Suppose there exists $j_0 \geq i_0$ such that for all $k \geq j_0$ , $v_k$ stabilizes v. We have $ v_k(1) = x_k = u_{i_0k}(x_0) = u_{j_0k}(u_{i_0j_0}(x_0)) = u_{j_0k}(x_{j_0}) = u_{j_0k}(v_{j_0}(1)), $ and so, $v_k = u_{j_0k} \circ v_{j_0}$ . Furthermore, $ u_k \circ v_k(1) = u_k(x_k) = u_k(u_{i_0k}(x_{i_0})) = u_{i_0}(x_{i_0}) = m = v(1). $ Thus, for all $k \geq j_0$ , $u_k \circ v_k = v$ . Since $v_k$ stabilizes v, one can now apply Lemma 3.1.5 (3) (with $f_k = v_k$ , $g = v$ and $i = j_0$ in the notation of the Lemma) to conclude that for all $k \geq j_0$ and for all R-modules N, $\operatorname {\mathrm {im}}(\operatorname {\mathrm {Hom}}_R(v_{j_0},N)) = \operatorname {\mathrm {im}}(\operatorname {\mathrm {Hom}}_R(v_k,N))$ .

Proof. Note that any stabilizer of v is a cyclic stabilizer of v that v dominates. Thus, $(3) \implies (1)$ follows by Theorem 3.4.17 (3) $\implies $ (1) since M is a flat R-module. The implication $(2) \implies (3)$ is clear.

It suffices to show $(1) \implies (2)$ . Since M is flat we can express it as a colimit of a filtered system of free R-modules of finite rank $\{(L_i, u_{ij})\colon i,j \in I, i \leq j\}$ . Then one can choose $i \in I$ and a lift $x_i$ of $v(1)$ to $L_i$ such that the unique map $u \colon R \to L_i$ that sends $1 \mapsto x_i$ stabilizes v by Proposition 3.4.25 (4).

Proof. The equivalences of (3), (4) and (5) follow from Theorem 3.4.17 and hold even when R is not local. The implication $(1) \implies (2)$ is clear while $(2) \implies (3)$ follows from Lemma 3.3.4 and the fact that free modules are Ohm-Rush (Example 3.4.7).

It remains to show $(5) \implies (1)$ . Since M is flat and $f \in c_M(f)M$ , v admits a stabilizer by Corollary 3.4.27. Then (1) follows by Proposition 3.1.6 applied to $g = v$ .