FROBENIUS ACTIONS ON LOCAL COHOMOLOGY MODULES AND DEFORMATION

Abstract

Let $(R,\mathfrak{m})$ be a Noetherian local ring of characteristic $p>0$. We introduce and study $F$-full and $F$-anti-nilpotent singularities, both are defined in terms of the Frobenius actions on the local cohomology modules of $R$ supported at the maximal ideal. We prove that if $R/(x)$ is $F$-full or $F$-anti-nilpotent for a nonzero divisor $x\in R$, then so is $R$. We use these results to obtain new cases on the deformation of $F$-injectivity.

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. [13] for its introduction), which was systematically introduced by Hochster and Huneke [9] 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 [25] 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. [13, 25]). It is well known that $F$ -rationality always deforms while $F$ -regularity and $F$ -purity do not deform in general [22, 23]. Whether $F$ -injectivity deforms is a long- standing open problem [6] (for recent developments, we refer to [11, 18]). 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 [20, 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 [18]. The second condition has been studied in [5, 16], 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 [11] 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. (1) if $R/(x)$ is $F$ -anti-nilpotent, then so is $R$ ;

2. (2) if $R/(x)$ is $F$ -full, then so is $R$ ;

3. (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 $\mathscr{F}_{R}(M)\rightarrow M$ , 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. 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 [12]. 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 [3] 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 [12, Lemma 2.2]. One can also characterize $F$ -injectivity using certain ideal closure operations (see [17, 20] 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. (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. (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 [5] and [16] 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 [5, Theorem 4.15] that this is equivalent to $R$ being $F$ -anti-nilpotent.

Remark 2.4.

1. (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 [16, Theorem 1.1]. In particular, $F$ -pure local rings are $F$ -full.

2. (2) We can construct many $F$ -anti-nilpotent (equivalently, stably FH-finite) rings that are not $F$ -pure [20, Sections 5 and 6].

3. (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 [5, 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. (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

   $$\begin{eqnarray}\left[\frac{(s^{3}t)^{2}}{s^{4}},-\frac{(st^{3})^{2}}{t^{4}}\right]\in R_{s^{4}}\oplus R_{t^{4}}\end{eqnarray}$$
   spans the local cohomology module $H_{\mathfrak{m}}^{1}(R)$ . 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. (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 [18, Examples 4.11 and 4.16]) shows that

   $$\begin{eqnarray}H_{\mathfrak{m}_{R}}^{2}(R)=[H_{\mathfrak{ m}_{R}}^{2}(R)]_{0}\cong [H_{\mathfrak{m}_{A}}^{2}(A)]_{0}=k.\end{eqnarray}$$
   Since $p\equiv 2$ 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 [16, Theorem 1.1].

Remark 2.6.

1. (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 [15, 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. (2) Suppose $R$ is a local ring essentially of finite type over $\mathbb{C}$ and $R$ is Du Bois (we refer to [21] or [18] 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 $J\subseteq I=\sqrt{J}$  [18, Lemma 3.3]. This is the main ingredient in proving singularities of dense $F$ -injective type deform [18, Theorem C].

3. (3) Since $F$ -injective singularity is the conjectured characteristic $p>0$ analog of Du Bois singularity [1, 21], 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 [18, Example 3.5]. However, constructing such examples seems hard. In fact, [5, Example 2.16] (or its variants like [18, 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 [16, 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 [2, 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 [7, 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 [11] and was the key tool in [11].

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 [11, Lemma 3.2], so we omit the proof.

Proposition 3.2. The following are equivalent:

1. (i) $x$ is a surjective element.

2. (ii) For all $0<h\leqslant k$ the multiplication map

   $$\begin{eqnarray}R/(x^{h})\overset{x^{k-h}}{\rightarrow }R/(x^{k})\end{eqnarray}$$
   induces an injection
   $$\begin{eqnarray}H_{\mathfrak{m}}^{i}(R/(x^{h}))\rightarrow H_{\mathfrak{ m}}^{i}(R/(x^{k}))\end{eqnarray}$$
   for each $i\geqslant 0$ .

3. (iii) For all $0<h\leqslant k$ the short exact sequence

   $$\begin{eqnarray}0\rightarrow R/(x^{h})\overset{x^{k-h}}{\rightarrow }R/(x^{k})\rightarrow R/(x^{k-h})\rightarrow 0\end{eqnarray}$$
   induces a short exact sequence
   $$\begin{eqnarray}0\rightarrow H_{\mathfrak{m}}^{i}(R/(x^{h}))\rightarrow H_{\mathfrak{ m}}^{i}(R/(x^{k}))\rightarrow H_{\mathfrak{ m}}^{i}(R/(x^{k-h}))\rightarrow 0\end{eqnarray}$$
   for each $i\geqslant 0$ .

Proposition 3.3. The following are equivalent:

1. (i) $x$ is a surjective element.

2. (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 [11, 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 [11, 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 [18, 24].

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 [11, 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 [11]:

Corollary 3.8. (Compare with [11], 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

1. (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.

2. (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.

3. (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 [20, 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:

1. (i) if $R/(x)$ is $F$ -anti-nilpotent, then so is $R$ ;

2. (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 [6] and Singh [22] for the deformation of $F$ -purity.

Example 4.3. (Compare with [20, 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 [12, Proposition 5.38]. So $R/(y)$ is $F$ -anti-nilpotent by [16, 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. [3, 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.

1. (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 [3, Corollary 7.3.3].

2. (ii) Suppose $(R,\mathfrak{m})$ is an image of a Cohen–Macaulay local ring. By the Grothendieck finiteness theorem (cf. [3, Theorem 9.5.2]) we have

   $$\begin{eqnarray}f_{\mathfrak{m}}(M)=\min \{\operatorname{depth}M_{\mathfrak{p}}+\dim R/\mathfrak{p}\,:\,\mathfrak{p}\in \text{Supp}(M)\setminus \{\mathfrak{m}\}\}.\end{eqnarray}$$

3. (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 [11, Appendix]. In the proof of [11, 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 [11, Lemma A.2] (and hence [11, 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. [11, 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. [21, 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 [7, Remark 13.6]. In particular, $R$ is universally catenary. ³ But if a universally catenary ring satisfies $(S_{2})$ , then it is equidimensional (see [10, 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. [4])

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 [11] 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 [19, 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 [11].

Corollary 5.13. [11, 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 [1, 21] 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 [18, Example 3.5], which is based on the construction of [5, 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.

Proof. We know $R$ is $F$ -injective by Corollary 3.8(iii). But we also know $R$ is $F$ -full by Theorem 4.2(ii). This shows that $R$ is strongly $F$ -injective.◻