# Generalized Depth and Associated Primes in the Perfect Closure $R^{\infty}$

George Whelan  George Whelan, Department of Arts and Sciences, George Mason
University Korea, Incheon, Korea, $21985$<EMAIL_ADDRESS>
###### Abstract.

For a reduced Noetherian ring $R$ of characteristic $p>0$, in this paper we
discuss an extension of $R$ called its perfect closure $R^{\infty}$. This
extension contains all $p^{e}$-th roots of elements of $R$, and is usually
non-Noetherian. We first define the generalized notions of associated primes
of a module over a non-Noetherian ring. Then for any $R$-module $M$, we state
a correspondence between certain generalized prime ideals of
$(R^{\infty}\otimes_{R}M)/N$ over $R^{\infty}$, and the union of associated
prime ideals of $F^{e}(M)/N_{e}$ as $e\in\mathbb{N}$ varies. Here $F$ refers
to the Frobenius functor, and in the paper we define an $F$-sequence of
submodules $\\{N_{e}\\}\subseteq\\{F^{e}(M)\\}$ as $e$ varies, while
$\underrightarrow{\lim}\ N_{e}=N$. Under the further assumptions that $M$ is
finitely generated and $(R,\mathfrak{m})$ is an $F$-pure local ring, we then
show that $\operatorname{depth}_{R}(F^{e}(M))$ is constant for $e\gg 0$, and
we call this value the stabilizing depth, or $\operatorname{s\ depth}_{R}(M)$.
Lastly, we turn to non-Noetherian measures of the depth of
$R^{\infty}\otimes_{R}M$ over $R^{\infty}$, which generalize as well. Two of
these values are the $\operatorname{k\ depth}$ and the $\operatorname{c\
depth}$, and we show $\operatorname{k\
depth}_{R^{\infty}}(R^{\infty}\otimes_{R}M)=\operatorname{s\
depth}_{R}(M)\geq\operatorname{c\
depth}_{R^{\infty}}(R^{\infty}\otimes_{R}M)$, while all three values are equal
under certain assumptions.

##  1\. Introduction

Let $R$ be a reduced commutative Noetherian ring of characteristic $p>0$ with
unity. Its perfect closure $R^{\infty}$ is the extension formed by adjoining
all $p^{e}$-th roots of elements of $R$. As we discuss in sections 2.2 and 2.3
respectively, $R^{\infty}$ is constructed as a direct limit of extensions of
$R$, and $R^{\infty}\otimes_{R}M$ is a direct limit of $F^{e}(M)$ for an
$R$-module $M$, where $F$ denotes the Frobenius functor. Over these iterations
$F^{e}(M)$, it is natural to investigate what happens to fundamental notions
such as their associated primes in $R$, and if $(R,\mathfrak{m})$ is local,
the depth of $F^{e}(M)$. However, if we compare any such findings to the same
measures of $R^{\infty}\otimes_{R}M$ as an $R^{\infty}$-module, the
investigation is immediately complicated by the fact that $R^{\infty}$ is
almost always a non-Noetherian ring. Both of these concepts are then not well
defined because characterizations which are equivalent in the Noetherian
context become distinct definitions.

Over a non-Noetherian ring $S$, for a module $M$ the definition of associated
primes of a module generalizes to the additional concepts of weakly associated
primes, strong Krull primes, and Krull primes, which have the respective
containments
$\operatorname{Ass}_{S}(M)\subseteq\operatorname{\widetilde{Ass}}_{S}(M)\subseteq\operatorname{sK}_{S}(M)\subseteq\operatorname{K}_{S}(M)$.
For an overview of this topic see [IR84], and we define them in section 2.5.

For $R$ and $M$ as above, in order to state the behavior of associated prime
ideals of $F^{e}(M)$ in full generality as $e$ grows, we must define in
section 2.4.2 an $F$-sequence of sub-modules
$\\{N_{e}\\}\subseteq\\{F^{e}(M)\\}$. The direct limit
$N=\underrightarrow{\lim}\ N_{e}$ of such modules is a sub-module of
$R^{\infty}\otimes_{R}M$. This notion is a generalization of $f$-sequences of
ideals as discussed in [NS04], and which we address in section 2.4.1. Then
using the fact that $\operatorname{Spec}(R)$ and
$\operatorname{Spec}(R^{\infty})$ are homeomorphic topological spaces under
the contraction map
$\varphi:\operatorname{Spec}(R^{\infty})\rightarrow\operatorname{Spec}(R)$, as
shown in section 2.2, our main result in section 3 states that as $e$ varies:

$\displaystyle\varphi^{-1}\bigg{(}\bigcup\operatorname{Ass}_{R}\big{(}F^{e}(M)/N_{e}\big{)}\bigg{)}$
$\displaystyle=\operatorname{\widetilde{Ass}}_{R^{\infty}}\big{(}(R^{\infty}\otimes_{R}M)/N\big{)}$
$\displaystyle=\operatorname{sK}_{R^{\infty}}\big{(}(R^{\infty}\otimes_{R}M)/N\big{)}.$

where $\\{N_{e}\\}$ is an $F$-sequence of sub-modules of $\\{F^{e}(M)\\}$. An
analogous result also holds for $f$-sequences of ideals.

Regarding non-Noetherian depths, see [Hoc74] and [Bar72] for a discussion of
these values. We define them in section 2.6, and for a local ring
$(S,\mathfrak{n})$ and $S$-module $M$, we have $\operatorname{r\
depth}_{S}(M)\geq\operatorname{k\ depth}_{S}(M)\geq\operatorname{c\
depth}_{S}(M)$. Again, these are all equal if $S$ is Noetherian and $M$ is
finitely generated.

Now letting $M$ be a finitely generated module over a Noetherian local
$(R,\mathfrak{m})$, in general
$\operatorname{depth}_{R}(M)\neq\operatorname{depth}_{R}\big{(}F^{e}(M)\big{)}$
as $e$ grows, as in example 4.6. Under the assumption that $R$ is an $F$-pure
ring, which we discuss in section 2.7, we find that this value is non-
increasing, and therefore must eventually be constant for $e\gg 0$. In section
4.1 we call this value the the stabilizing depth:

$\operatorname{s\
depth}_{R}(M):=\operatorname{depth}_{R}\big{(}F^{e}(M)\big{)}\ \text{for }\
e\gg 0$.

Since this measure of $M$ is defined over iterations $F^{e}(M)$, in section
4.2 we compare it to the non-Noetherian depth measures of the limit
$\underrightarrow{\lim}\ F^{e}(M)=R^{\infty}\otimes_{R}M$ as an
$R^{\infty}$-module. We find that in general:

$\operatorname{k\ depth}_{R^{\infty}}(R^{\infty}\otimes_{R}M)=\operatorname{s\
depth}_{R}(M)\geq\operatorname{c\
depth}_{R^{\infty}}(R^{\infty}\otimes_{R}M)$.

Under the assumption that $R$ satisfies countable prime avoidance (see [LW12,
Lemma 13.2], [Bur72, Lemma 3], [SV85], [HH00]), or if a specific application
of the prime avoidance lemma is satisfied, we have:

$\operatorname{k\ depth}_{R^{\infty}}(R^{\infty}\otimes_{R}M)=\operatorname{s\
depth}_{R}(M)=\operatorname{c\ depth}_{R^{\infty}}(R^{\infty}\otimes_{R}M)$.

###  1.1. Outline

Section 2 consists of preliminary discussions of relevant topics to this
paper: The perfect closure $R^{\infty}$, the Frobenius functor, Frobenius
sequences of ideals and modules, non-Noetherian associated prime ideals, non-
Noetherian depth, and $F$-pure rings. In section 3 we discuss our results
regarding non-Noetherian associated prime ideals of $R^{\infty}\otimes_{R}M$
over $R^{\infty}$ for an $R$-module $M$. In section 4 we assume $F$-purity for
our rings $R$, we define stabilizing depth for a finitely generated $R$-module
$M$, and we compare $\operatorname{s\ depth}_{R}(M)$ to other non-Noetherian
depth measures over $R^{\infty}$.

##  2\. Preliminaries

###  2.1. Notation and conventions

All rings are commutative with unity, reduced, and have characteristic $p>0$.
The reduced assumption for a ring $S$ means $\sqrt{0}=0$ in $S$. We write $q$
to refer to $p^{e}$ for $e\in\mathbb{N}$, and occasionally
$q^{\prime}=p^{e^{\prime}}$. $S$ will refer to an arbitrary ring, $R$ will
refer to an arbitrary Noetherian ring, and if either is local we will
respectively write $(S,\mathfrak{n})$ or $(R,\mathfrak{m})$. If $\bf x$ is a
finite sequence of elements in a ring $S$, $\bf|x|$ refers to the cardinality
of the set $\\{\bf x\\}$. For a ring $S$ the map $f:S\rightarrow S$ refers to
the Frobenius endomorphism $f(s)=s^{p}$, while $f^{e}$ refers to the $e^{th}$
iteration of $f$ for $e\in\mathbb{N}$. If $M$ is an $S$-module,
$\operatorname{Ass}_{S}(M)$ refers to the associated prime ideals of $M$ in
$\operatorname{Spec}(S)$, that is, prime ideals which are annihilators of
elements of $M$. If $I\subset S$ is an ideal, $I^{[q]}:=\\{(x^{q})\ |\ x\in
I\\}$ is the $q$-th Frobenius power of $I$,
$I^{[\frac{1}{q}]}:=\\{(x^{\frac{1}{q}})\ |\ x\in I,\ \text{and}\
x^{\frac{1}{q}}\in S\\}$, and $I^{F}$ refers to the Frobenius closure
$I^{F}:=\\{x\ |\ x^{q}\in I^{[q]}\ \text{for some}\ q\\}$ (alternately,
$I^{F}:=IR^{\infty}\cap R$ for $R^{\infty}$ defined below). If $I=I^{F}$, it
is Frobenius closed.

###  2.2. The perfect closure of a reduced ring of characteristic $p>0$

A ring $S$ of characteristic $p>0$ is _perfect_ if the Frobenius map is an
automorphism on $S$. Let $R$ be a reduced commutative Noetherian ring, whereby
$f:R\rightarrow R$ is injective since there exists no non-zero $x\in R$ such
that $x^{p}=0$, but the map is not surjective in general. The _perfect
closure_ $R^{\infty}$ is an extension of $R$ which will contain all $q$-th
roots of all elements of $R$. Hence $f:R^{\infty}\rightarrow R^{\infty}$ will
be an automorphism. Greenberg showed that such a ring always exists,
regardless of whether a ring is reduced [Gre65]. For an explicit construction
which also addresses the non-reduced case see [Jor82], and a description in
[NS04].

We include an explicit construction of $R^{\infty}$ using our reduced
hypothesis for $R$. Since $R$ is Noetherian and reduced, it has only finitely
many associated prime ideals $P_{1},\ldots P_{n}$, and $P_{1}\cap\ldots\cap
P_{n}=\sqrt{0}=0$. By virtue of $R$ being reduced, the natural map
$R\rightarrow\prod_{i=1}^{n}(R/P_{i})$ is injective. Moreover, the total ring
of quotients is $K=q(R)=\prod_{i=1}^{n}q(R/P_{i})$. For each $i$, let
$K_{i}=q(R/P_{i})$, the fraction field of $R/P_{i}$, and let
$\overline{K}_{i}$ be its algebraic closure. Let
$\bar{K}=\prod_{i=1}^{n}\overline{K}_{i}$. Let
$R^{\frac{1}{q}}:=\\{x\in\bar{K}\ |\ x^{q}\in R\\}$. Then
$R^{\infty}:=\underrightarrow{\lim}\ R^{\frac{1}{q}}=\bigcup R^{\frac{1}{q}}$.

$R^{\infty}$ will rarely be Noetherian. For example if $R:=\mathbb{F}_{p}[x]$,
then
$R^{\infty}=\mathbb{F}_{p}[x,x^{\frac{1}{p}},x^{\frac{1}{p^{2}}},\ldots]$.
Specifically, $R^{\infty}$ will be Noetherian if and only if $R$ is a direct
product of finitely many fields [NS04, Theorem 6.3]. However $R$ and
$R^{\infty}$ always share some key similarities which we will discuss
throughout this paper.

In [NS04, Theorem $6.1,\ i)$], the authors showed that there exists an order
isomorphism between $\operatorname{Spec}(R)$ and
$\operatorname{Spec}(R^{\infty})$. That is, the contraction map
$\varphi:\operatorname{Spec}(R^{\infty})\rightarrow\operatorname{Spec}(R)$ is
an order-preserving bijection. We show that the two spectra actually share a
stronger relationship in that they are homemorphic topological spaces.

######  Theorem 1.

Let $R$ be a reduced Noetherian ring of characteristic $p>0$, and let
$R^{\infty}$ be its perfect closure. Then the contraction map
$\varphi:\operatorname{Spec}(R^{\infty})\rightarrow\operatorname{Spec}(R)$ is
a homeomorphism with respect to the Zariski topology.

###### Proof.

$\varphi$ is already known to be a bijection by [NS04, Theorem $6.1,\ i)$].
$\varphi$ is always continuous, but we claim that
$\varphi^{-1}:\operatorname{Spec}(R)\rightarrow\operatorname{Spec}(R^{\infty})$
is continuous as well. Let $V(J)\subset\operatorname{Spec}(R^{\infty})$ be a
closed set for some ideal $J\subset R^{\infty}$. We claim
$\varphi(V(J))=V(J\cap R)$, and hence its pre-image under $\varphi^{-1}$ is
closed. If $J\subseteq P^{\infty}$, then clearly $J\cap R\subseteq
P^{\infty}\cap R=P$, whence $P\in V(J\cap R)$. Conversely, if $J\cap
R\subseteq P$, fix $x^{\frac{1}{q}}\in J$. Then $x\in(J^{[q]}R^{\infty}\cap
R)\subseteq(J\cap R)\subseteq P=(P^{\infty}\cap R)\subseteq P^{\infty}$. Hence
$x^{\frac{1}{q}}\in P^{\infty}$, and $J\subseteq P^{\infty}$. ∎

######  Notation 2.1.

Due to this relationship, we will use the notation $P$ and $P^{\infty}$ to
denote corresponding prime ideals in $R$ an $R^{\infty}$ respectively.

###  2.3. The Frobenius functor

Let $R^{\frac{1}{f}}$ denote an $R$-$R$ bimodule, which is isomorphic to the
ring $R^{\frac{1}{p}}$. However, for $a^{\frac{1}{p}}\in R^{\frac{1}{f}}$ and
$r,s\in R$, let the left and right actions by $R$ be $r\circ
a^{\frac{1}{q}}\cdot s=r^{\frac{1}{q}}a^{\frac{1}{q}}s$. Iterating, we have
$R^{\frac{1}{f^{e}}}\cong R^{\frac{1}{q}}$, and left and right action by $R$
is $r\circ a^{\frac{1}{q}}\cdot s=r^{\frac{1}{q}}a^{\frac{1}{q}}s$.
$R^{\frac{1}{f^{e}}}$ is additionally a right
$R^{\frac{1}{q^{\prime}}}$-module for all $q^{\prime}\leq q$ with action
$a^{\frac{1}{q}}\cdot
r^{\frac{1}{q^{\prime}}}=a^{\frac{1}{q}}r^{\frac{1}{q^{\prime}}}$, which is
compatible with the right action by $R$ since $R\subseteq R^{\frac{1}{q}}$.

######  Remark 2.2.

We discuss the module $R^{\frac{1}{f}}$ rather than the standard $R^{f}$ or
$R^{F}$ such as defined in [BH97] because under our construction the direct
limit $\underrightarrow{\lim}\ F^{e}(M)$ will be an $R^{\infty}$-module.

######  Definition 2.3.

If $M$ is an $R$-module, the _Frobenius functor_ is a right exact functor in
both the categories of right and left $R$-modules, defined by:

$F(M):=R^{\frac{1}{f}}\otimes_{R}M$

and if $\varphi:M\rightarrow N$ is an $R$-module map, then
$F(\varphi):F(M)\rightarrow F(N)$ is the map:

$F(\varphi):=1_{R^{1/f}}\otimes_{R}\varphi$

######  Notation 2.4.

$m_{e}=\sum_{i=1}^{n}(s_{i}^{\frac{1}{q}}\otimes_{R}m_{i})_{e}$ will refer to
an arbitrary element of $F^{e}(M)$. For example, let $\varphi:M\rightarrow M$
be the identity map on an $R$-module $M$. If $q<q^{\prime}$, and if
$\sum_{i=1}^{n}(s_{i}^{\frac{1}{q}}\otimes_{R}m_{i})_{e}=m_{e}\in F^{e}(M)$,
then its image under $F^{e^{\prime}-e}(\varphi)$ is
$m_{e^{\prime}}=\sum_{i=1}^{n}(s_{i}^{\frac{1}{q}}\otimes_{R}m_{i})_{e^{\prime}}$,
where each $m_{i}\in M$ does not change, and each $s_{i}^{\frac{1}{q}}\in
R^{\frac{1}{q}}\subseteq R^{\frac{1}{q^{\prime}}}$ does not change.

Similarly, $m_{\infty}$ will refer to elements of $R^{\infty}\otimes_{R}M$.

$F(M)$ is an $R-R^{\frac{1}{p}}$ bi-module (and hence an $R-R$ bi-module) with
left and right action (on a simple tensor):

$\displaystyle r\ \circ\ m_{1}\cdot s^{\frac{1}{p}}=$ $\displaystyle\
r\circ(a^{\frac{1}{p}}\otimes_{R}m)_{1}\cdot s^{\frac{1}{p}}$ $\displaystyle=$
$\displaystyle\ (r\circ a^{\frac{1}{p}}\cdot s^{\frac{1}{p}}\otimes_{R}m)_{1}$
$\displaystyle=$ $\displaystyle\
(r^{\frac{1}{p}}a^{\frac{1}{p}}s^{\frac{1}{p}}\otimes_{R}m)_{1}$
$\displaystyle=$ $\displaystyle\ r^{\frac{1}{p}}m_{1}s^{\frac{1}{p}}$

We now have a directed system
$\big{(}F^{e}(M),(\psi_{e,e^{\prime}}\otimes_{R}1_{M})\big{)}$, since we have
$F^{e}(M):=(R^{\frac{1}{f^{e}}}\otimes_{R}M)$, where the maps
$\psi_{e,e^{\prime}}:R^{\frac{1}{f^{e}}}\hookrightarrow
R^{\frac{1}{f^{e^{\prime}}}}$ are embeddings. Therefore,

$(R^{\infty}\otimes_{R}M):=\underrightarrow{\lim}\ F^{e}(M)$,

which is a right $R^{\infty}$-module (and hence a right $R\subseteq
R^{\infty}$ module), whose action (on a simple tensor) is consistent with each
$F^{e}(M)$:

$\displaystyle m_{\infty}\cdot r^{\frac{1}{q^{\prime}}}=$ $\displaystyle\
(a^{\frac{1}{q}}\otimes_{R}m)_{\infty}\cdot r^{\frac{1}{q^{\prime}}}$
$\displaystyle=$ $\displaystyle\ (a^{\frac{1}{q}}\cdot
r^{\frac{1}{q^{\prime}}}\otimes_{R}m)_{\infty}$ $\displaystyle=$
$\displaystyle\
(a^{\frac{1}{q}}r^{\frac{1}{q^{\prime}}}\otimes_{R}m)_{\infty}$
$\displaystyle=$ $\displaystyle\ m_{\infty}r^{\frac{1}{q^{\prime}}}$

######  Remark 2.5.

Since the left action by $R$ on $F^{e}(M)$ is identical to simple left
multiplication by the ring $R^{\frac{1}{q}}$, $F^{e}(M)$ will be finitely
generated as a left $R$-module, even though it may not be finitely generated
by right action , since $R$ may not be $F$-finite.

######  Definition 2.6.

If $M,N$ are $R$-modules with $N\subseteq M$ the _Frobenius closure_ of $N$ in
$M$ is defined $N^{F}_{M}:=\\{m\in M\ |\ (1\otimes_{R}m)_{e}\in
N^{[q]}\subseteq F^{e}(M)\ \text{for}\ e\gg 0\\}$. Here
$N^{[q]}:=F^{e}(i)(N)$, where $i:N\rightarrow M$ is the embedding map, and
$F^{e}(i)$ is its image under the $e$-th iteration of the Frobenius functor.

Equivalently $N^{F}_{M}:=\\{m\in M\ |\ \overline{(1\otimes_{R}m)}_{e}=0\in
F^{e}(M/N)\ \text{for}\ e\gg 0\\}$. Also note that in particular,
$0^{F}_{M}:=\\{m\in M\ |\ (1\otimes_{R}m)_{e}=0\in F^{e}(M)\ \text{for}\ e\gg
0\\}$.

######  Remark 2.7.

Note that since we are discussing a directed system, by definition
$0\in(R^{\infty}\otimes_{R}M)$ is precisely the image of the elements of
$0^{F}_{F^{e}(M)}$ as $e$ varies, or $\underrightarrow{\lim}\
0^{F}_{F^{e}(M)}=0\in R^{\infty}\otimes_{R}M$. That is,

$\displaystyle m_{e}\in 0^{F}_{F^{e}(M)}\ \text{for some}\ e$
$\displaystyle\Leftrightarrow m_{e^{\prime}}=0\in F^{e^{\prime}}(M)\ \text{for
some}\ e\leq e^{\prime}$ $\displaystyle\Leftrightarrow m_{\infty}=0\
\text{in}\ R^{\infty}$

###  2.4. Frobenius sequences

####  2.4.1. $f$-sequences of ideals

######  Definition 2.8.

In a reduced ring $S$ of characteristic $p>0$, an _$f$ -sequence of ideals in
$S$_ is a descending chain of ideals $\\{J_{e}\\}$ such that
$f^{-1}(J_{e+1})=J_{e}$ for every $e$.

######  Remark 2.9.

We do not require for $S$ to be reduced for this definition, because as
mentioned in [NS04, Remark $4.2,(ii)$], $\sqrt{0}\subseteq J_{e}$ for all $e$.
However all rings in this paper are reduced, so we include this hypothesis in
our definition.

For an extensive discussion about the properties of $f$-sequences, see [NS04].
We will not give the subject a similarly thorough treatment, however there are
some features relevant to this paper.

######  Remark 2.10.

Let $S$ be a ring of characteristic $p>0$, and let $\\{J_{e}\\}$ be an
$f$-sequence in $S$.

  1. (1)

For all $e$, $\operatorname{Ass}_{S}(S/J_{e})\subseteq$
$\operatorname{Ass}_{S}(S/J_{e+1})$ [NS04, Remark $4.2,\ iv)$].

  2. (2)

For all $e$, $J_{e}$ is Frobenius closed [NS04, Remark $4.1\ ii)$]. Hence
given an ideal $I\subset S$ we can view $\\{(I^{[q]})^{F}\\}$ as a minimal
$f$-sequence, since any $f$-sequence which contians $I$ at some stage must
therefore contain the entire sequence.

  3. (3)

$\sqrt{J_{e}}=\sqrt{J_{e^{\prime}}}$ for all $e,e^{\prime}$, or all ideals in
the sequence have the same radical [NS04, Remark $4.2,i)$].

  4. (4)

For a Notherian ring $R$ there exists a bijective correspondence between
$f$-sequences of $R$ and ideals in $R^{\infty}$ [NS04, Corollary $3.2$]. For
an ideal $J\subset R^{\infty}$, this correspondence is given explicitly by
$\Gamma\ :\ J\mapsto\\{J_{e}\\}$, where for every $e\in\mathbb{N}$,
$J_{e}:=\\{r\in R\ |\ r^{\frac{1}{q}}\in J\\}$. Alternatively, we can say
$J_{e}=f^{e}(J)\cap R$, and in particular note that $J_{0}=J\cap R$.

We include some examples of $f$-sequences.

######  Example 2.11.

Let $R$ be a Noetherian ring of characteristic $p>0$. The following sequences
$\\{J_{e}\\}\subset R$ are $f$-sequences with corresponding ideals $J\subset
R^{\infty}$. In each example, for every $e\in\mathbb{N}$, let $J_{e}$ be as
defined.

  1. (1)

$J_{e}=(I^{[q]})^{F}$, $J=IR^{\infty}$, where $I\subset R$ is any ideal.

  2. (2)

Letting $R=k[x,y]$, $J_{e}=(x,y)^{[q]}=(x^{q},y^{q})$, $J=(x,y)R^{\infty}$,
where $k$ is any field of characteristic $p>0$.

  3. (3)

Letting $R=k[x,y]$, $J_{e}=(x,y^{q})$,
$J=(x,x^{\frac{1}{p}},x^{\frac{1}{p^{2}}},\ldots,y)R^{\infty}$.

  4. (4)

$J_{e}=P$ for some $P\in\operatorname{Spec}(R)$, $J=P^{\infty}$.

Note that example $2$ is of the form $\\{I^{[q]}\\}$ for an $I\subset R$, but
examples $1,3$ and $4$ show that such a form does not characterize all
$f$-sequences. In general a sequence of Frobenius iterates $\\{I^{[q]}\\}$ for
some $I\subset R$ will not yield an $f$-sequence since $I^{[q]}$ may not be
Frobenius closed as $q$ varies. However if $R$ is an $F$-pure ring, as we
discuss below, all ideals are Frobenius closed, and hence $\\{I^{[q]}\\}$ will
always be an $f$-sequence.

Continuing with our assumption that $R$ is Noetherian, let $\\{J_{e}\\}$ be an
$f$-sequence. Fix $e$ and we have an ascending chain,

$\ldots J_{e-2}=f^{-2}(J_{e})\cap R\ \supseteq\ J_{e-1}=f^{-1}(J_{e})\cap R\
\supseteq\ J_{e}$

which must therefore stabilize, usually after the zero-th term in the sequence
$J_{0}$. Suppose the chain stabilizes to some ideal $J$ for which
$J=f^{-1}(J)$. Unsurprisingly, $J$ is the radical ideal from 2.10, fact $3$
above.

######  Proposition 2.12.

Let $R$ be a reduced Noetherian ring of characteristic $p>0$, and Let
$\\{J_{e}\\}_{e\in\mathbb{N}}$ be an $f$ sequence in $R$, and let let
$\mathfrak{a}=\sqrt{J_{e}}$ for all $e$. Let $J\subset R$ be the ideal such
that the ascending chain

$\ldots\supseteq J_{e-2}\supseteq J_{e-1}\supseteq J_{e}\supset\ldots$

stabilizes, for $e\in\mathbb{N}$. Then $J=\mathfrak{a}$.

###### Proof.

To show $J\subseteq\mathfrak{a}$, fix $r\in J$. Then for all $e$, there exists
a $q^{\prime}\geq q$ such that $r^{q^{\prime}}\in J_{e}$, and
$r\in\sqrt{J_{e}}=\mathfrak{a}$.

Conversely if $r\in\mathfrak{a}$, then for all $e$ there exists an $n$ such
that $r^{n}\in J_{e}$. Let $q^{\prime}>n$, and therefore $r^{q^{\prime}}\in
J_{e}$, whereby $f^{-e^{\prime}}(r^{q^{\prime}})=r\in J$. ∎

######  Remark 2.13.

Let $R$ be a reduced Noetherian ring of characteristic $p>0$, and Let
$\\{J_{e}\\}_{e\in\mathbb{N}}$ be an $f$ sequence in $R$. Then there exists
some $n$ such that $f^{-n}(J_{0})$ is the radical ideal for the sequence.
Hence any $f$-sequence can be uniquely extended upward, or shifted to a longer
$f$-sequence $\mathfrak{a}_{e}$ where $\mathfrak{a}_{0}=\sqrt{J_{e}}$ for all
$e$, $\mathfrak{a}_{e}:=f^{-n}(J_{e})$.

####  2.4.2. $F$-sequences of modules

For a module $M$ over a Noetherian ring $R$ of characteristic $p>0$, for every
$e$ we have a natural map $g_{e}:F^{e}(M)\rightarrow F^{e+1}(M)$, where
$g_{e}:=(\psi_{e,e+1}\ \otimes_{R}\ 1_{M})$ as discussed in section 2.3. In
particular $g_{0}:M\rightarrow F(M)$, and in general for $e\leq e^{\prime}$,
we write $g_{e,e^{\prime}}:F^{e}(M)\rightarrow F^{e^{\prime}}(M)$.

######  Notation 2.14.

For $m_{e}\in F^{e}(M)$, we will write $g_{e}(m_{e})=m_{e+1}$, and for $r\in
R$, $g_{e}(r\circ
m_{e})=g_{e}(r^{\frac{1}{q}}m_{e})=r^{\frac{1}{q}}m_{e+1}=r^{p}\circ m_{e+1}$.

######  Definition 2.15.

An _$F$ -sequence of sub-modules_ of $F^{e}(M)$ for some $R$-module $M$ is a
sequence of sub-modules $\\{N_{e}\\}\subseteq\\{F^{e}(M)\\}$ such that
$(g_{e})^{-1}(N_{e+1})=N_{e}$ for all $e$.

######  Remark 2.16.

Under the standard $R-R$ bi-module construction $R^{f^{e}}$ as mentioned in
remark 2.2, for every $e$, $g_{e}=f$ is the Frobenius map, and each $N_{e}$ is
a sub-module of $M$.

$F$-sequences of sub-modules are up to isomorphism a generalization of
$f$-sequences of ideals, where $M=R$. As such, all features from remark 2.10
generalize as well. Some proofs and comments justifying these facts follow
from similar techniques used by [NS04]. First we state a lemma which we use
for fact $3$.

######  Lemma 2.17.

Let $R$ be Noetherian commutative ring of characteristic $p>0$, let $M$ be an
$R$-module, and let $N_{e}$ be an $F$-sequence of submodules on $M$, and let
$J_{e}=\operatorname{ann}_{R}\big{(}F^{e}(M)/N_{e}\big{)}$ for every $e$. Then
$\\{J_{e}\\}$ is an $f$-sequence of ideals in $R$.

###### Proof.

For every $e$, we want to show $f^{-1}(J_{e+1})=J_{e}$. Fix $r\in
f^{-1}(J_{e+1})$, then $r^{p}\in J_{e+1}$. Now fix $m_{e}\in F^{e}(M)\setminus
N_{e}$. By remark 2.18, number $2$ below, $N_{e}=N_{e}^{F}$ and hence
$g_{e}(m_{e})=m_{e+1}\notin N_{e+1}$. Since $r^{p}\in J^{e+1}$, we have
$r^{p}\circ m_{e+1}=r^{\frac{p}{qp}}m_{e+1}=r^{\frac{1}{q}}m_{e+1}\in
N_{e+1}$. But $r\circ m_{e}=r^{\frac{1}{q}}m_{e}$, and thus $r\circ m_{e}\in
g_{e}^{-1}(r^{\frac{1}{q}}m_{e+1})\subseteq N_{e}$. Thus $r\in
J_{e}=\operatorname{ann}_{R}\big{(}F^{e}(M)/N_{e}\big{)}$.

Conversely, first note that $F^{e}(M)$ is generated by elements of the form
$(1\otimes_{R}m)_{e}$ for some $m\in M$, and their images under $g_{e}$
generate $F^{e+1}(M)$.

Now fix $r\in J_{e}$, and we want to show $r^{p}\in J_{e+1}$. Let
$(1\otimes_{R}m)_{e+1}$ be a generator of $F^{e+1}(M)$, and consider
$r^{p}\circ(1\otimes_{R}m)_{e+1}=(r^{\frac{1}{q}}\otimes_{R}m)_{e+1}$. Now,
any $(r^{\frac{1}{q}}\otimes_{R}m)_{e}\in
g_{e}^{-1}\big{(}(r^{\frac{1}{q}}\otimes_{R}m)_{e+1}\big{)}$ will be in
$N_{e}$ since it is equal to $r\circ(1\otimes_{R}m)_{e}$. Therefore,
$r^{p}\circ(1\otimes_{R}m)_{e+1}\in N_{e+1}$, and since this was an arbitrary
generator, we have $r^{p}\circ F^{e+1}(M)\subseteq N_{e+1}$, and $r^{p}\in
J_{e+1}$. ∎

######  Remark 2.18.

Let $R$ be a Noetherian ring of characteristic $p>0$, and let $\\{N_{e}\\}$ be
an $f$-sequence in $F^{e}(M)$.

  1. (1)

For all $e$, $\operatorname{Ass}_{R}\big{(}F^{e}(M)/N_{e}\big{)}\subseteq$
$\operatorname{Ass}_{R}\big{(}F^{e+1}(M)/N_{e+1}\big{)}$.

###### Proof.

Let $P\in\operatorname{Ass}_{R}\big{(}F^{e}(M)/N_{e}\big{)}$ with
$P=(N_{e}:_{R}m_{e})$, where $g_{e}(m_{e})=m_{e+1}\notin N_{e+1}$ by
definition of an $F$-sequence. For $r\in P$, $g_{e}(r\circ
m_{e})=g_{e}(r^{\frac{1}{q}}m_{e})=r^{\frac{1}{q}}m_{e+1}=r^{\frac{pq}{q}}\circ
m_{e+1}=r^{p}\circ m_{e+1}\in N_{e+1}$. Hence
$P^{[p]}\subseteq(N_{e+1}:_{R}m_{e+1})$.

On the other hand, if $s\in(N_{e+1}:_{R}m_{e+1})$, then certainly
$s^{p}\in(N_{e+1}:_{R}m_{e+1})$. We then have $g_{e}^{-1}(s^{p}\ \circ\
m_{e+1})=g_{e}^{-1}(s^{\frac{p}{qp}}m_{e+1})=g_{e}^{-1}(s^{\frac{1}{q}}m_{e+1})=s^{\frac{1}{q}}m_{e}=s\circ
m_{e}\in N_{e}$, again by definition of $F$-sequences, and $s\in P$.

Finally, since $P^{[p]}\subseteq(N_{e+1}:_{R}m_{e+1})\subseteq P$, and
$P=\sqrt{P^{[p]}}$, $P$ is minimal over $P^{[p]}$, and hence $P$ is minimal
over the annihilator of an element of $F^{e+1}(M)/N_{e+1}$, and since $R$ is
Noetherian (see sec. 2.5),
$P\in\operatorname{\widetilde{Ass}}_{R}\big{(}F^{e+1}(M)/N_{e+1}\big{)}=\operatorname{Ass}_{R}\big{(}F^{e+1}(M)/N_{e+1}\big{)}$.
∎

  2. (2)

For every $e$, $N_{e}$ is Frobenius closed.

###### Proof.

Let $N_{e}$ be a term in an $F$-sequence, and fix $n_{e}\in N_{e}^{F}$. Then
$g_{e,e^{\prime}}(n_{e})=n_{e^{\prime}}\in N^{[q^{\prime}]}\subseteq
N_{e^{\prime}}$ for $e^{\prime}\gg e$. But then $n_{e}\in
g_{e,e^{\prime}}^{-1}(n_{e^{\prime}})$, whereby $n_{e}\in
g_{e,e^{\prime}}^{-1}(N_{e^{\prime}})=N_{e}$. ∎

  3. (3)

$\sqrt{\operatorname{ann}_{R}\big{(}F^{e}(M)/N_{e}\big{)}}=\sqrt{\operatorname{ann}_{R}\big{(}F^{e^{\prime}}(M)/N_{e^{\prime}}\big{)}}$
for all $e,e^{\prime}$.

###### Proof.

By lemma 2.17,
$\\{J_{e}\\}=\\{\operatorname{ann}_{R}\big{(}F^{e}(M)/N_{e}\big{)}\\}$ is an
$f$-sequence. Then this statement is true by remark 2.10, fact $3$. ∎

  4. (4)

There exists a bijective correspondence between $F$-sequences of $F^{e}(M)$
and submodules of $R^{\infty}\otimes_{R}M$. This fact is true because any sub-
module of $N\subseteq R^{\infty}\otimes_{R}M$ is by definition the direct
limit of an $F$-sequence of modules.

For an submodule $N\subseteq R^{\infty}\otimes_{R}M$, this correspondence is
given explicitly by $\Gamma\ :\ N\mapsto\\{N_{e}\\}$, where for every
$e\in\mathbb{N}$, $N_{e}:=\\{n_{e}\in F^{e}(M)\ |\ n_{\infty}\in N\\}$.

######  Example 2.19.

  1. (1)

Recall the $f$-sequence of ideals $\\{J_{e}\\}=\\{(I^{[q]})^{F}\\}$ for some
ideal $I$. Here
$N_{e}=(R^{\frac{1}{f^{e}}}\otimes_{R}I)^{F}\cong(IR^{\frac{1}{p^{e}}})^{F}$,
where this last expression is isomorphic to $(I^{[q]})^{F}\subseteq R$. Notice
that $N=\underrightarrow{\lim}\ N_{e}\cong IR^{\infty}$, which as we stated
above was the ideal in $R^{\infty}$ corresponding to this $f$-sequence of
ideals.

  2. (2)

For $R=k[x,y]$, recall the $f$-sequence of ideals
$\\{J_{e}\\}=\\{(x,y^{q})\\}$. Here
$N_{e}\cong(x^{\frac{1}{q}},y)R^{\frac{1}{q}}$, which is isomorphic to
$(x,y^{q})\subseteq R$. Notice that $N=\underrightarrow{\lim}\
N_{e}\cong(x,x^{\frac{1}{p}},x^{\frac{1}{p^{2}}},\ldots,y)R^{\infty}$, which
again was the ideal in $R^{\infty}$ corresponding to this $f$-sequence of
ideals.

  3. (3)

Letting $R$ and $M$ be as above, then $\\{0^{F}_{F^{e}(M)}\\}$ is an
$F$-sequence of sub-modules of $F^{e}(M)$. This fact is easy to see because:

$\displaystyle m_{e}\in 0^{F}_{F^{e}(M)}$ $\displaystyle\Leftrightarrow
g_{e,e^{\prime}}(m_{e})=0\in F^{e^{\prime}}(M)\ \text{for}\ e^{\prime}\gg e$
$\displaystyle\Leftrightarrow g_{e}(m_{e})\in 0^{F}_{F^{e+1}(M)}$

As stated above in remark 2.7, $N=0=\underrightarrow{\lim}\ 0^{F}_{F^{e}(M)}$.

###  2.5. Generalized associated prime ideals

Associated prime ideals of a module become more subtle over a non-Noetherian
ring. If $S$ is a ring, we state four subsets of $\operatorname{Spec}(S)$
which are in general distinct.

######  Definition 2.20.

Let $S$ be any commutative ring with identity (not necessarily of prime
characteristic). Let $M$ be an $S$-module. Let $P\in\operatorname{Spec}(S)$ be
a prime ideal.

  1. (1)

$P\in\operatorname{Ass}_{S}(M)$ is an _associated prime ideal_ of $M$, if $P$
is the annihilator in $S$ of some non-zero element $m\in M$. In such a case we
write $P=\text{ann}_{S}(m)$. That is, $rm=0$ for all $r\in P$, and if
$r^{\prime}m=0$ then $r^{\prime}\in P$.

  2. (2)

$P\in\operatorname{\widetilde{Ass}}_{S}(M)$ is a _weakly associated prime_ ,
or weak Bourbaki prime of $M$ if it is minimal over some $\text{ann}_{S}(m)$
for some $m\in M$. That is, $\text{ann}_{S}(m)\subseteq P$, and if there
exists a prime ideal $Q\subset S$ such that $\text{ann}_{S}(m)\subseteq
Q\subseteq P$, then $Q=P$.

  3. (3)

$P\in\operatorname{sK}_{S}(M)$ is a _strong Krull prime_ of $M$ if for any
finitely generated sub-ideal $I\subseteq P$ we have
$I\subseteq\text{ann}_{S}(m)\subseteq P$ for some $m\in M$.

  4. (4)

$P\in\operatorname{K}_{S}(M)$ is a _Krull prime_ of $M$ if for any element
$r\in P$ we have $r\in\text{ann}_{S}(m)\subseteq P$ for some $m\in M$.

For any ring $S$ and $S$-module $M$, we always have the containments
$\operatorname{Ass}_{S}(M)\subseteq\operatorname{\widetilde{Ass}}_{S}(M)\subseteq\operatorname{sK}_{S}(M)\subseteq\operatorname{K}_{S}(M)$,
with none of the containments reversible in general. For an overview see
[IR84]. It is possible that $\operatorname{Ass}_{S}(M)$ is empty, and we
provide an example.

######  Example 2.21.

Let $R=\mathbb{F}_{p}[x]$, and hence
$R^{\infty}=\mathbb{F}_{p}[x,x^{\frac{1}{p}},x^{\frac{1}{p^{2}}},\ldots]$. We
claim $\operatorname{Ass}_{R^{\infty}}(R^{\infty}/(x)R^{\infty})=\emptyset$.

To show this claim, first note that $(x)^{\infty}=\sqrt{(x)R^{\infty}}$. Since
$(x)^{\infty}$ is maximal it is therefore the only prime ideal which contains
$(x)R^{\infty}$, and hence is the only possible element of
$\operatorname{Ass}_{R^{\infty}}(R^{\infty}/(x)R^{\infty})$.

Suppose $(x)^{\infty}=((x)R^{\infty}:_{R^{\infty}}r)$ for some $r\in
R^{\infty}\setminus(x)R^{\infty}$, where
$r=\Sigma_{i=0}^{n}a_{i}x^{\frac{i}{q}}$ for some $q$, and
$a_{i}\in\mathbb{F}_{p}$ for all $i$. Then
$x^{\frac{1}{pq}}r\notin(x)R^{\infty}$. We can see this fact by multiplying:

$\displaystyle x^{\frac{1}{pq}}r$
$\displaystyle=x^{\frac{1}{pq}}\Sigma_{i=0}^{n}a_{i}x^{\frac{i}{q}}$
$\displaystyle=\Sigma_{i=0}^{n}a_{i}x^{\frac{1}{pq}}x^{\frac{i}{q}}$
$\displaystyle=\Sigma_{i=0}^{n}a_{i}x^{\frac{ip+1}{qp}}$

But since $r\notin(x)R^{\infty}$, for at least one $i$, the corresponding
monomial is non-zero, and $i<q$. But then $\frac{ip+1}{qp}<1$, and we see that
$x^{\frac{1}{pq}}r\notin(x)R^{\infty}$.

Hence $(x)^{\infty}$ cannot be an associated prime of
$R^{\infty}/(x)R^{\infty}$, and thus
$\operatorname{Ass}_{R^{\infty}}(R^{\infty}/(x)R^{\infty})=\emptyset$.

While $\operatorname{Ass}_{S}(M)$ can be an empty set, if $S$ is a Noetherian
ring and $M\neq 0$, there always exists at least one associated prime of $M$.
In this case the first three sets are equal, i.e.
$\operatorname{Ass}_{S}(M)=\operatorname{\widetilde{Ass}}_{S}(M)=\operatorname{sK}_{S}(M)$.
For an example of why $\operatorname{K}_{S}(M)$ is omitted from this equality,
see [ES14, Remark $2.2$]. However, if $S$ is Noetherian and $M$ is finitely
generated, then all four sets are equal, and $\operatorname{Ass}_{S}(M)$ is a
finite set.

###  2.6. Generalized depth

Let $S$ be any ring, again not necessarily Noetherian, let $I\subset S$ be an
ideal, and let $M$ be an $S$-module. The _grade_ of $I$ on $M$ is another
fundamential notion which generalizes into multiple concepts in the non-
Noetherian context. Discussions of generalized grades can be found in [Hoc74]
and [Bar72], and we include their definitions.

######  Definition 2.22.

Let $S$ be any commutative ring with identity, let $M$ be an $S$-module, and
let $I\subset S$ be an ideal such that $M\neq IM$.

  1. (1)

$\operatorname{c\ gr}_{S}(I,M):=\text{sup}\\{|\mathbf{x}|\\}$ where
$\mathbf{x}\subset I$ is a finite $M$-regular sequence contained in $I$.

  2. (2)

$\operatorname{k\ gr}_{S}(I,M):=\text{sup}\\{n-h\\}$ where
$\mathbf{x}=x_{1},\ldots,x_{n}\subset I$ is a finite set, and $h$ is index of
the highest non-zero homology group of the Koszul complex
$K_{\bullet}(x_{1},\ldots,x_{n};M)$. If $\mathbf{x}$ is a finite sequence, we
can discuss the koszul grade on $\mathbf{x}$, $\operatorname{k\
gr}_{S}(\mathbf{x},M):=n-h$ with $n$ and $h$ as above.

  3. (3)

$\operatorname{r\ gr}_{S}(I,M):=\text{inf}\\{i\ |\
\operatorname{Ext}^{i}_{S}(S/I,M)\neq 0\\}$.

For arbitrary $S,I,\text{and}\ M$, $\operatorname{c\
gr}_{S}(I,M)\leq\operatorname{k\ gr}_{S}(I,M)\leq\operatorname{r\
gr}_{S}(I,M)$ is always true, while each inequality can be strict. See [ES14,
Remark $2.2$] for an example where $\text{c}\
\operatorname{gr}_{S}(I,M)=0<\operatorname{k\ gr}_{S}(I,M)$, in which the
inequality holds by Remark 2.24 below. See [Bar72, $2$] for an example where
$\operatorname{k\ gr}_{S}(I,M)<\operatorname{r\ gr}_{S}(I,M)$.

If $I\subseteq J$ are ideals in $S$, then for each notion  grade, we have
$\operatorname{gr}_{S}(I,M)\leq$ $\operatorname{gr}_{S}(J,M)$, [Bar72]. For
example, $\operatorname{c\ gr}_{S}(I,M)\leq\operatorname{c\ gr}_{S}(J,M)$.

######  Example 2.23.

These measures could be infinite. Let $S=k[x_{1},x_{2},\ldots]$ be the non-
Noetherian polynomial ring over some field $k$ with infinitely many variables,
let $I=(x_{1},x_{2},\ldots)$, and $M=S$. Then $x_{1},x_{2},\ldots$ is an
infinite regular sequence over $M$. Hence $\operatorname{c\ gr}_{S}(I,M)$ is
infinite, as are the other two measures since $\operatorname{c\
gr}_{S}(I,M)\leq\operatorname{k\ gr}_{S}(I,M)\leq\operatorname{r\
gr}_{S}(I,M)$.

If $S$ is a Noetherian ring and $M$ is a finitely generated $S$-module, then
all three concepts  $\operatorname{gr}_{S}(I,M)$ coincide, and are indeed
finite. In such a case, all maximal $M$-sequences in $I$ have the same length.

If $(S,\mathfrak{n})$ is local, the  _depth_ is, $\underline{\hskip 8.5359pt}\
\operatorname{depth}_{S}(M):=\underline{\hskip 8.5359pt}\
\operatorname{gr}_{S}(\mathfrak{n},M)$. The relationships between generalized
associated prime ideals and generalized depth are an exercise following from
their definitions.

######  Remark 2.24.

Let $(S,\mathfrak{n},l)$ be a local ring with quotient field
$l=S/\mathfrak{n}$, and let $M$ be an $S$-module such that
$M\neq\mathfrak{n}M$. Then,

  1. (1)

$\mathfrak{n}\in\operatorname{Ass}_{S}(M)\ \text{if and only if}\
\operatorname{r\ depth}_{S}(M)=0$

  2. (2)

$\mathfrak{n}\in\operatorname{sK}_{S}(M)\ \text{if and only if}\
\operatorname{k\ depth}_{S}(M)=0$

  3. (3)

$\mathfrak{n}\in\operatorname{K}_{S}(M)\ \text{if and only if}\
\operatorname{c\ depth}_{S}(M)=0$

###### Proof.

1):

$\begin{array}[]{ll}\mathfrak{n}\in\operatorname{Ass}_{S}(M)&\Leftrightarrow
0\neq\operatorname{Hom}_{S}(l,M)\cong\operatorname{Ext}_{S}^{0}(l,M)\\\
&\Leftrightarrow\operatorname{r\ depth}_{S}(M)=0\end{array}$

2):

$\begin{array}[]{ll}\mathfrak{n}\in\operatorname{sK}_{S}(M)&\Leftrightarrow\text{for
all finite}\
\mathbf{x}=x_{1},\ldots,x_{n}\subset\mathfrak{n},(\mathbf{x})\subseteq\text{ann}_{S}(m)\subseteq\mathfrak{n}\\\
&\ \ \ \ \text{for some non-zero}\ m\in M\\\ &\Leftrightarrow\ \text{for all
finite}\
\mathbf{x}=x_{1},\ldots,x_{n}\subset\mathfrak{n},H_{n}(\mathbf{x};M)\neq 0\\\
&\Leftrightarrow\text{sup}\\{\operatorname{k\ gr}_{S}(\mathbf{x};M)\ |\
\mathbf{x}\subset\mathfrak{n}\ \text{is a finite set}\\}=0\\\
&\Leftrightarrow\operatorname{k\ depth}_{S}(M)=0\end{array}$

3):

$\begin{array}[]{ll}\mathfrak{n}\in\operatorname{K}_{S}(M)&\Leftrightarrow\text{for
all}\ x\in\mathfrak{n},\ x\in\text{ann}_{S}(m)\subseteq\mathfrak{n}\ \text{for
some}\ m\in M,\ \text{i.e.,}\\\ &\ \ \ \ \text{there exist no}\
M$-$\text{regular elements in}\ \mathfrak{n}\\\
&\Leftrightarrow\text{sup}\\{|\mathbf{x}|\ |\ \mathbf{x}\subset\mathfrak{n}\
\text{is a regular sequence}\\}=0\\\ &\Leftrightarrow\operatorname{c\
depth}_{S}(M)=0\\\ \end{array}$

∎

Note that we have no equivalence for $\operatorname{\widetilde{Ass}}_{S}(M)$.
These generalized depths do not relate to weakly associated primes in an
obvious way.

Also note that if $S$ is Noetherian and $M$ is finitely generated, then all
concepts coincide, and simply $\mathfrak{n}\in\operatorname{Ass}_{S}(M)\
\text{if and only if}\ \operatorname{depth}_{S}(M)=0$.

###  2.7. F-pure rings

Let $S$ be any ring, and let $A$ and $B$ be $S$-modules. An $S$-module map
$\psi:A\rightarrow B$ is _pure_ if the map
$\psi\otimes_{S}1_{M}:A\otimes_{S}M\rightarrow B\otimes_{S}M$ is injective for
all $S$-modules $M$. Alternately we say that $A$ is a pure submodule of $B$.

We say that a Noetherian ring $R$ of characteristic $p>0$ is _$F$ -pure_ if
the embedding $\psi:R\hookrightarrow R^{\frac{1}{p}}$ is a pure map.
Equivalently, for any $R$-module $M$ the induced map $M\rightarrow
F(M):=R\otimes_{R}M\rightarrow R^{\frac{1}{p}}\otimes_{R}M$ is an injective
map (all modules act flat). Note that if $R$ is $F$-pure, then for any $q\leq
q^{\prime}$ the embeddings $\psi_{e,e^{\prime}}:R^{\frac{1}{q}}\hookrightarrow
R^{\frac{1}{q^{\prime}}}$, as well as the embeddings
$\psi_{e,e^{\infty}}:R^{\frac{1}{q}}\hookrightarrow R^{\infty}$ are pure maps
as well.

If a ring $R$ is $F$-pure, then for any $R$-module $M$ we have that
$M\subseteq F(M)$ is a pure submodule as well. This statement holds since
given $M\hookrightarrow
F(M):=(R\otimes_{R}M)\hookrightarrow(R^{\frac{1}{f}}\otimes_{R}M)$, and given
any $R$-module $N$, we have
$\big{(}M\otimes_{R}N\big{)}\rightarrow\big{(}F(M)\otimes_{R}N\big{)}$:

$\begin{array}[]{cl}&=\big{(}(R\otimes_{R}M)\otimes_{R}N\big{)}\rightarrow\big{(}(R^{\frac{1}{f}}\otimes_{R}M)\otimes_{R}N\big{)}\\\
&=\big{(}R\otimes_{R}(M\otimes_{R}N)\big{)}\rightarrow\big{(}R^{\frac{1}{f}}\otimes_{R}(M\otimes_{R}N)\big{)}\\\
&=\big{(}M\otimes_{R}N\big{)}\rightarrow\big{(}F(M\otimes_{R}N)\big{)}\end{array}$

which is an injection by $F$-purity of $R$.

######  Remark 2.25.

If $R$ is an $F$-pure ring, then every ideal $I\subset R$ is Frobenius closed,
since if $r^{q}\in I^{[q]}$, then $r\in IR^{\frac{1}{q}}$, and hence
$\overline{r}=\overline{0}$ in
$R^{\frac{1}{q}}/IR^{\frac{1}{q}}\cong(R^{\frac{1}{q}}\otimes_{R}R/I)$. But
since $R$ is $F$-pure,
$R/I\cong(R\otimes_{R}R/I)\rightarrow(R^{\frac{1}{q}}\otimes_{R}R/I)$ is
injective, $\overline{r}=\overline{0}$ in $R/I$, and $r\in I$.

In particular all $F$-pure rings are reduced, since the $0$ ideal is Frobenius
closed. Hence we need not mention the reduced assumption in section 4.

######  Remark 2.26.

A particular case of characteristic $p>0$ rings is _$F$ -finite_ rings $R$,
for which $R^{\frac{1}{p}}$ is finitely generated as a right $R$-module. In
such a case $F$-purity is equivalent to _$F$ -splitness_ [HR76, Corollary
5.3], meaning rings such that the embedding $R\hookrightarrow R^{\frac{1}{p}}$
splits. We will not discuss this characterization. However the reader will be
advised that under the assumption of $F$-finiteness, $F$-splitness is a
sufficient condition in order for all results in section 4 to hold.

######  Example 2.27.

Some examples of $F$-pure rings are

  1. (1)

Regular rings such as the polynomial ring $S=k[x_{1},\ldots,x_{n}]$, where $k$
is a field of characteristic $p>0$.

  2. (2)

If $(S,\mathfrak{n})$ is any $F$-finite regular local ring and $R=S/I$ is any
quotient, $R$ is $F$-pure if and only if
$(I^{[q]}:_{R}I)\nsubseteq\mathfrak{n}^{[q]}$ for all $q$. This result is
known as _Fedder’s criterion_ [Fed83].

  3. (3)

A non-example of an $F$-pure ring is $R:=[x,y,z]/(x^{p}-yz^{p})$. We can see
that the ideal $(z)$ is not Frobenius closed. Certainly $x\notin(z)$, but
$x^{p}=yz^{p}\in(z)^{[p]}$, and hence $x\in(z)^{F}\setminus(z)$.

Alternatively, we can see $x=y^{\frac{1}{p}}z\in(z)R^{\infty}\cap R=(z)^{F}$.

##  3\. Results for generalized associated prime ideals

Given a module $M$ over a regular local ring $R$, Epstein and Shapiro showed a
characterization of the strong Krull primes of $R^{\infty}\otimes_{R}M$.

######  Theorem 2.

[ES14, Corollary 4.9] Let $R$ be a regular Noetherian ring of characteristic
$p>0$. Let $M$ be any $R$-module. Then

$\operatorname{sK}_{R^{\infty}}(R^{\infty}\otimes_{R}M)=\bigcup_{\mathfrak{q}\in\operatorname{Ass}_{R}M}\operatorname{sK}_{R^{\infty}}(R^{\infty}/\mathfrak{q}R^{\infty}).$

Omitting the hypothesis of the regularity of $R$, and for an $F$-sequence
$N_{e}\subseteq F^{e}(M)$, we find a more general result.

######  Theorem 3.

Let $R$ be a reduced Noetherian ring of characteristic $p>0$, and let
$R^{\infty}$ be its perfect closure. Let $P$ and $P^{\infty}$ be corresponding
prime ideals in $\operatorname{Spec}(R)$ and $\operatorname{Spec}(R^{\infty})$
respectively. Let $\\{N_{e}\\}\subseteq\\{F^{e}(M)\\}$ be an $F$-sequence of
sub-modules of $F^{e}(M)$ for an $R$-module $M$. Let
$N=\underrightarrow{\lim}\ N_{e}$ be the corresponding sub-module of
$(R^{\infty}\otimes_{R}M)$. Then the following are all equivalent:

  1. (1)

$P\in\bigcup\operatorname{Ass}_{R}\big{(}F^{e}(M)/N_{e}\big{)}$

  2. (2)

$P^{\infty}\in\operatorname{\widetilde{Ass}}_{R^{\infty}}\big{(}(R^{\infty}\otimes_{R}M)/N\big{)}$

  3. (3)

$P^{\infty}\in\operatorname{sK}_{R^{\infty}}\big{(}(R^{\infty}\otimes_{R}M)/N\big{)}$

###### Proof.

$1\Rightarrow 2$: Suppose $P\in\bigcup\operatorname{Ass}_{R}(F^{e}(M)/N_{e})$
with $P=\big{(}N_{e}:_{R}m_{e}\big{)}$. I.e. for all $r\in P$, we have $r\circ
m_{e}=r^{\frac{1}{q}}m_{e}\in N_{e}$. Equivalently,
$r^{\frac{1}{q}}m_{\infty}=0$ in $(R^{\infty}\otimes_{R}M)/N$. Thus
$PR^{\infty}\subset
P^{[\frac{1}{q}]}R^{\infty}\subseteq(N:_{R^{\infty}}m_{\infty})$.

On the other hand, fix some
$s^{\frac{1}{q^{\prime}}}\in(N:_{R^{\infty}}m_{\infty})$. If $q^{\prime}\leq
q$, then $s^{\frac{q}{q^{\prime}}}\in R$ and $s^{\frac{q}{q^{\prime}}}\circ
m_{e}=s^{\frac{1}{q^{\prime}}}m_{e}\in N_{e}$. Hence
$s^{\frac{q}{q^{\prime}}}\in P$, and $s^{\frac{1}{q^{\prime}}}\in P^{\infty}$.

However if $q^{\prime}>q$, consider the image of $s\circ
m_{e}={s}^{\frac{1}{q}}m_{e}$ under the natural map
$g_{e,e^{\prime}}:F^{e}(M)\rightarrow F^{e^{\prime}}(M)$:

$\displaystyle g_{e,e^{\prime}}({s}^{\frac{1}{q}}m_{e})$
$\displaystyle={s}^{\frac{1}{q}}m_{e^{\prime}}$
$\displaystyle=s^{\prime\frac{q^{\prime}}{q}}\circ m_{e^{\prime}}$
$\displaystyle=s^{\prime(\frac{q^{\prime}}{q}-1)}s^{\prime}\circ
m_{e^{\prime}}$
$\displaystyle=s^{\prime(\frac{q^{\prime}}{q}-1)}\circ{s}^{\frac{1}{q^{\prime}}}m_{e^{\prime}}$

But ${s}^{\frac{1}{q^{\prime}}}m_{e^{\prime}}\in
N_{e^{\prime}}=N_{e^{\prime}}^{F}$ (see remark 2.18 fact $2$), because
$s^{\frac{1}{q^{\prime}}}\in(N:_{R^{\infty}}m_{\infty})$. Therefore
$g_{e,e^{\prime}}({s}\ \circ\mathfrak{m}_{e})\in N_{e^{\prime}}$, and $s\
\circ\ m_{e}\in N_{e}$ because $N_{e}=N_{e}^{F}$. Thus
$s\in(N_{e}:_{R}m_{e})$, whereby $s\in P$ and $s^{\frac{1}{q^{\prime}}}\in
P^{\infty}$.

We have now shown that
$PR^{\infty}\subseteq(N:_{R^{\infty}}m_{\infty})\subseteq P^{\infty}$. Since
$\sqrt{PR^{\infty}}=P^{\infty}$, $P^{\infty}$ is minimal over $PR^{\infty}$.
Hence $P^{\infty}$ is minimal over $(N:_{R^{\infty}}m_{\infty})$ as well. This
fact shows that $P^{\infty}$ is minimal over the annihilator of an element of
$(R^{\infty}\otimes_{R^{\infty}}M)/N$, and therefore
$P^{\infty}\in\operatorname{\widetilde{Ass}}_{R}\big{(}(R^{\infty}\otimes_{R}M)/N\big{)}$.

$2\Rightarrow 3$: Always true, as stated above.

$3\Rightarrow 1$: Suppose
$P^{\infty}\in\operatorname{sK}_{R}\big{(}(R^{\infty}\otimes_{R}M)/N\big{)}$,
then $PR^{\infty}$ is a finitely generated sub-ideal, and we have
$PR^{\infty}\subseteq(N:_{R^{\infty}}m_{\infty})\subseteq P^{\infty}$ for some
$m_{\infty}$. If $P=(r_{1},\ldots,r_{n})$, then for each $i$ there exists a
$q_{i}$ such that $r_{i}^{\frac{1}{q_{i}}}m_{e_{i}}\in N_{e_{i}}$. Letting
$q=\text{min}\\{q_{i}\\}$, we have $P\subseteq(N_{e}:_{R}m_{e})$.

On the other hand, fix $s\in(N_{e}:_{R}m_{e})$, whereby $s\circ
m_{e}=s^{\frac{1}{q}}m_{e}\in N_{e}$. Again by definition,
$s^{\frac{1}{q}}\in(N:_{R^{\infty}}m_{\infty})\subseteq P^{\infty}$. Thus
$s\in P^{\infty}\cap R=P$. Therefore $(N_{e}:_{R}m_{e})\subseteq P$.

We now have $P=(N_{e}:_{R}m_{e})$, and
$P\in\operatorname{Ass}_{R}\big{(}F^{e}(M)/N_{e}\big{)}$. ∎

######  Remark 3.1.

In the proof we use the fact that $P\in\operatorname{Ass}_{R}(F^{e}(M)/N_{e})$
under left action. However, consider the localized module
$(F^{e}(M)/N_{e})_{P}$ over $R_{P}$, then this module has left depth $0$, i.e.
there exist no left-regular elements for this module in $R_{P}$. Fix $r\in
PR_{P}$, then $r$ is left-regular over $(F^{e}(M)/N_{e})_{P}$ if and only if
$r^{q}$ is left-regular if an only if $r$ is right-regular. The first
statement is true by properties of regular sequences (specifically here of
length $1$) [Nor76, Theorem 5.1.3], and the second statement is true because
left action by $r^{q}$ is equivalent to right action by $r$. Therefore
$(F^{e}(M)/N_{e})_{P}$ over $R_{P}$ also has no right-regular elements, and
$\operatorname{Ass}_{R}\big{(}F^{e}(M)/N_{e}\big{)}$ is a two-sided concept.

As stated in example 2.19, $\\{0^{F}_{F^{e}(M)}\\}$ is an $F$-sequence. We
therefore have a particular case of theorem 3 which more directly addresses
theorem 2.

######  Corollary 3.2.

Let $R$ be a reduced Noetherian ring of characteristic $p>0$, and let
$R^{\infty}$ be its perfect closure. Let $P$ and $P^{\infty}$ be corresponding
prime ideals in $\operatorname{Spec}(R)$ and $\operatorname{Spec}(R^{\infty})$
respectively. Let $M$ be an $R$-module. Then the following are all equivalent:

  1. (1)

$P\in\bigcup\operatorname{Ass}_{R}\big{(}F^{e}(M)/0^{F}_{F^{e}(M)}\big{)}$

  2. (2)

$P^{\infty}\in\operatorname{\widetilde{Ass}}_{R^{\infty}}(R^{\infty}\otimes_{R}M)$

  3. (3)

$P^{\infty}\in\operatorname{sK}_{R^{\infty}}(R^{\infty}\otimes_{R}M)$

###### Proof.

Let $\\{N_{e}\\}=\\{0^{F}_{F^{e}(M)}\\}$, and $N=0=\underrightarrow{\lim}\
0^{F}_{F^{e}(M)}$. The result follows from theorem 3. ∎

Another particular case of theorem 3 is when $\\{J_{e}\\}$ is an $f$-sequence
of ideals in a Noetherian ring.

######  Corollary 3.3.

Let $R$ be a reduced Noetherian ring of characteristic $p>0$, and let
$R^{\infty}$ be its perfect closure. Let $\\{J_{e}\\}$ be an $f$-sequence in
$R$, and let $J\subset R^{\infty}$ be its corresponding ideal in $R^{\infty}$.
Let $P$ and $P^{\infty}$ be corresponding prime ideals in
$\operatorname{Spec}(R)$ and $\operatorname{Spec}(R^{\infty})$ respectively.
Then the following are all equivalent:

  1. (1)

$P\in\bigcup\operatorname{Ass}_{R}(R/J_{e})$

  2. (2)

$P^{\infty}\in\operatorname{\widetilde{Ass}}_{R^{\infty}}(R^{\infty}/J)$

  3. (3)

$P^{\infty}\in\operatorname{sK}_{R^{\infty}}(R^{\infty}/J$)

###### Proof.

Letting $M=R$, $\\{N_{e}\\}=\\{J_{e}\\}$, and $N=J$ we see that this corollary
is a special case of theorem 3. ∎

Lastly for an ideal $I\subseteq R$, we have the particular case for the
minimal $f$-sequence $\\{(I^{[q]})^{F}\\}$.

######  Corollary 3.4.

Let $R$ be a reduced Noetherian ring of characteristic $p>0$, let $I\subset R$
be an ideal, and let $R^{\infty}$ be its perfect closure. Let $P$ and
$P^{\infty}$ be corresponding prime ideals in $\operatorname{Spec}(R)$ and
$\operatorname{Spec}(R^{\infty})$ respectively. Then the following are all
equivalent:

  1. (1)

$P\in\bigcup\operatorname{Ass}_{R}(R/(I^{[q]})^{F})$

  2. (2)

$P^{\infty}\in\operatorname{\widetilde{Ass}}_{R^{\infty}}(R^{\infty}/IR^{\infty})$

  3. (3)

$P^{\infty}\in\operatorname{sK}_{R^{\infty}}(R^{\infty}/IR^{\infty})$

###### Proof.

Letting $J=IR^{\infty}$, and $J_{e}=(I^{[q]})^{F}$ for all $e$, we see this
corollary is a special case of corollary 3.3.

Alternately, letting $N_{e}=(I^{[q]})^{F}$ and $N=IR^{\infty}$, we see the
corollary is a special case of theorem 3. ∎

##  4\. Results over $F$-pure rings

###  4.1. Stabilizing depth

In an $F$-pure ring, since all ideals $I\subseteq R$ are Frobenius closed (see
Remark 2.25), $\\{(I^{[q]})^{F}\\}=\\{I^{[q]}\\}$ is always an $f$-sequence.
As stated in remark 2.10, $\operatorname{Ass}_{R}(R/I^{[q]})$ can gain
elements as $q$ grows. We ask: is depth${}_{R}(R/I^{[q]})$ non-increasing as
$q$ grows? Or considering that

$\displaystyle R/I^{[p]}$ $\displaystyle\cong
R^{\frac{1}{q}}/IR^{\frac{1}{q}}$
$\displaystyle\cong(R^{\frac{1}{p}}\otimes_{R}R/I)$ $\displaystyle=F(R/I)$

we can more generally ask: is depth${}_{R}\big{(}F^{e}(M)\big{)}$ non-
increasing as $e$ grows?

We must first state a lemma addressing the differing left and right actions of
$R$ on $F(M)$.

######  Lemma 4.1.

Let $R$ be an $F$-pure Noetherian ring of characteristic $p>0$. Let $M$ be an
$R$-module, and let $\mathbf{x}=x_{1},\ldots,x_{n}\subset R$ be a finite
sequence. Then for any $e\in\mathbb{N}$, $\mathbf{x}$ is a regular left
$F^{e}(M)$-sequence if and only if $\mathbf{x}$ is a regular right
$F^{e}(M)$-sequence.

###### Proof.

Fix $e\in\mathbb{N}$. It is easy to see that $\mathbf{x}$ is a left-regular
$F^{e}(M)$-sequence if and only if
$\mathbf{x}^{[q]}=x_{1}^{q},\ldots,x_{n}^{q}$ is a left-regular
$F^{e}(M)$-sequence if and only if $\mathbf{x}$ is a right-regular
$F^{e}(M)$-sequence.

The first equivalence is true by properties of regular sequences [Nor76,
Theorem 5.1.3], whereby $\mathbf{x}$ is a left-regular $F^{e}(M)$ if and only
if $\mathbf{x}^{[q]}$ is a left-regular $F^{e}(M)$ sequence.

The second equivalence is true because for any $m_{e}\in F^{e}(M)$, the left
action by any $x_{j}^{q}$ for $j=1,\ldots,n$ is equivalent to right action by
$x_{j}$. That is,

$\begin{array}[]{cl}x_{j}^{q}\circ m_{e}&=x_{j}m_{e}\\\ &=m_{e}x_{j}\\\
&=m_{e}\cdot x_{j}\end{array}$

∎

We then have a corollary.

######  Lemma 4.2.

Let $(R,\mathfrak{m})$ be an $F$-pure Noetherian local ring of characteristic
$p>0$. Let $M$ be an $R$-module, and let $F^{e}(M)$ denote the image of $M$
under the $e$-th iteration of the Frobenius functor for $e\in\mathbb{N}$. Then
$\operatorname{left\ depth}_{R}F^{e}(M)=\operatorname{right\
depth}_{R}F^{e}(M)$. Two-sided $\operatorname{depth}_{R}F^{e}(M)$ is therefore
a well-defined concept.

###### Proof.

By lemma 4.1, $\mathbf{x}\subset R$ is left $F^{e}(M)$-regular if and only if
it is right $F^{e}(M)$-regular. Therefore all maximal such sequences have the
same length under right and left action. ∎

Our next lemma is a fact regarding all pure submodule inclusions.

######  Lemma 4.3.

Let $(R,\mathfrak{m})$ be a reduced Noetherian local ring. Let $M\subseteq N$
be a pure inclusion of $R$-modules such that $M\neq 0$ is finitely generated
over $R$. Let the finite set $\mathbf{x}\subset R$ be an $N$-regular sequence.
Then $\mathbf{x}$ is regular over $M$ as well.

###### Proof.

We proceed by induction on $|\mathbf{x}|$. Let $\mathbf{x}=x$ be a singleton.
Then $M/xM\neq 0$ by Nakayama’s lemma since $x\in\mathfrak{m}$ and $M\neq 0$.
Furthermore, $x$ clearly annihilates no element of $M\subset N$.

Now assume the statment is true for $|\mathbf{x}|=n-1$. Let $|\mathbf{x}|=n$,
and let $\mathbf{x}^{\prime}=x_{1},\ldots,x_{n-1}$ denote the first $n-1$
elements of $\mathbf{x}$. Again, $M/\mathbf{x}^{\prime}M\neq 0$ by Nakayama’s
lemma since $\mathbf{x}^{\prime}\subset\mathfrak{m}$ and $M\neq 0$.

Additionally, $M/\mathbf{x}^{\prime}M$ embeds into $N/\mathbf{x}^{\prime}N$.
In order to see this fact, recall that $M/\mathbf{x}^{\prime}M\cong
M\otimes_{R}R/(\mathbf{x}^{\prime})$, and $N/\mathbf{x}^{\prime}N\cong
N\otimes_{R}R/(\mathbf{x}^{\prime})$, and since $M$ is a pure sub $N$-module
we have the embedding $M\otimes_{R}R/(\mathbf{x}^{\prime})\hookrightarrow
N\otimes_{R}R/(\mathbf{x}^{\prime})$.

Thus since $x_{n}$ annihilates no element of $N/\mathbf{x}^{\prime}N$, it will
therefore annihilate no element of the submodule $M/\mathbf{x}^{\prime}M$.
Therefore, $\mathbf{x}$ is an $M$-regular sequence. ∎

In this context, we now have an answer to our question above.

######  Lemma 4.4.

Let $(R,\mathfrak{m})$ be a local Noetherian $F$-pure ring of characteristic
$p>0$, and let $M$ be a finitely generated $R$-module. If
$\mathbf{x}=x_{1},\ldots,x_{n}$ is regular over $F(M)$, then it is regular
over $M$.

###### Proof.

$M\subseteq F(M)$ is a pure submodule. Hence by 4.3, if $\mathbf{x}$ is
regular over $F(M)$, it is also regular over $M$. ∎

So for an $F$-pure ring, we do have non-increasing depth under the Frobenius
functor.

######  Theorem 4.

Let $(R,\mathfrak{m})$ be a local Noetherian $F$-pure ring of characteristic
$p>0$, and let $M$ be a finitely generated $R$-module. Then,

$\operatorname{depth}_{R}(M)\geq\operatorname{depth}_{R}\big{(}F(M)\big{)}$.

###### Proof.

If $\operatorname{depth}_{R}\big{(}F(M)\big{)}=d$ and $x_{1},\ldots,x_{d}$ is
a maximal $F(M)$-sequence, then by lemma 4.4 $x_{1},\ldots,x_{d}$ is
$M$-regular, and $\operatorname{depth}_{R}(M)\geq d$. ∎

Recall that $F^{e}(M)$ is always finitely generated as left $R$-module (see
remark 2.5). The identical argument shows
$\operatorname{depth}_{R}\big{(}F^{e}(M)\big{)}\geq\operatorname{depth}_{R}\big{(}F^{e+1}(M)\big{)}$
for every $e$. I.e., $\operatorname{depth}_{R}\big{(}F^{e}(M)\big{)}$ is non-
increasing $e$ grows. Since it is bounded below by $0$, the value must
stabilize, meaning there must exist some $d\in\mathbb{N}$ for which
$\operatorname{depth}_{R}\big{(}F^{e}(M)\big{)}=d$ for $e\gg 0$. This fact
yields a definition.

######  Definition 4.5.

Let $(R,\mathfrak{m})$ be an F-pure Noetherian local ring of characteristic
$p>0$, and let $M$ be a finitely generated $R$-module. The _stabilizing depth_
is defined:

$\operatorname{s\
depth}_{R}(M):=\operatorname{depth}_{R}\big{(}F^{e}(M)\big{)}\ \text{for }\
e\gg 0$.

######  Example 4.6.

For any $t\geq 0$, we can construct a class of examples of modules $M$ for
which $\operatorname{s\ depth}_{R}(M)>\operatorname{s\ depth}_{R}(M)=t$.

Let $S$ be an $F$-pure Noetherian ring of characteristic $p>0$ and $J\subset
S$ be an ideal such that $\operatorname{Ass}_{S}(S/J^{[q]})$ grows as $q$
grows. For an example for such cyclic modules, see [SS04]. Let
$P\in\operatorname{Ass}_{S}(S/J^{[q]})\setminus\operatorname{Ass}_{S}(S/J)$
for some $q$, and let $A=S_{P}$ be $S$ localized at $P$. Letting
$\mathfrak{a}=PA$ we have a local ring $(A,\mathfrak{a})$. Now
$\operatorname{depth}_{A}(A/JA)>0=\operatorname{s\
depth}_{A}(A/JA)=\operatorname{depth}_{A}(A/J^{[q]}A)$.

Now let $R:=A\llbracket x_{1},\ldots,x_{t}\rrbracket$ for variables $x_{i}$,
which is local with maximal ideal
$\mathfrak{m}=(x_{1},\ldots,x_{t})+\mathfrak{a}R$ [Mat86, Discussion on Page
4]. Let $I=JR$. Then for all $q$,
$\operatorname{depth}_{R}(R/I^{[q]})=\operatorname{depth}_{A}(A/J^{[q]})\ +\
t$ (follows from [BH97, 1.2.16]). Hence
$\operatorname{depth}_{R}(R/I)=\operatorname{depth}_{A}(A/J)+t>t$, but
$\operatorname{depth}_{R}(R/I^{[q]})=\operatorname{depth}_{A}(A/J^{[q]})+t=0+t=t$
for $q\gg 0$. Thus $\operatorname{depth}_{R}(R/I)>\operatorname{s\
depth}_{R}(R/I)=t$.

###  4.2. Depth type comparisons

We investigate how the stabilizing depth of a finitely generated module $M$
over an $F$-pure ring $R$ relates to the generalized depths of
$(R^{\infty}\otimes_{R}M)$ over the usually non-Noetherian extension
$R^{\infty}$. We first compare $\operatorname{s\ depth}_{R}(M)$ with
$\operatorname{c\ depth}_{R^{\infty}}(R^{\infty}\otimes_{R}M)$, but we begin
with a lemma regarding regular sequences from $R$.

######  Lemma 4.7.

Let $R$ be an $F$-pure Noetherian ring of characteristic $p>0$, and let
$R^{\infty}$ be its perfect closure. Let $M$ be an $R$-module, and let
$\mathbf{x}=x_{1},\ldots,x_{n}\subset R$ be a finite sequence. Then,
$\mathbf{x}$ is a regular $F^{e}(M)$-sequence for all $e\in\mathbb{N}$ if and
only if $\mathbf{x}$ is a regular $(R^{\infty}\otimes_{R}M)$-sequence.

###### Proof.

If $\mathbf{x}$ is a regular $(R^{\infty}\otimes_{R}M)$-sequence, then by 4.3,
$\mathbf{x}$ is a regular right $F^{e}(M)$-sequence since
$F^{e}(M)\hookrightarrow(R^{\infty}\otimes_{R}M)$ is a pure submodule by
$F$-purity of $R$, and $\mathbf{x}$ is left-regular if and only it is right-
regular.

Conversely, we proceed by induction on $|\mathbf{x}|$. If $\mathbf{x}=x$ is a
singleton, suppose $x$ is not $(R^{\infty}\otimes_{R}M)$-regular, with
$m_{\infty}x=0$. Then $0=m_{e}x$ in some $F^{e}(M)$, and $0\neq m_{e}$, since
it was not zero in $(R^{\infty}\otimes_{R}M)=\underrightarrow{Lim}\ F^{e}(M)$.

Now suppose $\mathbf{x}=x_{1},\ldots,x_{n}\subset R$ is $F^{e}(M)$-regular for
all $e$, in particular right-regular, and let
$\mathbf{x}^{\prime}=x_{1},\ldots,x_{n-1}$, which is by hypothesis regular
over $(R^{\infty}\otimes_{R}M)$. We must show that $x_{n}$ is regular over

$\displaystyle(R^{\infty}\otimes_{R}M)/(R^{\infty}\otimes_{R}M)\mathbf{x}^{\prime}$
$\displaystyle\cong(R^{\infty}\otimes_{R}M)\otimes_{R}R/\mathbf{x}^{\prime}R$
$\displaystyle\cong R^{\infty}\otimes_{R}(M\otimes_{R}R/\mathbf{x}^{\prime}R)$
$\displaystyle\cong R^{\infty}\otimes_{R}M/\mathbf{x}^{\prime}M$

We then apply the identical argument to the regular element $x_{n}$ over the
module $F^{e}(M/\mathbf{x}^{\prime}M)$, which is therefore is also regular
over the module $(R^{\infty}\otimes_{R}M/\mathbf{x}^{\prime}M)$. ∎

If $\bf x$ is a maximal regular sequence over $F^{e}(M)$ for some specific
$e$, and $\bf y$ is a maximal regular sequence over $F^{e}(M)$ for all $e$,
clearly $|\mathbf{x}|\geq|\mathbf{y}|$.

######  Theorem 5.

Let $R$ be an $F$-pure Noetherian ring of characteristic $p>0$, let
$R^{\infty}$ be its perfect closure, and let $M$ be a finitely generated
$R$-module. Then $\operatorname{c\
depth}_{R^{\infty}}(R^{\infty}\otimes_{R}M)$ is finite, and more specifically

$\operatorname{s\ depth}_{R}(M)\geq\operatorname{c\
depth}_{R^{\infty}}(R^{\infty}\otimes_{R}M)$

###### Proof.

Let $\bf x^{\prime}$
$=x_{1}^{\frac{1}{q_{1}}},\ldots,x_{n}^{\frac{1}{q_{n}}}\subset R^{\infty}$ be
an $(R^{\infty}\otimes_{R}M)$-regular sequence. Then by properties of regular
sequences [Nor76, Theorem 5.1.3], $\bf x$ $:=x_{1},\ldots,x_{n}\subset
R\subseteq R^{\infty}$ is $(R^{\infty}\otimes_{R}M)$-regular as well. By lemma
4.7, $\bf x$ is $F^{e}(M)$-regular for all $e$. Hence any maximal
$(R^{\infty}\otimes_{R}M)$-sequence can have at most the length of
$\operatorname{s\
depth}_{R}(M)=\operatorname{depth}_{R}\big{(}F^{e}(M)\big{)}$ for $e\gg 0$. ∎

After c depth, the next largest depth value of $(R^{\infty}\otimes_{R}M)$ over
$R^{\infty}$ is the Koszul depth. We find that it coincides with the
stabilizing depth.

######  Notation 4.8.

Before we discuss the next result, we make note of some notation used
throughout its proof. If $\mathbf{y}\subset R^{\frac{1}{q}}\subseteq
R^{\infty}$ is some finite sequence for some $q$ (perhaps even
$\mathbf{y}\subset R$), we can construct the koszul complex
$K_{\bullet}(\mathbf{y};F^{e^{\prime}}(M))$ for any $e^{\prime}\geq e$, since
$R^{\frac{1}{q}}\subseteq R^{\frac{1}{q^{\prime}}}$. But we can also construct
$K_{\bullet}(\mathbf{y};R^{\infty}\otimes_{R}M)$. Since both such complexes
are built from the same finite sequence $\mathbf{y}$, the matrices defining
the differentials $d_{j}$ are identical in each version. When we discuss the
differentials for the Koszul complex over $F^{e}(M)$ we write $d_{j}^{q}$,
while the differentials over $R^{\infty}\otimes_{R}M$ are denoted
$d_{j}^{\infty}$.

Lastly we require a lemma.

######  Lemma 4.9.

Let $(R,\mathfrak{m})$ be an F-pure Noetherian local ring of characteristic
$p>0$, let $(R^{\infty},\mathfrak{m}^{\infty})$ be its perfect closure, and
let $M$ be a finitely generated $R$-module. Then,

$\operatorname{k\ depth}_{R^{\infty}}(R^{\infty}\otimes_{R}M)=\
\sup\\{\operatorname{k\
gr}_{R^{\infty}}\big{(}\mathfrak{m}^{[\frac{1}{q}]}R^{\infty},(R^{\infty}\otimes_{R}M)\big{)}\\}$
for all $e\in\mathbb{N}$.

###### Proof.

We know that $\operatorname{k\ depth}_{R^{\infty}}(R^{\infty}\otimes_{R}M)=$
sup$\\{\operatorname{k\
gr}_{R^{\infty}}\big{(}I,(R^{\infty}\otimes_{R}M)\big{)}\\}$, where $I\subset
R^{\infty}$ is _any_ finitely generated sub-ideal. We therefore must show that
the ideals $\mathfrak{m}^{[\frac{1}{q}]}R^{\infty}$ determine the Koszul
depth.

Recall for ideals $I\subseteq J\subset R^{\infty}$, we have $\operatorname{k\
gr}_{R^{\infty}}(I,M)\leq\operatorname{k\ gr}_{R^{\infty}}(J,M)$.

Let $I=(y_{1}^{\frac{1}{q_{1}}},\ldots,y_{n}^{\frac{1}{q_{n}}})\subset
R^{\infty}$ be a finitely generated ideal. Then
$I\subseteq\mathfrak{m}^{\infty}$ since $(R^{\infty},\mathfrak{m}^{\infty})$
is a local ring. Letting $q\geq\text{max}\\{q_{i}\\}$, we see that
$I\subseteq\mathfrak{m}^{[\frac{1}{q}]}R^{\infty}$. Hence for _any_ finitely
generated $I\subset R^{\infty}$, we have $\operatorname{k\
gr}_{R^{\infty}}(I,M)\leq\operatorname{k\
gr}_{R^{\infty}}(\mathfrak{m}^{[\frac{1}{q}]}R^{\infty},M)\big{)}$ for $q\gg
0$, since it is a sub-ideal. Therefore, ideals of the form
$\mathfrak{m}^{[\frac{1}{q}]}R^{\infty}$ achieve sup$\\{\operatorname{k\
gr}_{R^{\infty}}\big{(}I,(R^{\infty}\otimes_{R}M)\big{)}\\}$. ∎

######  Theorem 6.

Let $(R,\mathfrak{m})$ be an F-pure Noetherian local ring of characteristic
$p>0$, let $R^{\infty}$ be its perfect closure, and let $M$ be a finitely
generated $R$-module. Then $\operatorname{k\
depth}_{R^{\infty}}(R^{\infty}\otimes_{R}M)$ is finite, and more specifically

$\operatorname{k\ depth}_{R^{\infty}}(R^{\infty}\otimes_{R}M)=\operatorname{s\
depth}_{R}(M)$

###### Proof.

To show $\operatorname{k\
depth}_{R^{\infty}}(R^{\infty}\otimes_{R}M)\geq\operatorname{s\
depth}_{R}(M)$, let $\mathbf{x}=x_{1},\ldots,x_{n}$ be minimal system of
generators for $\mathfrak{m}$. We claim that we have the comparison
$\operatorname{k\
gr}_{R^{\infty}}\big{(}(\mathbf{x})R^{\infty},(R^{\infty}\otimes_{R}M)\big{)}\geq\operatorname{s\
depth}_{R}(M)$, which will prove the inequality, since $\operatorname{k\
depth}_{R^{\infty}}(R^{\infty}\otimes_{R}M)=\operatorname{k\
gr}_{R^{\infty}}\big{(}\mathfrak{m}^{\infty},(R^{\infty}\otimes_{R}M)\big{)}\geq\operatorname{k\
gr}_{R^{\infty}}\big{(}(\mathbf{x})R^{\infty},(R^{\infty}\otimes_{R}M)\big{)}$.

To prove this claim, let $\operatorname{depth}_{R}\big{(}F^{e}(M)\big{)}\geq
d$ for all $e$, and let $j>n-d$. Fix $e$. Recall that since $R$ is Noetherian,
$\operatorname{k\ depth}_{R}(F^{e}(M))=\operatorname{depth}_{R}(F^{e}(M))$.
Thus $H_{j}\big{(}\mathbf{x};F^{e}(M)\big{)}=0$, and we must also show
$H_{j}\big{(}\mathbf{x};(R^{\infty}\otimes_{R}M)\big{)}=0$. Fix
$z_{\infty}\in\text{ker}(d_{j}^{\infty})$. Then
$z_{e}\in\text{ker}(d_{j}^{q})$. By exactness at this $j$-th homology group,
there exists a non-zero $y_{e}\in\big{(}F^{e}(M)\big{)}^{n\choose j+1}$ such
that $d_{j+1}^{q}(y_{e})=z_{e}$. By $F$-purity of $R$, we have an injection
$F^{e}(M)^{n\choose j+1}\hookrightarrow(R^{\infty}\otimes_{R}M)^{n\choose
j+1}$, hence the image $y_{\infty}$ in $(R^{\infty}\otimes_{R}M)^{n\choose
j+1}$ is non-zero. But now $d_{j+1}^{\infty}(y_{\infty})=z_{\infty}$.

Since $z_{\infty}\in\text{ker}(d_{j}^{\infty})$ was chosen arbitrarily,
therefore $\text{ker}(d_{j}^{\infty})=\text{image}(d_{j+1}^{\infty})$, and
$H_{j}\big{(}\mathbf{x};(R^{\infty}\otimes_{R}M)\big{)}=0$.

To show the reverse inequality, again let $\mathbf{x}=x_{1},\ldots,x_{n}$ be
minimal system of generators for $\mathfrak{m}$, and recall by lemma 4.9,
$\operatorname{k\
depth}_{R^{\infty}}(R^{\infty}\otimes_{R}M)=\sup\\{\operatorname{k\
gr}_{R^{\infty}}\big{(}(\mathbf{x})^{[\frac{1}{q}]}R^{\infty},(R^{\infty}\otimes_{R}M)\big{)}\\}$.
We claim that for any $e$, $\operatorname{k\
gr}_{R^{\infty}}\big{(}(\mathbf{x})^{[\frac{1}{q}]}R^{\infty},(R^{\infty}\otimes_{R}M)\big{)}\leq\operatorname{depth}_{R}\big{(}F^{e}(M)\big{)}$.
Hence we know $\operatorname{k\ depth}_{R^{\infty}}(R^{\infty}\otimes_{R}M)$
is finite, and since the statement is true for $e\gg 0$, the inequality is
true.

To prove this claim, fix $e$. Let $\operatorname{k\
gr}_{R^{\infty}}\big{(}(\mathbf{x})^{[\frac{1}{q}]}R^{\infty},(R^{\infty}\otimes_{R}M)\big{)}=d$,
and $j>n-d$, whereby
$H_{j}\big{(}\mathbf{x}^{[\frac{1}{q}]};(R^{\infty}\otimes_{R}M)\big{)}=0$. We
must show that $H_{j}\big{(}\mathbf{x};F^{e}(M)\big{)}=0$ as well, where
$\mathbf{x}$ acts on the left, which is equivalent to right action by
$\mathbf{x}^{\frac{1}{q}}\in R^{\frac{1}{q}}$. Therefore, we can equivalently
ask whether $H_{j}\big{(}\mathbf{x}^{[\frac{1}{q}]};F^{e}(M)\big{)}=0$ via
right action. Fix $z_{e}\in\text{ker}(d_{j}^{q})$, whereby
$z_{\infty}\in\text{ker}(d_{j}^{\infty})$. Then
$\overline{z_{\infty}}=0\in\text{coker}(d_{j+1}^{\infty})$ by exactness of
$H_{\bullet}\big{(}\mathbf{x}^{\frac{1}{q}};(R^{\infty}\otimes_{R}M)\big{)}$
at the $j$-th position. But
$\text{coker}(d_{j+1}^{\infty})=\text{coker}(d_{j+1}^{q}\otimes_{R}1_{R^{\infty}})=\text{coker}(d_{j+1}^{q})\otimes_{R}R^{\infty}$,
and since $R$ is $F$-pure we have an injection
$\text{coker}(d_{j+1}^{q})\hookrightarrow\text{coker}(d_{j+1}^{q})\otimes_{R}R^{\infty}$.
Since $\overline{z_{\infty}}=0$ in the image of this map, therefore its
preimage $\overline{z_{e}}=0$ as well.

Since $z_{e}\in\text{ker}(d_{j}^{q})$ was chosen arbitrarily,
$H_{\bullet}\big{(}\mathbf{x};F^{e}(M)\big{)}$ is exact at the $j$-th
position. ∎

Combining these two results, for $R$ and $M$ as above:

$\operatorname{k\ depth}_{R^{\infty}}(R^{\infty}\otimes_{R}M)=\operatorname{s\
depth}_{R}(M)\geq\operatorname{c\ depth}_{R^{\infty}}(R^{\infty}\otimes_{R}M)$

In order to find a condition under which this last inequality is an equality,
currently we require a further supposition on our rings. Namely, we require
rings which either satisfy countable prime avoidance, or a specific
application of the prime avoidance lemma is sufficient.

######  Definition 4.10.

A Noetherian ring $R$ satisfies the countable prime avoidance lemma if for any
ideal $I\subset R$, and countable collection of prime ideals $\\{P_{e}\\}$, if
$I\subseteq\bigcup_{e\in\mathbb{N}}P_{e}$, then $I\subseteq P_{e}$ for some
$e$.

######  Remark 4.11.

Countable prime avoidance is satisfied by any complete Noetherian local ring,
or any ring which contains uncountably many elements
$\\{u_{\lambda}\\}_{\lambda\in\Lambda}$ for which $u_{\lambda}-u_{\mu}$ is a
unit for $\lambda\neq\mu$. In particular, any ring which contains an
uncountable field is an example of this second condition (see [LW12, Lemma
13.2], [Bur72, Lemma 3], [SV85], [HH00]).

######  Theorem 7.

Let $(R,\mathfrak{m})$ be an F-pure Noetherian local ring of characteristic
$p>0$, and let $M$ be a finitely generated $R$-module. If either of the
following two conditions hold:

  1. (1)

$R$ satisfies countable prime avoidance

  2. (2)

$\displaystyle\bigcup\bigcup_{e}\operatorname{Ass}_{R}\big{(}F^{e}(M)/F^{e}(M)\mathbf{y}\big{)}$
contains finitely many prime ideals, where $\bf y$ $\subset R$ is a maximal
$(R^{\infty}\otimes_{R}M)$ sequence in $R^{\infty}$

then,

$\operatorname{s\ depth}_{R}(M)=\operatorname{c\
depth}_{R^{\infty}}(R^{\infty}\otimes_{R}M)$

###### Proof.

We have already shown that $\operatorname{s\ depth}_{R}(M)\geq\operatorname{c\
depth}_{R^{\infty}}(R^{\infty}\otimes_{R}M)$.

Conversely, suppose that $\operatorname{s\ depth}_{R}(M)>\operatorname{c\
depth}_{R^{\infty}}(R^{\infty}\otimes_{R}M)=d$. Let
$\mathbf{y}=y_{1},\ldots,y_{d}\subset R$ be a maximal
$(R^{\infty}\otimes_{R}M)$ sequence. Then for all $e$,
$\operatorname{depth}_{R}\big{(}F^{e}(M)/F^{e}(M)\mathbf{y}\big{)}>0=\operatorname{c\
depth}_{R^{\infty}}\big{(}(R^{\infty}\otimes_{R}M)/(R^{\infty}\otimes_{R}M)\mathbf{y}\big{)}$.
We claim that
$\mathfrak{m}=\displaystyle\bigcup\bigcup_{e}\operatorname{Ass}_{R}\big{(}F^{e}(M)/F^{e}(M)\mathbf{y}\big{)}$.
Note that this union consists of at most countably many prime ideals since it
is a countable union of finite sets.

Clearly the union is contained in $\mathfrak{m}$, since $(R,\mathfrak{m})$ is
local. Conversely, suppose that there exists some
$z\in\mathfrak{m}\setminus\displaystyle\bigcup\bigcup_{e}\operatorname{Ass}_{R}\big{(}F^{e}(M)/F^{e}(M)\mathbf{y}\big{)}$.
Then $z$ is not contained in any associated prime of
$F^{e}(M)/F^{e}(M)\mathbf{y}$ for any $e$. Hence $z$ is regular over
$F^{e}(M)/F^{e}(M)\mathbf{y}$ for all $e$. Then $\\{\mathbf{y},z\\}$ is a
right regular $F^{e}(M)$-sequence for all $e$, and hence it is
$(R^{\infty}\otimes_{R}M)$-regular by lemma 4.7. But this fact contradicts the
maximality of $\mathbf{y}$.

Now suppose either condition of the theorem holds. Since $\mathfrak{m}$ is
contained in this union of prime ideals, then $\mathfrak{m}\subseteq P$ for
some $P\in\operatorname{Ass}_{R}\big{(}F^{e}(M)/F^{e}(M)\mathbf{y}\big{)}$ and
for some $e$. But since $\mathfrak{m}$ is maximal, $\mathfrak{m}=P$. We now
have the maximal ideal of $R$ shown to be an associated prime of
$F^{e}(M)/F^{e}(M)\mathbf{y}$, and hence
$\operatorname{depth}_{R}\big{(}F^{e}(M)/F^{e}(M)\mathbf{y}\big{)}=0$. We have
contradicted our assumption that
$\operatorname{depth}_{R}\big{(}F^{e}(M)/F^{e}(M)\mathbf{y}\big{)}>0$. ∎

## Acknowledgments

The author would like to thank his thesis advisor Neil Epstein for his
guidance throughout this project. And though our formal mentorship is over, he
has still been incredibly generous with his time while proofreading rough
drafts of this paper. In this latest role, he can particularly be thanked for
discussions of $F$-sequences, and the suggestion of lemma 2.17.

## References

  * [Bar72] Floyd Barger, _A theory of grade for commutative rings_ , Proc. Amer. Math. Soc. 36 (1972), no. 2, 365–368.
  * [BH97] Winfried Bruns and Jürgen Herzog, _Cohen-Macaulay rings_ , revised ed., Cambridge Studies in Advanced Mathematics, no. 39, Cambridge Univ. Press, Cambridge, 1997.
  * [Bur72] Lindsay Burch, _Codimension and analytic spread_ , Proc. Cambridge Philos. Soc. 72 (1972), 369–373.
  * [ES14] Neil Epstein and Jay Shapiro, _Strong Krull primes and flat modules_ , J. Pure Appl. Algebra 218 (2014), no. 9, 1712–1729.
  * [Fed83] R. Fedder, _F-purity and rational singularity_ , Trans. Amer. Math. Soc. 278 (1983), no. 2, 461–480.
  * [Gre65] Marvin J. Greenberg, _Perfect closures of rings and schemes_ , Proc. Amer. Math. Soc. 16 (1965), 313–317.
  * [HH00] Melvin Hochster and Craig Huneke, _Localization and test exponents for tight closure_ , Michigan Math. J. 48 (2000), 305–329.
  * [Hoc74] Melvin Hochster, _Grade-sensitive modules and perfect modules_ , Proc. London Math. Soc. (3) 29 (1974), 55–76.
  * [HR76] Melvin Hochster and Joel L. Roberts, _The purity of Frobenius and local cohomology_ , Adv. Math. 21 (1976), no. 2, 117–172.
  * [IR84] Juana Iroz and David E. Rush, _Associated primes in non-Noetherian rings_ , Canad. J. Math. 36 (1984), no. 2, 344–360.
  * [Jor82] D.A. Jordan, _Bijective extensions of injective ring endomorphisms_ , J. London Math. Soc. (2) 25 (1982), 435–448.
  * [LW12] G. Leuschke and R. Wiegand, _Cohen-Macaulay representations_ , second ed., Mathematical Surveys and Monographs, vol. 181, American Mathematical Society, Providence, RI, 2012.
  * [Mat86] Hideyuki Matsumura, _Commutative ring theory_ , Cambridge Studies in Advanced Mathematics, no. 8, Cambridge Univ. Press, Cambridge, 1986, Translated from the Japanese by M. Reid.
  * [Nor76] Douglas G. Northcott, _Finite free resolutions_ , Cambridge Tracts in Mathematics, no. 71, Cambridge Univ. Press, Cambridge, 1976.
  * [NS04] Nicole Nossem and Rodney Y. Sharp, _Ideals in a perfect closure, linear growth of primary decompositions, and tight closure_ , Trans. Amer. Math. Soc. 356 (2004), no. 9, 3687–3720.
  * [SS04] Anurag K. Singh and Irena Swanson, _Associated primes and local cohomology modules and of Frobenius powers_ , Int. Math. Res. Not. 33 (2004), 1703–1733.
  * [SV85] Rodney Y. Sharp and Peter Vámos, _Baire’s category theorem and prime avoidance in complete local rings_ , Arch. Math. (Basel) (1985), no. 44, 243–248.