Local cohomology over Steenrod algebras

Introduction

Let $\mathbb{F}_q$ be a finite field with $q=p^s$ elements. Let $S=\bigoplus_{n\geq0}S_n$ be a non-negatively graded Noetherian $\mathbb{F}_q$-algebra with $S_0=\mathbb{F}_q$. Assume $S$ has a structure of an unstable Steenrod algebra. Recall that this means there exists $\mathbb{F}_q$ linear maps $\mathcal{P}^i:S\to S$ with $\mathcal{P}^0$ to be the identity and satisfy the Cartan identity:

We have for homogeneous $u$, $\mathcal{P}^{\deg u}(u)=u^q$ and $\mathcal{P}^i(u)=0$ for $i>\deg u$. We also have that $\mathcal{P}^r(S_n)\subseteq S_{n+r(q-1)}$ for all $n\in\mathbb{Z}$.

A large class of examples of unstable Steenrod algebras are rings of invariants $S$ of a finite group $G\subseteq GL_d(\mathbb{F}_q)$. We note that $GL_d(\mathbb{F}_q)$ is a finite group and so $S$ contains $D^*(d)$ the ring of invariants of $GL_d(\mathbb{F}_q)$. Dickson proved that $D^*(d)$ is a polynomial ring with $D^*(d)=\mathbb{F}_q[d_{d,0},d_{d,1},\ldots,d_{d,d-1}]$ with $\deg d_{d,i}=q^d-q^i$. We follow the exposition given in [10].

One of the landmark results in modular invariant theory is the proof by D. Bourguiba and S. Zarati, see [1], of the Landweber-Stong conjecture [5]. This result states that $\operatorname{depth}S\geq r$ if and only if $d_{d,d-1},\ldots,d_{d,d-r}$ is an $S$-regular sequence. In our paper [9] we gave an alternate and considerably simpler proof of this conjecture.

D. Bourguiba and S. Zarati prove a considerably more general result than the Landweber-Stong conjecture, see [1 1.3] A natural question was whether our techniques generalize. In this paper we show that our techniques from [9] generalize and give an alternate and considerably simpler proof of D. Bourguiba and S. Zarati’s result.

The crucial point is the notion of a Cartan $S$-module which was introduced in [9].

It is easily seen that if $M$ is a Cartan $S$-module then $\operatorname{ann}_S M$ is a $\mathcal{P}^*$-invariant ideal of $S$, see 3.10. The main result which enables all our applications is the following:

Applications:

I. We give a considerably simpler and a direct proof of Bourguiba and Zarati’s result.

II. Associate primes of Cartan modules

An ideal $I$ in $S$ is said to be $\mathcal{P}^*$-invariant if $\mathcal{P}^*(I)\subseteq I$ for all $i\geq0$. We prove

Proof of Theorem 1.5 requires that if $M$ is a Cartan $S$-module then for any ideal $J$ of $S$ we have $H_J^0(M)$ is a Cartan submodule of $M$. This follows from Theorem 1.2. However we also give a direct proof. Theorem 1.5 is not true if $M$ is not finitely generated. We give a counter-example, see 8.5.

$S_r$ property

Let $T$ be a Noetherian ring $T$ and let $M$ be a finitely generated $T$-module. Set depth $0=+\infty$. Recall $M$ satisfies $S_r$ property of Serre if

A natural question is what happens if $d_{d,0}^{q^t},d_{d,1}^{q^t},\cdots d_{d,r-1}^{q^t}$ is an $M$-regular sequence. We extend our result in [9 1.6] to prove

Cohomological annihilators

1.7. If $M$ is a finitely generated Cartan $S$-module then $H^i_{S_+}(M)$ is a Cartan $S$-module. It is not difficult to prove that $\operatorname{height}\operatorname{ann} H^i_{S_+}(M)\ge d-i$, see 6.5. As $\operatorname{ann} H^i_{S_+}(M)$ is $P^*$-invariant it follows that $d_{d,0}^{q^t}\in\sqrt{\operatorname{ann} H^i_{S_+}(M)}$ for $i<d=\dim S$. Thus some power $d_{d,0}^{q^t}$ works as a cohomological annihilator. We show

Proof. This follows from an exercise problem [3 8.1.6] and the discussion in 1.7. □

In [10 11.5.5] it is proved that if $S$ is a ring of invariants then $S_{d_{d,0}}$ is Cohen-Macaulay. We extend this result.

Remark 1.10. If $\dim M<d$ and $M$ is a Cartan module then by 3.12, $d_{d,0}^{q^t}\in\sqrt{\operatorname{ann}\overline{M}}$. So $M_{d_{d,0}^{q^t}}=0$. Thus Corollary 1.9 has content only if $\dim M=\dim S=d$.

Monomial ideals

Let $R=\mathbb{F}_q[X_1,\ldots,X_d]$ with the natural Steenrod algebra structure. Let $I$ be a monomial ideal in $R$. It is not difficult to show that $I$ is a $P^*$-invariant ideal of $R$, see 2.3. As an application of this observation we prove:

Finite generation of cohomology

Let $S$ be an unstable Steenrod algebra. If $\dim S\ge2$ then there are infinitely many homogeneous prime ideals. Note there exists only finitely many $\mathcal{P}^{*}$-invariant prime ideals. For a prime ideal $P$ of $S$ define

We prove

We now describe in brief the contents of this paper. In section two we discuss a few preliminary results that we need. In Section three we discuss our notion of Cartan modules and discuss a few preliminary properties. In section four we give an alternate description of Cartan $S$-modules which is convenient for us. In the next section discuss a Cartan $S$-module structure on $M_x$ where $x\in S$ is homogeneous and $M$ is a Cartan $S$-module. In section six we prove Theorem 1.2. In the next section we give a proof of Theorem 1.4. In section eight we give proof of Theorem 1.5. In the next section we give proof of our applications III. Finally in section ten we give a proof of Corollary 1.9. In the next section we give a proof of Theorem 1.11. Finally in section twelve we give a proof of Theorem 1.12.

Preliminaries

Steenrod operators

We give the following well-known construction of Steenrod operators on polynomial rings and ring of invariants. We follow the exposition given in [10].

Let $p$ be a prime, $q=p^s$ and let $V$ be a $\mathbb{F}_q$-vector space of dimension $d$. Define $\mathcal{P}(\xi):\mathbb{F}_q[V]\to\mathbb{F}_q[V][\xi]$ (with $\deg\xi=1-q$) by the rules:

1. $\mathcal{P}(\xi)(v)=v+v^q\xi$ for all $v\in V^*$.

2. $\mathcal{P}(\xi)(u+w)=\mathcal{P}(\xi)(u)+\mathcal{P}(\xi)(w)$ for all $u,w\in\mathbb{F}_q[V]$.

3. $\mathcal{P}(\xi)(uw)=\mathcal{P}(\xi)(u)\mathcal{P}(\xi)(w)$ for all $u,w\in\mathbb{F}_q[V]$.

4. $\mathcal{P}(\xi)(1)=1$.

We note that $\mathcal{P}(\xi)$ is a ring homomorphism of degree zero.

By separating out homogeneous components we obtain $\mathbb{F}_q$ linear maps $\mathcal{P}^i:\mathbb{F}_q[V]\to\mathbb{F}_q[V]$ by the requirement

Note $\mathcal{P}^0$ is the identity. The operations $\mathcal{P}^i$ are called the Steenrod reduced power operations over $\mathbb{F}_q$ and when $p=2$, also denoted by $Sq^i$ and referred to as Steenrod squaring operations. We will use the following Cartan formulae:

We have for homogeneous $u$, $\mathcal{P}^{\deg u}(u)=u^q$ and $\mathcal{P}^i(u)=0$ for $i>\deg u$. We also have that $\mathcal{P}^r(\mathbb{F}_q[V]_n)\subseteq\mathbb{F}_q[V]_{n+r(q-1)}$ for all $n\in\mathbb{Z}$.

##

Let $G \subseteq GL(V)$ be a finite group. Let $S = \mathbb{F}_q[V]^G$. We define an action of $G$ on $\mathbb{F}_q[V][\xi]$ by giving the trivial action on $\xi$. We note that $\sigma(P(\xi)(r)) = P(\xi)(\sigma(r))$ for all $\sigma\in G$. Thus the maps $\mathcal{P}^i$ restrict to maps from $S$ to $S$. So $S$ is an unstable Steenrod algebra.

We will need the following easily proved result.

We now give

Proof of Proposition 2.3. (1) By using 2.4 repeatedly it suffices to prove that $(X_i)$ is $\mathcal{P}^*$-invariant. But this is obvious.

(2) This follows from (1). $\square$

Cartan Modules

In this section we describe our notion of Cartan modules. We work with the setup as in 1.3. Let $\mathcal{P}^i:S \to S$ be the $i^{\text{th}}$-Steenrod operator on $S$.

Next we define morphisms of Cartan $S$-modules.

3.3. Let $M,N$ be Cartan $S$-modules. We denote

It is evident that $\operatorname{Hom}_{\mathcal{C}(S)}(M,N)$ is an abelian group. The following result follows easily from the definition of morphism of Cartan $S$-modules.

We will need a few preliminaries.

We also need the following easily proved result.

3.7. Let $C(S)$ denote the category of Cartan $S$-modules and Cartan morphisms. Then by 3.4, 3.5, 3.6 it follows that $C(S)$ is an abelian category. As a consequence we obtain

3.9. An ideal $I$ of $S$ is said to be $\mathcal{P}^*$-invariant if $\mathcal{P}^i(I)\subseteq I$ for all $i\geq0$. We note that if $I$ is $\mathcal{P}^*$-invariant then so is $\sqrt{I}$, see [10 11.2.5].

The following Lemma is critical. We proved it in [9 3.5] for rings of invariants. Same proof works in general.

3.11. The non-zero $\mathcal{P}^*$-prime ideals of $D(d)^{q^t}$ are $(d_{d,0}^{q^t},\cdots,d_{d,i}^{q^t})$ for $i=0,\ldots,d-1$; see [8 9.2.7]. We need the following result. We proved it for ring of invariants, see [9 2.8]. The same proof works in general.

An alternate description of Cartan modules

In this section we give an alternate description of Cartan modules which will be useful later. Throughout $S$ is an unstable Steenrod $\mathbb{F}_q$-algebra.

The total Steenrod operation

4.1. The total Steenrod operation. This is the map defined by $\mathcal{P}(\xi): S \to S[\xi]$ where $\mathcal{P}(\xi)(u) = \sum_{i\geq0} P^{i}(u)\xi^{i}$. We note that $\mathcal{P}(\xi)$ is a ring homomorphism. Furthermore if we set $\deg(\xi)=1-q$ then $\mathcal{P}(\xi)$ is a homogeneous ring homomorphism. We note that $S[\xi]$ has elements of negative degrees (as $\deg(\xi)=1-q$) and so is a $\mathbb{Z}$-graded ring.

Important point

4.2. Important point: (with setup as above). As $\mathcal{P}(\xi): S \to S[\xi]$ is a ring homomorphism, it induces a new $S$ algebra structure on $S[\xi]$.

Definition 4.3. Let $M$ be an $S$-module and let $N$ be an $S[\xi]$-module and let $f:M\to N$ be an additive homogeneous graded map. We say $f$ is a Steenrod map if $f$ is $S$-linear with $N$ given an $S$-module structure via $\mathcal{P}(\xi)$, i.e., $f(um)=\mathcal{P}(\xi)(u)f(m)$ for all $u\in S$ and for all $m\in M$.

Construction of the graded subring

4.4. We need to construct a graded subring $S[[\xi]]^*$ of $S[[\xi]]$. For a homogeneous element $s\in S$ set $|s|=\deg(s)$. For $n\in\mathbb{Z}$, set

It is readily verified that $S[[\xi]]^*_{n}$ is an additive subgroup of $S[[\xi]]$. Furthermore it is easy to check that

Set $S[[\xi]]^*=\bigoplus_{n\in\mathbb{Z}}S[[\xi]]^*_{n}$. It is a graded ring. It also contains $S[\xi]$ as a graded subring.

Construction of the graded submodule

4.5. Let $M$ be a graded $S$-module (not necessarily finitely generated). We need to construct a graded $S[[\xi]]^*$ submodule $M[[\xi]]^*$ of $M[[\xi]]$. For a homogeneous element $m\in M$ set $|m|=\deg(m)$. For $n\in\mathbb{Z}$, set

It is readily verified that $M[[\xi]]^*_{n}$ is an additive subgroup of $M[[\xi]]$. Furthermore it is easy to check that

Set $M[[\xi]]^*=\bigoplus_{n\in\mathbb{Z}}M[[\xi]]^*_{n}$. It is a graded $S[[\xi]]^*$-module. It also contains $M[\xi]$ as $S[\xi]$-submodule.

We now give our characterization of Cartan modules.

Proof. (a) $\Longrightarrow$ (b).

Define $Q_M(\xi): M \to M[[\xi]]^*$ by

with $\deg\xi=1-q$. We note that $Q_M$ takes homogeneous elements of degree $r$ to degree $r$ elements in $M[[\xi]]^*$. Let $u\in S$ and $m\in M$ be homogeneous. We have

The result follows.

(b) $\Longrightarrow$ (a).

Set

Then $Q_M^r:M\to M$ are $\mathbb{F}_q$-linear maps. Furthermore as $Q_M$ is a homogeneous map and $|\xi|=1-q$ it follows that $Q_M^r(M_n)\subseteq M_{n+r(q-1)}$ for all $n\in\mathbb{Z}$. By assumption $Q_M^0(m)=m$ for all $m\in M$. As $Q_M$ is a Steenrod map we have for $u\in S$, $m\in M$ homogeneous

The Cartan identity follows. $\square$

4.7. Let $x$ be a non-nilpotent homogeneous element in $S$. Then we can form the $S[[\xi]]^*$-graded module $S_x[[\xi]]^*$. It is readily verified that $S_x[[\xi]]^*$ is a graded ring. Furthermore the natural homogeneous map $S[[\xi]]^*\to S_x[[\xi]]^*$ is a ring homomorphism. Similarly we can show that if $M$ is a graded $S$-module then $M_x[[\xi]]^*$ is a graded $S_x[[\xi]]^*$-module.

We need the following result.

Proof. (1) and (2). Let $|x|=c$. We show that $\mathcal{P}(\xi)(x)^{-1}\in S_x[[\xi]]^{*}_{-c}$. Set

We prove by induction that $|\alpha_n|=n(q-1)-c$ for all $n\geq0$. We have $\mathcal{P}(\xi)(x)\mathcal{P}(\xi)(x)^{-1}=1$. So we have

Comparing coefficients we have $x\alpha_0=1$. So $\alpha_0=1/x$ has degree $-c$. This also verifies (2).

Let $n\geq1$. By induction suppose $|\alpha_r|=r(q-1)-c$ for $r=0,1,\ldots,n-1$. We have $\sum_{i+j=n}\mathcal{P}^{i}(x)\alpha_j=0$. So

We have

So $|\alpha_n|=n(q-1)-c$. The result follows by induction. $\square$

The Cartan $S$-module structure on $M_x$

In this section we prove two critical results. Throughout $S$ is a Steenrod $\mathbb{F}_q$-algebra, $M$ is a Cartan $S$-module and $x\in S$ is homogeneous. We first prove

The next result is also critical.

We first give

Proof of Theorem 5.1. We may assume that $M_x\ne0$, otherwise there is nothing to prove. In particular $x$ is a non-nilpotent element of $S$. We note that $M_x[[\xi]]$ is a graded $S_x[[\xi]]^*$-module and so an $S[\xi]$-module via the natural map $S[\xi]\to S_x[[\xi]]$. Let $Q_M(\xi):M\to M[[\xi]]^*$ be the total Steenrod operation on $M$. We construct a Steenrod map $Q_{M_x}(\xi):M_x\to M_x[[\xi]]^*$ as

We note that the right hand side always belongs to $M_x[[\xi]]^*$, see 4.8 and 4.7. We first show that the formula is well defined. Suppose $m_1/x^n=m_2/x^l$. Then there exists $r\geq0$ such that $x^{l+r}m_1=x^{n+r}m_2$. As $M$ is a Cartan $S$-module we get $\mathcal{P}(\xi)(x)^{l+r}Q_M(m_1)=\mathcal{P}(\xi)(x)^{n+r}Q_M(m_2)$. So we obtain

Thus we have a well-defined map $Q_{M_x}: M_x \to M_x[[\xi]]^*$. We note that

using the definition of Cartan modules and 4.8. Also note that $Q_{M_x}(\xi)$ is homogeneous of degree zero. For $u \in S$ we have

Thus $Q_{M_x}(\xi)$ is a Steenrod map. The assertion follows from Theorem 4.6. □

Next we give

Proof of Theorem 5.2. We have nothing to show if $M_x=0$ or $M_{xy}=0$. So assume that both are non-zero. So $x$, $xy$ (and hence $y$) are non-nilpotent elements of $S$. We let $i:M_x \to M_{xy}$ denote the natural map $i(m/x^n)=y^nm/(xy)^n$. This induces a natural map $\eta:M_x[[\xi]]^* \to M_{xy}[[\xi]]^*$. We have

From this it easily follows that the map $M_x \to M_{xy}$ is a Cartan morphism. □

Proof of Theorem 1.2

We now give

Proof of Theorem 1.2. Let $C_M=0\to C^0\to C^1\to\cdots\to C^m\to0$ be the modified Čech complex on $\mathbf{x}$ with coefficients in $M$, see [2 5.1.5]. Here

where the differential $\partial^t:C^t\to C^{t+1}$ is given by

if $\{i_1,\ldots,i_t\}=\{j_1,\ldots,\widehat{j_s},\ldots,j_{t+1}\}$ and 0 otherwise. It follows from 5.1 and 3.5 that $C^t$ is a Cartan $S$-module for all $t$. By 5.2 and 3.6 it follows that $\partial^t$ is a morphism of Cartan modules. By 3.8 it follows that $H^i_I(M)$ has a structure of a Cartan $S$-module. □

Remark 6.1. Suppose $\mathbf{x}$ and $\mathbf{y}$ be two different sets of homogeneous generators of $I$. Then we can give a Cartan $S$-module structure on $H^i_I(M)$ which depends on $\mathbf{x}$ and another Cartan $S$-module structure which depends on $\mathbf{y}$. We do not know whether these structures are the same. For our applications it suffices to know there is at least one Cartan $S$-module structure on local cohomology modules.

Remark 6.2. We note that in our proof of Theorem 1.2 the Cartan module $M$ need not be finitely generated as an $S$-module. As a corollary we get that for any homogeneous ideals $I_1,\ldots,I_t$ the iterated local cohomology module

has a structure of a Cartan $S$-module.

Remark 6.3. In the study of local cohomology the so called Lyubeznik functors plays an important role, see [6 p. 44]. Iterated local cohomology modules forms a large subclass of Lyubeznik functors. We do not know whether if $\mathcal{T}$ is a graded Lyubeznik functor on $\operatorname{Mod}(S)$ then $\mathcal{T}(S)$ is a Cartan module. We however suspect that it is a Cartan $S$-module.

6.4. Let $E$ be the $\ast$-injective hull of $\mathbb{F}_q$ as an $S$-module. Set $(-)^\vee= {}^\ast\operatorname{Hom}_S(-,E)$. If $U$ is a finitely generated graded $S$-module then $U^\vee$ is a $\ast$-Artinian $S$-module. If $U$ is a $\ast$-Artinian $S$-module then $U^\vee$ is a finitely generated graded $S$-module. Furthermore the natural map $U \to U^{\vee\vee}$ is an isomorphism if either $U$ is finitely generated graded or when $U$ is a $\ast$-Artinian $S$-module.

6.5. If $M$ is a finitely generated graded $S$-module then $H^i_{S_+}(M)$ are $\ast$-Artinian $S$-module for all $i \ge0$. Furthermore it is well known that $\dim H^i_{S_+}(M)^\vee$ is $\le i$, for instance see [9 10.1].

Corollary 6.6. Let $M$ be a finitely generated graded Cartan $S$-module. Set $J_i=\operatorname{ann} H^i_{S_+}(M)$. Then $d^q_{d,0},\ldots,d^q_{d,d-i-1}\in\sqrt{J_i}$ for $i<d$.

Proof. By 1.2 we get that $H^i_{S_+}(M)$ is a Cartan-module. So $J_i$ is a $\mathcal{P}^\ast$-invariant ideal of $S$. Therefore $\sqrt{J_i}$ is also a $\mathcal{P}^\ast$-invariant ideal of $S$, (see 3.9). Thus $J_i=\operatorname{ann} H^i_{S_+}(M)^\vee$ and the later module has dimension $\le i$, see 6.5. The result follows from 3.12 $\square$

Proof of Theorem 1.4

We make the convention that the depth of the zero module i $+\infty$. In this section we give

Proof of Theorem 1.4. The assertion (ii) $\Longrightarrow$ (i) is trivial. We prove the converse. We note that $d^q_{d,0},\ldots,d^q_{d,d-1}$ is a system of parameters of $S$. So if $M$ is Cohen-Macaulay then we have nothing to prove.

Assume $\operatorname{depth} M=d-g$ where $g>0$. Let $D=D^\ast(d)^{q}$ be the (fractal) Dickson algebra. We consider $M$ as a $D$-module. Set $J=(d^q_{d,d-1},\ldots,d^q_{d,g})$. We note that it suffices to prove $\operatorname{depth} M_P\ge d-g$ for all primes $P$ in $D$ which contain $J$, see [3 1.2.10]. The assertion holds if $P=D_+$ by our assumption. Now let $P\supseteq J$ and $P\ne D_+$. Let $D(u)$ be the $\ast$-canonical module of $D$. We note that $\dim H^i_{S_+}(M)^\vee=\operatorname{Ext}^{d-i}_D(M,D(u))$ has dimension $\le i$ (see 6.6). Set $K_{d-i}=\operatorname{ann}_D\operatorname{Ext}^{d-i}_D(S,D(u))$. So $d^q_{d,0},\ldots,d^q_{d,d-i-1}\in\sqrt{K_{d-i}}$ by 6.6 for $i<d$.

Let $P\supseteq J$ be minimal. Then $P=J$. As $d^q_{d,0}\notin P$ we get that $\operatorname{Ext}^i_D(M,D(u))_P=0$ for $i>0$. So $M_P$ is either Cohen-Macaulay of depth $=d-g$ or $M_P$ is zero.

Let $P\supseteq J$ be such that $\operatorname{height} P=d-g+r$ with $r\ge1$ and that $P\ne D_+$. We assert that $\operatorname{depth} M_P\ge d-g$. If not then $\operatorname{Ext}^{d-g+r-(d-g-i)}_D(M,D(u))_P\ne0$ for some $i > 0$. So we have $\operatorname{Ext}^{r+i}(M,D(u))_{P} \ne0$. We have $U=\operatorname{Ext}^{r+i}(M,D(u))=H^{d-r-i}_{S_{+}}(M)^{\vee}$. So $d_{d,0}^{q^{t}},\cdots,d_{d,r+i-1}^{q^{t}}\in\sqrt{\operatorname{ann}_{D}U}$ by 6.6. As $U_{P}\ne0$ we get that $d_{d,0}^{q^{t}},\cdots,d_{d,r+i-1}^{q^{t}}\in P$. If $t+i-1\ge g$ then note that $D_{+}\subseteq P$ which is a contradiction. So $r+i-1<g$. Thus $P$ contains $d-g+r+i$ variables where $i\ge1$. This is a contradiction as $\operatorname{height}P=d-g+r$. The result follows. □

Proof of Theorem 1.5

8.1. In this section we give a proof of Theorem 1.5. Throughout $S$ is an unstable $\mathbb{F}_{q}$-Steenrod algebra and $M$ is a Cartan $S$-module. The following result follows from Theorem 1.2. We give a direct proof.

We need the following version of a result in [4 3.13].

We prove Theorem 1.5 by induction on $|\operatorname{Ass}_{S}M|$ (here $|W|$ denotes the cardinality of a set $W$). The following result enables our induction.

We now give Proof of Theorem 1.5. We prove the result by induction on $|\operatorname{Ass}_S M|$. We first consider the case when $|\operatorname{Ass}_S M|=1$. Let $\{P\}=\operatorname{Ass}_S M$. Then trivially $P$ is a maximal element of $\operatorname{Ass}_S M$ and so by 8.4 we get that $P$ is $P^*$-invariant.

Now consider the case when $|\operatorname{Ass}_S M|=r$ and the result is known for all finitely generated Cartan $S$-modules $N$ with $|\operatorname{Ass}_S N|=r-1$. Let $P$ be a maximal element of $\operatorname{Ass}_S M$. Then by 8.4 we get $P$ is a $P^*$-invariant prime ideal of $S$. By 8.2 we get that $H_P^0(M)$ is a Cartan submodule of $M$. So $N=M/H_P^0(M)$ is also a finitely generated Cartan $S$-module. By 8.3 we get that $\operatorname{Ass}_S N=\operatorname{Ass}_S M\setminus\{P\}$. So $|\operatorname{Ass}_S N|=r-1$ and by our induction hypotheses we get that $Q$ is a $P^*$-invariant prime ideal of $S$ for every associate prime $Q$ of $N$. Our assertion follows. □

We give an example which shows the result in Theorem 1.5 need not hold if $M$ is not a finitely generated Cartan $S$-module.

Example 8.5. Let $R=\mathbb{F}_q[X_1,\ldots,X_d]$ with $d\geq2$ and the standard Steenrod algebra structure. As discussed earlier $R$ has infinitely many homogeneous prime ideals which are not $P^*$-invariant. Choose one such prime $P$ with height $g\geq1$. Then $M=H_P^g(S)$ is a not finitely generated $R$-module with $\operatorname{Ass} M=\{P\}$. By 1.2 we get that $M$ is a Cartan $S$-module.

Proof of Applications III

Proofs of result stated in our Application III

In this section we give all the proofs of result stated in our Application III. Throughout $S$ is a $\mathbb{F}_q$-Steenrod algebra and $M$ is a finitely generated Cartan $S$-module.

We will need the following result proved in [9 8.2] for invariant rings. The same proof works in general.

The following result follows easily from Theorem 1.2 and 9.2.

The following result is crucial for our applications.

Proof. The assertion (i) $\Longrightarrow$ (ii) is trivial. We now prove (ii) $\Longrightarrow$ (i). We have nothing to show when $r=0$. So assume $r\geq1$.

Let $P$ be a homogeneous prime ideal of $S$. Assume that $\operatorname{height} P=t\geq1$. Set

Set $s_P=\min\{\operatorname{height} P,r\}$. We make the following

Claim: If $P$ is a homogeneous prime ideal then $f_P(M)\geq s_P$. We first prove that the Claim proves the result. Let $P$ be not a $\mathcal{P}^{*}$-invariant homogeneous ideal. Then $H^i_{\mathcal{P}}(M)$ are finitely generated $S$-modules for $i < s_{\mathcal{P}}$. By 9.3 it follows that $H^i_{\mathcal{P}S_{P}}(M_{P})=H^i_{\mathcal{P}}(M)_{P}=0$ for $i < s_{P}$. Thus depth $M_{P} \ge s_{P}$. The result follows.

We note that the assertion $f_{P}(M) \ge s_{P}$ holds trivially when $r=1$.

Next consider the case when $r \ge2$ and the result holds for $\le r-1$. By induction hypothesis, $f_{P}(M) \ge\min\{\operatorname{height} P,r-1\}$ for all homogeneous primes. In particular depth $M_{q} \ge\min\{\operatorname{height} q,r-1\}$ for all homogeneous non $\mathcal{P}^{*}$-invariant prime ideals $q$ of $S$. We also have depth $M_{Q} \ge s_{Q}$ for all $\mathcal{P}^{*}$-invariant prime ideals $Q$ by our assumption. We may assume height $P \ge r$, otherwise there is nothing to show. By [2 13.1.17] we have

If height $q\le r-1$ then depth $M_{q}=\operatorname{height}q$. We have

The second equality holds as $S$ is equi-dimensional and catenary, see [7 Lemma 2, p. 250] for the local version; same proof works for the $*$-local case. If height $q\ge r$ then

Thus $f_{P}(M)\ge r$. So Claim follows. □

Surprisingly the $S_{r}$ property of a Cartan-module only depends on $\mathcal{P}^{*}$-invariant prime ideals in the following sense.

Proof. The assertion (i) $\Longrightarrow$ (ii) is clear.

We now prove (ii) $\Longrightarrow$ (i). It is well-known that it suffices to prove depth $M_{P}\ge\min\{\operatorname{height}P,r\}$ for homogeneous primes in $S$, for instance see [9 10.3] for the case when $M=S$. By 9.4 it follows that depth $M_{P}\ge\min\{\operatorname{height}P,r\}$ for all homogeneous primes $P$ of $S$. The result follows. □

Next we give

Proof of Theorem 1.6. We first prove (ii) $\Longrightarrow$ (i). Let $P$ be a graded $\mathcal{P}^{*}$-invariant prime ideal of $S$. We note that if height $P=g$ then $\dim S/P=d-g$, as $S$ is equi-dimensional and universally catenary, see [7 Lemma 2, p. 250] for the local version; same proof works for the $*$-local case. By 3.12 we get that if $g\le r$ then $P$ contains $d_{d,0}^{q^{t}},\ldots,d_{d,g-1}^{q^{t}}$. It follows that $M_{P}$ is Cohen-Macaulay. If height $P\ge r+1$ then by 3.12 we get that $P$ contains $d_{d,0},\ldots,d_{d,r-1}$. So depth $M_{P}\ge r$. The result follows from 9.5.

Next we prove (i) $\Longrightarrow$ (ii). The extension $D^{*}(d)^{q^{t}}\subseteq S$ is a finite extension. As $(d_{d,0}^{q^{t}},\ldots,d_{d,r-1}^{q^{t}})$ is a prime ideal of height $r$ in $D^{*}(d)$, it follows that $(d_{d,0}^{q^t},\ldots,d_{d,r-1}^{q^t})S$ has height $r$ as an ideal in $S$. As $M$ satisfies $S_r$ it follows from [3 1.2.10] that $\operatorname{grade}((d_{d,0},\ldots,d_{d,r-1})S,M)=r$. The result follows. $\square$

Proof of Corollary 1.9

In this section we give

Proof of Corollary 1.9. By 1.10 we may assume $\dim M=\dim S=d$. We work with the setup as in 1.3. Set $D=D(d)^{q^t}$. Let $D(u)$ be the canonical module of $D$. By local duality we get that $H_{\mathfrak{m}}^{i}(M)^{\vee}=\operatorname{Ext}_{D}^{d-i}(M,D(u))$. So $d_{d,0}^{q^t}\in\sqrt{\operatorname{ann}\operatorname{Ext}_{D}^{d-i}(M,D(u))}$ for $i<d$. We note that we have a finite extension $D_{d,0}^{q^t}\hookrightarrow S_{d,0}^{q^t}$. So it suffices to show that $M_{d,0}^{q^t}$ is Cohen-Macaulay as a $D_{d,0}^{q^t}$-module. Let $P$ be any homogeneous prime in $D$ not containing $d_{d,0}^{q^t}$. Then it follows that $\operatorname{Ext}_{D}^{i}(M,D(u))_P=0$ for $i>0$. It follows that $M_P$ is Cohen-Macaulay. The result follows. $\square$

Proof of Theorem 1.11

In this section we prove Theorem 1.11. To prove this result we need Theorem 23.3 from [7]. Unfortunately there is a typographical error in the statement of Theorem 23.3 in [7]. So we state it here.

Remark 11.2. In [7] $\operatorname{Ass}_{A}(E\otimes G)$ is typed instead of $\operatorname{Ass}_{B}(E\otimes G)$. Also note that $\varphi(\mathfrak{P})=\mathfrak{P}\cap A$ for $\mathfrak{P}\in\operatorname{Spec} B$.

Proof. The inclusion $R\to T$ is faithfully flat. Also note that as $I$ is a monomial ideal it is extended from $R$. Say $I=LT$ for a monomial ideal $L$ of $R$. Set $N=H_{\mathfrak{m}}^{j}(R/L)$. Then $M=N\otimes_R T$. As $R\to T$ is faithfully flat we get that $N$ is a finitely generated $R$-module. By 2.3 we get that $L$ is a $P^*$-invariant ideal of $R$. So $R/L$ is a Cartan $R$-module. By 1.2 we get that $N$ is a Cartan $R$-module. It is also finitely generated as an $R$-module. Thus if $P\in\operatorname{Ass}_{R}N$ then $P$ is a $P^*$-invariant ideal in $R$. So by [10 11.4.1] we get that $P$ is generated by some linear homogeneous polynomials of $R$. In 11.1 we take $A=R$, $B=T$ and $G=T$. We note that if $P\in\operatorname{Spec}(R)$ is homogeneous and generated by linear polynomials then $PT$ is a prime in $T$. The result follows from 11.1. $\square$

Proof of Theorem 1.12

In this section we give Proof of Theorem 1.12. Set $t = \operatorname{depth} S_P$. Suppose if possible $f_P > t$. Then note that $H_P^t(S)$ is finitely generated $S$-module. Then by 9.3 we get that $H_P^t(S)_P = 0$. But $H_{P S_P}^t(S_P) = H_P^t(S)_P = 0$. This is a contradiction as $t = \operatorname{depth} S_P$. □

We give an example which shows that the bound in Theorem 1.12 need not hold for $P^*$-invariant ideals.

Example 12.1. Let $I$ be a monomial ideal in $R = \mathbb{F}_q[X_1,\ldots,X_n]$ such that $S = R/I$ is generalized Cohen-Macaulay but not Cohen-Macaulay. We note that $S$ is an unstable Steenrod $\mathbb{F}_q$-algebra, see 2.3. We note that $f_{S_+} = d = \dim S$ while $\operatorname{depth} S_{S_+} < \dim S_{S_+} = d$. For a specific example see [11 3.5].

Department of Mathematics, Indian Institute of Technology Bombay, Powai, Mumbai 400 076, India

Email address: [email protected]