Divided power algebra is artinian as a module over the polynomial ring

Question

I already asked this on math.stackexchange.com, but did not receive much responses. I hope this is also appropriate for mathoverflow.

In the paper Homological algebra on a complete intersection, with an application to group representations by Eisenbud I found the following argument that I do not understand:

Suppose $B$ is a local artinian ring (in fact in the context it is a quotient of a regular ring by a maximal regular sequence, i.e a complete intersection). Then we can take the divided power algebra $D^n(B)$ for $n \in \mathbb{N}_0$, which is a free $B$-module, generated by elements $z^{(\alpha)}$ where $\alpha \in \mathbb{N}_0^n$.

This is actually a module over the polynomial ring $B[t_1,\dots,t_n]$, with the action defined by $t_i . z^{(\alpha)} := z^{(\alpha - e_i)}$. Here $e_i = (0,\dots,0,1,0,\dots,0)$ denotes the $i$-th basis vector.

The algebra structure is defined by $z^{(\alpha)}\cdot z^{(\beta)} = \frac{(\alpha + \beta)!}{\alpha!\beta!}z^{(\alpha + \beta)}$, though that is not relevant to my question.

Eisenbud now writes $$D^n(B) = \bigoplus_{k \in \mathbb{N}_0} \text{Hom}_B(B[t_1,\dots,t_n]_k, B)$$ which is clear, because the $\{z^{(\alpha)} : |\alpha|=k\}$ might be considered as the dual basis to $\{t^\alpha : |\alpha|=k\}$.

The argument I do not understand is that Eisenbud writes

$D^n(B)$ is artinian, because it is the dual of a noetherian $B[t_1,\dots,t_n]$-module.

Some thoughts I had about this:

1. This claim is actually false if $B$ itself is not artinian, because I might choose an infinite descending chain $I_0 \subset I_1 \subset \dots$ and then $I_0 \cdot D^n(B) \subset I_1 \cdot D^n(B) \subset \dots$ would be an infinite descending sequence in $D^n(B)$. So I would expect some argument using that $B$ is artinian.
2. To show that $D^n(B)$ is artinian, would it be enought to consider only graded submodules? I already asked this in a different question, but here lies my motivation, see https://math.stackexchange.com/questions/2978049/ascending-descending-chain-condition-on-graded-modules
3. If we had an inclusion reversing bijection $$\{N \subset D^n(B) \text{ submodule}\} \rightarrow \{I \subset B[t_1,\dots,t_n] \text{ ideal}\}$$ this would proof the claim, but I don't see any. Taking annihilators is an inclusion reversing map, but does not seem to be bijective, as a fellow student of me found the following counterexample:

   Let $B = \mathbb{Z}/8, n = 1$ and let $M\subset D^1(B)$ be the submodule generated by all elements of the form $2 z^{(k)} + z^{(k - 1)}$ for $k \in \mathbb{N}$. This is not the full module, because $M \cap D^1(B)_0 = (2) \subset B$, but its annihilator is $(0) \subset B[t_1]$: For any $0\neq f\in B[t_1]$, let $d = \text{deg}(f)$, then $f\cdot (2z^{(d+1)} + z^{(d)})$ can not be $0$, because $t_1$ maps $2z^{(k)} + z^{(k-1)}$ to $2z^{(k - 1)} + z^{(k - 2)}$.

Answer 1

You state no question, so I am not sure what you are asking. The argument Eisenbud seemed to have in mind is flawed, as you noticed. However the result is true:

Let $I$ be the maximal ideal of $B$, and $k = B / \mathfrak{m}$. The $B[t_1,\dots,t_n]$-module $D^n(B)$ has a finite filtration by the submodules $I^r D^n(B)$, with successive quotients $D^n(k) \otimes_k I^r/I^{r+1}$ where $I^r/I^{r+1}$ is a finite dimensional $k$-vector space, so it is enough to show that $D^n(k)$ is an artinian $k[t_1,\dots,t_n]$-module. Any sub- $k[t_1,\dots,t_n]$-module of $D^n(k)$ is also a sub-$k[[t_1,\dots,t_n]]$-module, and the annihilator map \begin{align*} \{ k[t_1,\dots,t_n]\text{-submodules of } D^n(k) \} & \rightarrow \{ \text{ideals of } k[[t_1,\dots,t_n]] \} \end{align*} is injective and inclusion reversing. Since $k[[t_1,\dots,t_n]]$ is a noetherian ring, we get the conclusion.

Answer 2

So we solved this today, in a way similar to how the Hilbert basis theorem might be proved. First show the following lemma, then the claim that $D^n(B)$ is artinian as a module over $B[t_1,\dotsc,t_n]$ is a direct consequence.

Lemma If $M$ is an artinian $R$-module, then $D^1 \otimes M$ is an artinian $R[t]$ module.

Proof: We can think of elements in $D^1 \otimes M$ as polynomials with coefficients in $M$. So each element $f$ has a degree $\deg f$ and a leading coefficient $lc(f)$. Now consider a descending chain $$ N_1 \supset N_2 \supset \dots$$ an set $M_{k, l } := \{ lc(f) : f \in N_k, \deg(f) = l \} \subset M$. Clearly $M_{k,l}\supset M_{k+1,l}$ and $M_{k,l}\supset M_{k,l+1}$.

Now observe that $N_k = N_{k+1}$ if and only if $M_{k,l} = M_{k+1,l}$ for all $l$. "$\Rightarrow$" is clear, so for "$\Leftarrow$" consider $f \in N_k \setminus N_{k+1}$ with minimal degree. Then there is some $g \in N_{k+1}$ with $lc(f) = lc(g)$ and $\deg(f) = \deg(g)$. So $f - g \in N_k\setminus N_{k+1}$ has lower degree, which is a contradiction.

Next consider the diagram \begin{align*} \vdots && \vdots && \vdots \\ \cap && \cap && \cap\\ M_{3,1} &\supset& M_{3,2} &\supset& M_{3,3} &\supset& \dots\\ \cap && \cap && \cap\\ M_{2,1} &\supset& M_{2,2} &\supset& M_{2,3} &\supset& \dots\\ \cap && \cap && \cap \\ M_{1,1} &\supset& M_{1,2} &\supset& M_{1,3} &\supset& \dots \end{align*} Now the sequence $M_{1,1} \supset M_{2,2} \supset M_{3,3} \supset \dots$ has to stabilize at some point, because $M$ is artinian, say at $M_{k,k}$. Now it is easy to see that $M_{n,m} = M_{k,k}$ for all $n,m\geq k$. Thus there are only finitely many rows in the diagram which are not yet equal, but they will equal after finitely many steps. So the criterion above applies and we see that the sequence $N_k$ has to stabilize, proving the claim.

This directly proves the following

Corollary If $B$ is artinian, then $D^n(B)$ is artinian as a module over $B[t_1,\dotsc,t_n]$.

Proof: Apply the lemma inductively.