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:

- 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.
- 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
- 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 $2z^{(k - 1)} + z^{(k - 2)}$.