On finite generation of certain $\operatorname{Ext}$'s

Question

All rings below are commutative.

I have the following situation: $A$ is a commutative ring, $B=A/I$, and I know that $B$ is noetherian. I have a $B$-module $M$ which is finitely generated as a $B$-module, and I consider the following module:

$$\operatorname{Ext}^n_A(B,M).$$

I give it a $B$-module structure via the first argument. My question is if in this generality these $\operatorname{Ext}$'s are finitely generated?

Motivation: Consider a field $k$, and a noetherian $k$-algebra $A$ which might not be finitely generated over $k$. Then I look at

$$\operatorname{Ext}^n_{A\otimes_k A}(A,M).$$

This gives Hochschild cohomology of $A$ over $k$, and I wonder if this is always finitely generated, even if $A$ is really big over $k$.

Thanks

Answer 1

It seems the extension modules need not be finitely generated over $B$. For example, let $k$ be a field, and take $A=k[x_1, x_2, \ldots]$ to be the polynomial ring in countably many variables. Let $I=(x_1, x_2, \ldots)$, so that $B=A/I=k$. Then $B$ is Noetherian, and $M=B=k$ is a finitely generated $B$-module.

To compute $\operatorname{Ext}^1_A(B,M) = \operatorname{Ext}^1_A(k,k)$, we first exhibit the first terms of a projective resolution of $B$ as an $A$-module: $$ \ldots \to A^{(\mathbb{N}\times\mathbb{N})} \stackrel{\alpha}{\to} A^{(\mathbb{N})} \stackrel{\beta}{\to} A \stackrel{\gamma}{\to} B \to 0, $$ where

1. $\gamma$ is the natural projection;
2. $\beta$ is given in matrix form by the row vector $(x_1, x_2, \ldots)$;
3. $\alpha$ sends the element $1_A$ in coordinates $(i,j)$ to the element of $A^{(\mathbb{N})}$ having $x_j$ in coordinate $i$, $-x_i$ in coordinate $j$, and zero everywhere else.

Now, applying the functor $\operatorname{Hom}_A(-, k)$, we get $$ \operatorname{Hom}_A(A,k) \stackrel{\beta^*}{\to} \operatorname{Hom}_A(A^{(\mathbb{N})}, k) \stackrel{\alpha^*}{\to} \operatorname{Hom}_A(A^{(\mathbb{N}\times\mathbb{N})}, k), $$

and the extension module is $\operatorname{Ext}_A^1(B,k) = \operatorname{Ker}(\alpha^*)/\operatorname{Im}(\beta^*)$. Note that $\operatorname{Hom}_A(A,k)$ is isomorphic to $k$, and that any morphism from $A$ to $k$ sends all variables $x_i$ to zero. Since the images of $\alpha$ and $\beta$ only involve polynomials with zero constant term, this implies that the morphisms $\alpha^*$ and $\beta^*$ are actually zero.

Thus $\operatorname{Ext}_A^1(B,k) = \operatorname{Ker}(\alpha^*)/\operatorname{Im}(\beta^*) = \operatorname{Hom}_A(A^{(\mathbb{N})}, k) \cong \operatorname{Hom}_A(A, k)^{\mathbb{N}}\cong k^\mathbb{N}$ is an infinite-dimensional $k$-vector space, and it is not finitely generated over $B=k$.