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I am sorry if this is a naive question.

Let $k$ be a field, and let $A$ be a finitely generated commutative $k$-algebra.

Let $M$ be a finite $A$-module.

Consider the Hochschild cohomologies of $A$ with coefficients in $M$.

Obviously, they are finitely generated modules over the enveloping algebra $A\otimes_k A$. But are they finitely generated over $A$ as well?

Thanks for any comments or remarks.

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I do not think it is true. However, in the case of smooth A, we have the Hochschild-Konstant-Rosenberg isomorphism, which tells that there is an isomorphism between $HH^i(A)$ and $\Omega^i_A$. In particular, $HH^i(A)$ is finitely generated. – Marci Aug 15 '13 at 12:35
Seems like maybe you can just use the adjunction and $Ext_{A\otimes A}(A,M)=Ext_A(\Delta^*\Delta_{*}(A),M).$ So it seems like you further have the fact that $\Delta^*\Delta_{*}(A)= \oplus Sym^p(L_{A/k})$, due to Quillen (L_{A/k} is the cotangent complex). So the question is does this $\oplus Sym^p(L_{A/k})$ have coherent cohomology in each degree. It seems the answer is yes, since the cotangent complex should also have coherent cohomology(getting confused with gradings...), but either way the answer should be pretty easy to see... – user36931 Aug 15 '13 at 12:54
But let me ask an even more naive question, I had always assumed that that the above $A\otimes_k A$ module structure should factor through the diagonal A. If you know that that automatically implies what you want, so perhaps your saying that is false? P.S. A nice reference for this Quillen fact is the following paper Also in Marci's comment the result should be $\wedge^i T_A$ for cohomology. – user36931 Aug 15 '13 at 13:49
If $M$ and $N$ are symmetric $A$-bimodules, then $\hom_{A^e}(M,N)$ is also a symmetric $A$-bimodule. IOW, the action of $A^e$ on this $\hom$ space factors through the multiplication map $A\otimes A\to A$, and the answer to your question follows at once, no? – Mariano Suárez-Alvarez Aug 15 '13 at 15:18
Yes, it should be $\wedge^i T_A$. Btw, if I remember well, there is an dg-alg. A (finitely generated over k) s.t. its Hochschild cohomology is not finitely generated over A (which does not answer the question). – Marci Aug 15 '13 at 16:52

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