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Let $A$ be an unital, associative, graded algebra over a base ring $k$. I'm happy to assume that $k$ is a field if need be, and will insist that $A$ free and of finite rank in each degree (locally finite). Further, $A$ is connected: it vanishes in negative degrees and is of rank 1 (generated by the unit) in degree 0; then the projection onto $k = A/A_{>0}$ makes $k$ a graded $A$-module.

Write $H^{q, r}(A) = Ext^{q, r}_A(k, k)$ for the cohomology of the ring $A$. This is bigraded: if we compute this as the cohomology of the cobar complex for $A$, then classes in $H^{q, r}(A)$ arise as the dual of tensors of the form $[a_1 | \dots | a_q]$ with the sum of the degrees of the $a_i$ being $r$. Notice that since non-unit $a_i$ are in positive degrees, $r \geq q$.

The whole of the cohomology $H^{*, *}(A)$ is a bigraded algebra, and so for any constant $j \geq 0$, the summand

$$H_j := \bigoplus_{q=0}^\infty H^{q, q+j}(A)$$

is a module of the subring $H_0 \leq H^{*, *}(A)$.

My question is: Are there known conditions on $A$ which ensure that $H_j$ is finitely generated over $H_0$?

Of course $H_0$ is always finitely generated over itself, so I'm really interested in $j>0$. In that vein, I know one condition which ensures finite generation: if $A$ is generated in degree $1$ and is Koszul, then by definition $H_j = 0$ for $j>0$. Is there anything less stringent than this?

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  • $\begingroup$ I may just be confused with the gradings/notation, but doesn't it seem like this should follow directly from the cobar complex being of finite type (since A is)? EDIT: never mind, I was indeed just confused. $\endgroup$ Sep 17, 2015 at 8:11

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