# Computing ext over graded rings

This question came up as I was reading Beilinson, Ginzburg, Soergel paper Koszul Duality Patterns in Representation Theory.

Suppose that $A$ is a Koszul ring (for the definition of Koszul ring see page 2 of BGS) and let $N$ be a graded $A$-module. As I understand it should be true that $$\operatorname{ext}_A ^{i+1} (A_0, N) = \hom_A (K^i,N)$$ where $K^i = \ker(P^i \rightarrow P^{i-1})$.

However I haven't been able to show it. The way I have been trying to show it is by induction since in the case were $i = 0$ maybe its possible to use the fact that $ext^1 _A (A_0, N) = \hom_A (K^0, N)/ Im(i^*)$ where $K^0 = \ker ( P^0 \rightarrow A_0)$ and $i: K^0 \hookrightarrow P^0$. But in order to even start here I have to show that $Im(i^*)$ is zero and I haven't been able to convince myself that it is.

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I believe your question come from understanding proof of (3)$\Rightarrow$(1) of Prop 2.1.3, which actually require more than $N$ just to be graded module, it also require $N$ to be pure. Have you tried using that condition as well? – Aaron Sep 3 '13 at 11:42
@Aaron, ack! I have an older version of the paper and they don't state that N is pure. It seems in the version on the AMS website this is corrected. I think it should work now, thanks! :) – Anette Sep 3 '13 at 14:10
@Anette Please consider self-answering your question, so that it gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. – Julian Kuelshammer Jun 3 '14 at 9:02