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Let $G$ be a finite group of order $n$ and $\Lambda={\mathbb{Z}}[G]$ the group ring of $G$. Let $A$ be a finitely generated free abelian group on which $G$ acts. Let $B$ be a finitely generated $\Lambda$-module.

Question. How can one prove that $n\cdot {\rm Ext}_\Lambda^1(A,B)=0$?

The idea must be that ${\rm Ext}_{\mathbb{Z}}^1(A,B)=0$ because $A$ is ${\mathbb{Z}}$-free. There should be some exact sequence $$ H^1(G,\dots)\to {\rm Ext}^1_\Lambda(A,B)\to {\rm Ext}^1_{\mathbb{Z}}(A,B)=0, $$ and $n\cdot H^1(G,\dots)=0$.

What is the exact sequence in question and from what spectral sequence does it come?

Mikhail Borovoi
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1 Answers1

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There's a spectral sequence $$H^i(G, \operatorname{Ext}^j_\mathbb Z (A,B)) \Rightarrow \operatorname{Ext}_{\Lambda}^{i+j}(A,B)$$ for general $A$ and $B$ which, since $\operatorname{Ext}_{\mathbb{Z}}^i=0$ for $i>1$, gives a long exact sequence $$0\to H^1(G,\operatorname{Hom}_{\mathbb{Z}}(A,B))\to \operatorname{Ext}_{\Lambda}^1(A,B) \to H^0(G,\operatorname{Ext}_{\mathbb{Z}}^1(A,B))\to H^2(G,\operatorname{Hom}_{\mathbb{Z}}(A,B))\to\dots$$ and hence isomorphisms $$H^i(G,\operatorname{Hom}_{\mathbb{Z}}(A,B))\cong \operatorname{Ext}_{\Lambda}^i(A,B)$$ if $A$ is $\mathbb{Z}$-free.

Jeremy Rickard
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