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I need some help regarding an argument from the proof of Proposition 4.2.1 in John Rognes article Galois Extensions of Structured Ring Spectra. We are supposed to prove that $\text{Ext}_{R[G]}^s(R,T)=0$ for $s \neq 0$ if $R \rightarrow T$ is a $G$-Galois extension where $G$ is finite. His argument is as follows:

Furthermore, $T$ is finitely generated projective (of constant rank 1) as an $R[G]$-module by Proposition 2.3.4(c). There is an isomorphism of left $R[G]$-modules $R[G] \cong \text{Hom}_R(R[G],R)$, since $G$ is finite, so $\text{Ext}^s_{R[G]}(R,R[G]) \cong \text{Ext}^s_{R}(R,R) = 0$ for $s \neq 0$. Therefore $\text{Ext}^s_{R[G]}(R,T)=0$ for $s \neq 0$, by the finite additivity of Ext in its second argument.

I'm confused about the use of the fact that $T$ is invertible as an $R[G]$-module. Can anyone help me to see how is it used in the argument?

Cheers!

Nelly L
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1 Answers1

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I don't think invertibility is used; the key is that $T$ is finitely generated projective. Given that, there is a module $U$ so that $T \oplus U$ is finitely generated free as an $R[G]$-module, which means that $\text{Ext}_{R[G]}^s(R, T \oplus U) = 0$ for $s \neq 0$.

John Palmieri
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  • Aha, since we have $\text{Ext}^s_{R[G]}(R,T \oplus U) \cong \text{Ext}^s_{R[G]}(R,\oplus R[G]) \cong \oplus \text{Ext}^s_{R[G]}(R,R[G])$, right? And then both $\text{Ext}^s_{R[G]}(R,T)=0$ and $\text{Ext}^s_{R[G]}(R,U)=0$? – Nelly L Jan 30 '16 at 19:43
  • Yes, exactly right. – John Palmieri Jan 31 '16 at 16:38