# Homology of infinite Koszul complex tensored with an injective module vanishes

Given a commutative noetherian ring $$A$$ and an element $$a\in A$$, we denote by $$A[a^{-1}]$$ the localisation of $$A$$ at $$\{1,a,a^2,\cdots\}$$. The infinite Koszul complex of $$A$$ and $$a$$ refers to the cochain complex $$K^\infty(A;a):=(\cdots\to 0\to A\xrightarrow{d} A[a^{-1}]\to 0\to\cdots),$$ concentrated at degree $$0$$ and $$1$$, where $$d$$ is given by the localisation map $$x\mapsto x/1$$.

Show that if $$E$$ is an injective $$A$$-module, then $$H^n(K^\infty(A;a)\otimes_A E)=0$$ for all $$n>0$$.

The tensored complex is $$\cdots\to 0\to E\xrightarrow{d\otimes 1}E[a^{-1}]\to 0\to \cdots,$$ and I have thought of proving the natural isomorphism $$H_{(a)}(M)\cong H^n(K^\infty(A;a)\otimes_A M),\ \forall M\in \mathrm{Mod}_A,\ \forall n\ge 0.$$ Knowing that $$H_I^n(E)=0$$ for any $$n>0$$, any ideal $$I$$, and any injective module $$E$$, this will yield the result. However, to prove the natural isomorphism I need some homological work. So before proceeding to that I would like to search for a more direct proof to the problem, possibly by using the characterisation of injective modules.

Let me suppress $$A$$ and $$a$$ from the notation and call your infinite Koszul complex $$K^\bullet$$ for simplicity. Notice first that you only care about $$n = 1,$$ as all other $$H^n(K^\bullet\otimes E)$$ are trivially $$0.$$ Second, notice that $$H^1(K^\bullet\otimes E) = E[a^{-1}]/\operatorname{im}(E\to E[a^{-1}]),$$ so it suffices to prove that the localization map $$E\to E[a^{-1}]$$ is surjective.
This is true in general: if $$A$$ is a Noetherian commutative ring, and $$E$$ is an injective $$A$$-module, then the natural map $$E\to E[a^{-1}$$ is surjective for any $$a\in A.$$
The proof goes as follows: there exists an $$r$$ such that $$\operatorname{Ann}(a^r)=\operatorname{Ann}(a^{r+i})$$ for any $$i\geq 0$$ by Noetherianity. Then, given $$e/a^n\in E[a^{-1}]$$ (with $$e\in E$$), define a map \begin{align*} a^{n+r}A&\to E\\ a^{n+r}b&\mapsto a^r b e. \end{align*} This is well-defined because the sequence of annihilators of powers of $$a$$ stabilize. Then injectivity of $$E$$ allows you to extend this map to a map $$f:A\to E,$$ and if $$f(1)=x,$$ then $$a^{n+r}x=a^r e.$$ Then it follows that $$e/a^n$$ is the image of $$x$$ under the localization map, and we are done.