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Homology of infinite Koszul complex tensored with an injective module vanishes

Mathematics Asked on December 3, 2021

Given a commutative noetherian ring $A$ and an element $ain A$, 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
$$K^infty(A;a):=(cdotsto 0to Axrightarrow{d} A[a^{-1}]to 0tocdots),$$
concentrated at degree $0$ and $1$, where $d$ is given by the localisation map $xmapsto x/1$.

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
$$cdotsto 0to Exrightarrow{dotimes 1}E[a^{-1}]to 0to cdots,$$
and I have thought of proving the natural isomorphism
$$H_{(a)}(M)cong H^n(K^infty(A;a)otimes_A M), forall Min mathrm{Mod}_A, forall nge 0.$$
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.

Thank you very much in advance for your help!

One Answer

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 $H^n(K^bulletotimes E)$ are trivially $0.$ Second, notice that $$H^1(K^bulletotimes E) = E[a^{-1}]/operatorname{im}(Eto E[a^{-1}]),$$ so it suffices to prove that the localization map $Eto E[a^{-1}]$ 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 $Eto E[a^{-1}$ is surjective for any $ain A.$

The proof goes as follows: there exists an $r$ such that $operatorname{Ann}(a^r)=operatorname{Ann}(a^{r+i})$ for any $igeq 0$ by Noetherianity. Then, given $e/a^nin E[a^{-1}]$ (with $ein E$), 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 $$ f:Ato E, $$ 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.

Answered by Stahl on December 3, 2021

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