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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.

Thank you very much in advance for your help!

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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.

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  • $\begingroup$ Thank you very much for the answer! $\endgroup$
    – Ivon
    Jul 23 '20 at 1:41

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