2
$\begingroup$

A basic question about Koszul homology from Matsumura's Commutative Ring Theory

In Theorem 16.5(ii) it is assumed that $(A,m)$ is a local ring and $x_1,\ldots,x_n \in m$, and $M$ is a finite $A$-module. Then it is claimed without much explanation that the Koszul homology groups $H_p(X,M)$ are finite $A$-modules for all $p$. Why is this so obviously true?

$\endgroup$
2
$\begingroup$

If $A$ is Noetherian, then this is indeed obvious: the homology group $H_p(X,M)$ is defined as $\ker g/\operatorname{im}f$ for certain maps $$M^i\stackrel{f}\to M^j\stackrel{g}\to M^k$$ and certain $i,j,k\in\mathbb{N}$. Since $M$ is finitely generated, so is $M^j$, and thus so is $\ker g$ since $A$ is Noetherian, and thus so is $H_p(X,M)$.

If $A$ is not assumed to be Noetherian, then this is not true. For instance, $A$ could be $k[t_1,t_2,t_3,\dots]/(t_1,t_2,t_3,\dots)^2$ for a field $k$. Then for $M=A$, $n=1$, and $x_1=t_1$, the Koszul complex is just $$0\to A\stackrel{t_1}\to A\to 0.$$ So $H_1(X,A)$ is just the kernel of $t_1:A\to A$ which is the maximal ideal $(t_1,t_2,t_3,\dots)$, which is not finitely generated.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.