$A$ a local ring and $a_{1}$, ..., $a_{n}$ elements in its maximal ideal, $M$ a finitely generated $A$-module. In this case apparently the homologies from the Koszul complex are finitely generated as $A$-module. Is there a simple explanation for this? Or is it some deep result? Thanks!
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3 Answers
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If your local rings are Noetherian it's obvious. The Koszul complex consists of finitely generated free modules and the homology modules are subquotients of it so also finitely generated.
For non-Noetherian rings it's not true, even when $n=1$. In this case the $H_1$ is the annihilator of $a_1$ which may not be finitely generated. As an example consider the quotient of $k[x_1,x_2,\ldots]$ by the ideal generated by all the $x_i x_j$. The annihilator of $x_1$ is the non-finitely generated maximal ideal.
answered May 3, 2010 at 19:58
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Take $A=k[x_1,x_2,\dots]/(x_ix_j:1\leq i,j)$ be the quotient of the ring of polynomials in countably many indeterminates by the ideal generated by quadratic monomials, and let $M=A$ and $a_1=x_1$.
answered May 3, 2010 at 20:00
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At least for the definition I know, it seems to be because all the groups in the Koszul complex are finitely generated as $A$-modules, being explicitely constructed as sums of tensor products of $M$!
answered May 3, 2010 at 19:56
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$\begingroup$ Unless you assume the ring is neotherian, there is something to prove, no? $\endgroup$ Commented May 3, 2010 at 19:57
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$\begingroup$ Thanks! I misread the question as being about finitely presented modules. As Robin Chapman correctly says, the question is stated needs some additional assumptions! $\endgroup$ Commented May 3, 2010 at 20:01