Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm studying on Cohen-Macaulay Rings of Bruns-Herzog.

Let $(R,\mathfrak{m},k)$ be a Noetherian local ring and $H_{\bullet}(R)$ its Koszul algebra. I found on the book (page 75) that "since $H_{\bullet}(R)$ has finite length, one has $H_{\bullet}(R)\otimes\hat{R}\cong H_{\bullet}(R)$".

I'd like to know why $H_{\bullet}(R)$ has finite length and why then we have $H_{\bullet}(R)\otimes\hat{R}\cong H_{\bullet}(R)$.

Could you help me please?

share|cite|improve this question
up vote 2 down vote accepted

If $M$ is any finite length $R$-module, then $\mathfrak m^n M = 0$ for sufficiently large $n$. (This follows from the definition of finite length.)

For any finitely gen. $R$-module $M$, we have that $M\otimes_R \hat{R} = \hat{M}$, the $\mathfrak m$-adic completion of $M$. (This follows from Artin--Rees.)

Combining these two statements gives the isomorphism $H_{\bullet}(R)\otimes \hat{R} \cong H_{\bullet}(R)$ that you asked about.

As for why the Koszul algebra has finite length: it would help if you recalled the definition, but my guess in this context is that $H_{\bullet}(R)$ means something like the homology of a Koszul complex built out of some system of parameters for $\mathfrak m$. If so, the homology consists of finitely generated $R$-modules (since the terms of the complex are free $R$-modules), which are also supported at $\mathfrak m$ (as follows from the fact that this is a Koszul complex made from a system of parameters for $\mathfrak m$); thus the homology modules are of finite length.

[More details on the last paragraph added at the request of the OP]

If we localize at a prime ideal other than the maximal ideal, then the system of parameters defining the Koszul complex now generate the unit ideal. So one has to prove that any Koszul complex built from a collection of generators of the unit ideal has trivial cohomology. This is a standard fact, I believe; if you like, it is a computation from the Cech cohomology point of view of the vanishing of higher cohomology of quasi-coherent sheaves on Spec of a ring.

This will show that the cohomology of the Koszul complex is supported at the maximal ideal. It also consists of finitely generated modules. Together these facts show that each cohomology module is of finite length (because any f.g. module supported just at the maximal ideal is of finite length).

share|cite|improve this answer
I don't understand why the Koszul algebra has finite length, the definition that you have is the same as mine. Could you explain it with more details, please? – Jacob Fox Jun 19 '11 at 13:07
@Jacob: Dear Jacob, More details added as per your request. Regards, – Matt E Jun 19 '11 at 15:38
@Matt: Dear Matt, when you say "module supported just at the maximal ideals" do you mean that $\mathrm{Supp}\;M=\{\mathfrak{m}\}$? – Jacob Fox Jun 19 '11 at 15:48
@Jacob: Dear Jacob, Yes. Regards, – Matt E Jun 19 '11 at 16:05
@Matt: Dear Matt, the definition of finite length is "having a finite composition series", am I right? So I don't understand why $\mathfrak{m}^nM=0$ for large $n$; can you give me at least some reference if the answer is long? I still don't understand why the modules of Koszul homology have finite length. P.S: I see that you are so good with this things, could you answer this question please? It's really important. – Jacob Fox Jun 21 '11 at 10:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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