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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?

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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).

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    @Jacob: Dear Jacob, You seem to be asking many questions about the arguments in Bruns--Herzog, while being unfamiliar with basic techniques of commutative algebra. Perhaps you should be working through a text like Atiyah--Macdonald first, or getting more precise direction from your thesis advisor (if this is for work on your thesis). Regards,2011-06-21