Lemma 10.52.13. Let $A \to B$ be a flat local homomorphism of local rings. Then for any $A$-module $M$ we have

$\text{length}_ A(M) \text{length}_ B(B/\mathfrak m_ AB) = \text{length}_ B(M \otimes _ A B).$

In particular, if $\text{length}_ B(B/\mathfrak m_ AB) < \infty$ then $M$ has finite length if and only if $M \otimes _ A B$ has finite length.

Proof. The ring map $A \to B$ is faithfully flat by Lemma 10.39.17. Hence if $0 = M_0 \subset M_1 \subset \ldots \subset M_ n = M$ is a chain of length $n$ in $M$, then the corresponding chain $0 = M_0 \otimes _ A B \subset M_1 \otimes _ A B \subset \ldots \subset M_ n \otimes _ A B = M \otimes _ A B$ has length $n$ also. This proves $\text{length}_ A(M) = \infty \Rightarrow \text{length}_ B(M \otimes _ A B) = \infty$. Next, assume $\text{length}_ A(M) < \infty$. In this case we see that $M$ has a filtration of length $\ell = \text{length}_ A(M)$ whose quotients are $A/\mathfrak m_ A$. Arguing as above we see that $M \otimes _ A B$ has a filtration of length $\ell$ whose quotients are isomorphic to $B \otimes _ A A/\mathfrak m_ A = B/\mathfrak m_ AB$. Thus the lemma follows. $\square$

There are also:

• 2 comment(s) on Section 10.52: Length

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).