Lemma 10.52.5 (00IX)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.52: Length
Lemma 10.52.5 (cite)

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Lemma 10.52.5. Let $R \to S$ be a ring map. Let $M$ be an $S$ -module. We always have $\text{length}_ R(M) \geq \text{length}_ S(M)$ . If $R \to S$ is surjective then equality holds.

Proof. A filtration of $M$ by $S$ -submodules gives rise a filtration of $M$ by $R$ -submodules. This proves the inequality. And if $R \to S$ is surjective, then any $R$ -submodule of $M$ is automatically an $S$ -submodule. Hence equality in this case. $\square$

Comments (2)

Comment #4878 by Peng DU on January 12, 2020 at 05:56

In the proof, it says "And if R→S is surjective, then any R-submodule of M is automatically an S-submodule", but I think that may need to assume the kernel of R→S kills the submodule N: $ker(R\to S)\subset Ann_R(N)$ (otherwise, take $I$ bigger thant $Ann_R(N)$ and $S=R/I$ , then we can't make N an S-submodule of M).

Comment #4879 by Aise Johan de Jong on January 12, 2020 at 09:13

But $M$ is assumed to be an $S$ -module so the kernel of $R \to S$ already annihilates all of $M$ .

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