The Stacks project

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

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: (otherwise, take bigger thant and , then we can't make N an S-submodule of M).

Comment #4879 by on

But is assumed to be an -module so the kernel of already annihilates all of .

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  • 2 comment(s) on Section 10.52: Length

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