Lemma 10.51.5. 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. $\square$

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: $ker(R\to S)\subset Ann_R(N)$ (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 on

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

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).