0
$\begingroup$

I am trying to prove the rank-nullity theorem for short exact sequences; if $R$ is an integral domain, and $M',\,M,\,M''$ are all $R$-modules with $$0\rightarrow M' \xrightarrow{\psi} M\xrightarrow{\phi} M''\rightarrow 0$$ being a short exact sequence then by the first isomorphism theorem: $$ M/M'\simeq M'' \implies \mathrm{rank}(M/M')=\mathrm{rank}(M'') $$ Morover $M'\simeq \ker \phi$, and so it's clear from this that $\mathrm{rank}(M)\geq\mathrm{rank}(M')+\mathrm{rank}(M'')$.

I was having more difficulty with the other direction. Namely, if $X\subseteq M$ is maximally $R$-linearly dependent set, since we can write $M=M'\sqcup M\backslash M'$ we have that we can split $X$ into $X_{M'}\sqcup X_{M\backslash M'}$ where $X_{M'}:=\{x\in X:x\in\ker\phi\}$, and $X_{M\backslash M'}:=\{x\in X:x\not\in\ker\phi\}$. In particular since $X$ is $R$-linearly independent in $M$, it must be true that $X_{M'}$ be $R$-linearly independent in $M'$ -- so that $\lvert X_{M'}\rvert \leq \mathrm{rank}(M')$. Similarly, $\lvert X_{M\backslash M'}\rvert\leq \mathrm{rank}(M'')$. Thus we have: $$ \mathrm{rank}(M) = \lvert X\rvert = \lvert X_{M'}\sqcup X_{M\backslash M'}\rvert = \lvert X_{m'}\rvert + \lvert X_{M\backslash M'}\rvert \leq \mathrm{rank}(M')+\mathrm{rank}(M'') $$ The only problem I see is that we could have $\phi(X_{M\backslash M'})\subseteq M''$ being $R$-linearly dependent since all we get from exactness is that $\phi$ is surjective -- but at the same time $M/M'\simeq M''$ by the first isomorphism theorem, so we should be okay? I looked at Short Exact Sequences & Rank Nullity, but those solutions seemed a little clunky or used tools I am not familiar enough with, and so I was wondering if there is a cleaner way to do it. Is there a way to justify $\lvert X_{M\backslash M'}\rvert\leq \mathrm{rank}(M'')$ using what I have or will my proof method not work?

$\endgroup$
  • 2
    $\begingroup$ I'd say that the easy way to do this would be to tensor with $K$, the fraction field of $R$, which is a flat $R$-module. $\endgroup$ – Lord Shark the Unknown Apr 23 at 17:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.