$R$ is a commutative ring, $M$ and $N$ finitely generated $R$-modules, $\alpha, \beta\in \operatorname{Hom}_{R}(M,N)$, $\mathfrak a\subset \operatorname{rad}(R)$ and $\alpha $ is surjective while $\beta(M) \subseteq\mathfrak aN$, prove that $\gamma=\alpha + \beta $ is an isomorphism between $M$ and $N$.

I understand why it is a surjection according to Nakayama's lemma but I don't know how to prove injectivity since $M$ might be way bigger than $N$ with regard to their generating sets.

  • $\begingroup$ This is clearly false as stated ($\beta$ could be zero). What is the source? $\endgroup$ – Matthew Towers May 26 '13 at 0:01
  • $\begingroup$ @YACP do you mean that if we assume that $M=N$ then the assertion holds? Then are there any conditions that are unnecessary? $\endgroup$ – Alex May 26 '13 at 0:15
  • $\begingroup$ @mt_: it is Chap10.12 of a term of commutative algebra by Altman and Kleinman $\endgroup$ – Alex May 26 '13 at 0:18
  • $\begingroup$ @Alex Yes, when $M=N$ the assertion holds. This is stated as Corollary 10.4 in your book and (in order to answer your second question) I can't see any superfluous condition. $\endgroup$ – user26857 May 26 '13 at 6:19

For $\beta=0$, the statement is that every surjective homomorphism between f.g. modules is an isomorphism. Of course, this is wrong. However, every surjective endomorphism of a f.g. module is an isomorphism, this is an application of the general form of Nakayama's lemma applied to the corresponding $R[x]$-module where $x$ acs as the endomorphism (details). This shows that the claim is true for $M=N$, and hence also when $M \cong N$.


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