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Suppose $R$ is a local ring with maximal ideal $m$ and suppose $M$ and $N$ are $R$-finitely generated modules. Let $f,g: M \rightarrow N$ be an $R$-module homomorphism. If $f$ is an isomorphism and $g(M) \subset mN$ why is $f+g$ in fact an isomorphism? I think this might follow by Nakayama's lemma but I don't see this. Can you please help?

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There are three good answers here, but no votes (other than mine). Surely something is wrong with our voting habits? – Jack Schmidt Mar 15 '11 at 1:48
@Jack I have said it before on meta and I will say it again here: if you worry too much about voting on this site, you will lose your sleep. It's not worth it! – Alex B. Mar 15 '11 at 2:09
up vote 5 down vote accepted

The hypotheses ensure that $g \otimes R/\mathfrak{m}: M/\mathfrak{m}M \rightarrow N/\mathfrak{m} N$ is zero and thus that $(f+g) \otimes R/\mathfrak{m}$ is an isomorphism. It is a standard consequence of Nakayama's Lemma that this implies that $f+g$ is surjective: see e.g. the end of Section 3.8.1 of my commutative algebra notes.

We still need to show that $f+g: M \rightarrow N$ is injective. Let $K$ be the kernel of this map. If we may assume that $K$ is a finitely generated $R$-module, then again this follows from Nakayama's Lemma since the hypotheses give $K/\mathfrak{m} K = \ker ((f+g) \otimes R/\mathfrak{m}) = 0$.

However, the assumption of the previous paragraph does seem to be an additional assumption, albeit a mild one. It is automatic if $M$ is not just finitely generated but finitely presented and this in turn is automatic if the ring $R$ is Noetherian. (In the olden days, "local rings" were required to be Noetherian. I wonder if that's what you meant?) Whether the result is still true without any additional assumptions I'm not sure off the top of my head. Perhaps some context would be helpful: why do you think this result is true?

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Dear Pete, it is a theorem (which follows from the Cayley-Hamilton theorem; see that any surjective endomorphism of a finitely generated module over any commutative ring is an isomorphism. So since $M$ and $N$ are known to be isomorphic already (under $f$), I do not think we need noetherianness. – Akhil Mathew Mar 15 '11 at 2:00
@Akhil: thanks for this. Like Johan de Jong, I was blissfully unaware of this (although unlike him, I have a well-thumbed copy of Matsumura at my bedside). The best way I know to remember this fact in the future is to write it up, so I will incorporate it into my commutative algebra notes in the near future. – Pete L. Clark Mar 15 '11 at 3:27
(In case people are wondering "how does Pete L. Clark know what's at Johan de Jong's bedside?" The answer is that I don't. It happens that I know Johan and his wife pretty well from my graduate school days, but not that well, of course. But in the blog post Akhil links to, Brian Conrad points out that Johan must have seen the result in question from Matsumura's book and Johan replies that he does not own that particular book of Matsumura.) – Pete L. Clark Mar 15 '11 at 3:30

You need to apply Nakayama to both the kernel of $f+g$ and to the cokernel of $f+g$. I suspect that you need to suppose $R$ to be Noetherian local, in order to know that the kernel is again finitely generated.

Now if $K$ is either the kernel or the cokernel, you must argue that $K = mK$. Give it a try...

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Composing with $f^{-1}$ allows us to assume $M=N$, $f=\mathrm{Id}$.

Let $M'=(\mathrm{Id}+g)(M)$. Then $P=M/M'$ satisfies $P=mP$ (if $x \in M$, $x=x+g(x)-g(x) \in M' + mM$) and is finitely generated, so that $P=0$.

Moreover, for all $x \in K = \ker (\mathrm{Id}+g)$, $x=-g(x)=g^2(x)=\ldots=(-1)^n g^n(x)$ for any integer $n$, so that $K \subset \bigcap_{n \geq 0} m^n M$. Now if $R$ is noetherian, this last module is finitely generated and satisfies the same condition as $P$, so Nakayama's lemma can be applied again and $K=0$.

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Actually the fact that $M'' = \bigcap_{n \geq 0} m^n M$ satisfies $M''=mM''$ is not really obvious, but follows from Artin-Rees lemma: – Plop Mar 15 '11 at 1:15

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