Sum is an isomorphism, finitely generated modules 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?
 A: 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?
A: 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...
A: 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 $K=0$.
