Module homomorphism question

Question

I am trying to work on this problem:

Let $R$ be a commutative ring, $I$ a nilpotent ideal of $R$, and $M,N$ $R$-modules. Let $\phi :M \to N$ be an $R$-module homomorphism. Show that if the induced map $\bar{\phi} : M/IM \to N/IN$ is surjective, then $\phi$ is surjective.

If I can show that $N = \phi(M) + IN$ then I get a recursion: $N = \phi(M) + I(\phi(M) + IN) = \phi(M) + I^{2}N = \cdots = \phi(M) + I^{k}N = \phi(M)$

I am just having trouble with the logic leading to the fact that $N = \phi(M) + IN$. The natural definition of $\bar{\phi}$ is $\bar{\phi}(M/IM) = \phi(M) + IN = N/IN$. I have seen a solution where they wrote $\bar{\phi}(M/IM) = (\phi(M) + IN)/IN = N/IN$ and used a isomorphism theorem to say that $\phi(M) + IN = N$ but I don't understand where that comes from.

Any help clearing this up would be appreciated!

Answer 1

Hint. $M\to N\to C\to 0$ is exact. so $ M/IM \to N/IN\to C/IC\to 0 $ is exact.
you will have $C=IC $ so $C=I^nC =0 $

Answer 2

The elementary way works very well: Take any $n\in N$ and look at its class $\overline{n}\in N/IN$. By assumption there exists $\overline{m}\in M/IM$ with $\overline{\phi(m)} = \overline{\phi}(\overline{m}) = \overline{n}$, thus $\varphi(m)-n\in IN$. But this means that $n = \phi(m)+n'$ for suitable $m\in M,n'\in IN$.

Answer 3

Here is an alternate proof using Nakayama's lemma.

Nakayama's Lemma: Let $M$ be a finitely generated $A$-module and $\mathfrak a$ an ideal of $A$ contained in the Jacobson radical $\mathfrak R$ of $A$. Then $\mathfrak aM = M$ implies $M = 0$.

Corollary: Let $M$ be a finitely generated $A$-module, $N$ a submodule of $M$, $\mathfrak a \subseteq \mathfrak R$ an ideal. Then $M = \mathfrak aM + N$ implies $M = N$.

Claim: Let $A$ be a ring, $\mathfrak a$ an ideal contained in the Jacobson radical of $A$; let $M$ be an $A$-module and $N$ a finitely generated $A$-module, and let $u : M \to N$ be a homomorphism. If the induced homomorphism $M / \mathfrak aM \to N / \mathfrak aN$ is surjective, then $u$ is surjective.

Proof: Note that $u$ is surjective if $u(M) = N$. If we show that $u(M) + \mathfrak aN = N$, then by the corollary to Nakayama's lemma, we can conclude that $u(M) = N$ and hence $u$ is surjective. It is trivial that $u(M) + \mathfrak aN \subseteq N$, so it suffices to show $N \subseteq u(M) + \mathfrak aN$. Let $n \in N$ and notice that $n \in n' + \mathfrak aN$ for some $n' \in N$ since cosets partition $N$. Since $\overline u$ is surjective, there exists $m \in M$ such that $$\overline u(\overline m) = \overline u(m + \mathfrak aM) = u(m) + \mathfrak aN = n' + \mathfrak aN.$$ Thus $n \in n' + \mathfrak aN = u(m) + \mathfrak aN \in N / \mathfrak aN = u(M) + \mathfrak aN$.