$\overline\phi: M/IM \to N/IN$ is surjective, then $\phi$ is surjective.

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

I have proceeded in this way $\overline\phi: M/IM \to N/IN \Rightarrow \hat \phi:(M/IM)^n=M^n/(IM)^n \to (N/IN)^n=N^n/(IN)^n$ is surjective. Now from here how do I conclude the claim?
OR
Do I have to show that $M/I^nM \to N/I^nN$ is surjective?

I am completely stuck here. Need help.

• What is $M^n$ when $M$ is module? – Bernard Jan 31 '16 at 17:53

‘$\overline\phi$ is surjective’ means $N=\phi(M)+IN$. From this you deduce $$N=\phi(M)+I(\phi(M)+IN)=\phi(M)+I\phi(M)+I^2N=\phi(M)+I^2N,$$ and by a silly induction: $$N=\phi(M)+I^kN\quad\text{for all}\enspace k\ge 1.$$ Now choose for $k$ the nilpotency index of $I$, and you get $\;N=\phi(M)$, i.e. you get the surjectivity of $\phi$.
• @Pedro Tamaroff: $\phi(M)$$is not necessarily a finitely generated module. – Bernard Jan 31 '16 at 18:31 • Ah, sure. Pity. – Pedro Tamaroff Jan 31 '16 at 18:57 • +1 Nice solution. I am just learning Algebra and I would've never thought of this. Does this sort of problem (or this difficulty level of problem) become very routine/easy to solve after studying algebra a while, or is it something novel that you did not know how to solve right away and had to think about a little bit? – Ovi Sep 20 '18 at 18:24 • Actually, it's standard reasoning. – Bernard Sep 20 '18 at 18:57 Maybe this is another description of Bernard's answer; so if it dont help you I will delete: Let I^n=0. As above: N=Im(\phi)+IN. So I(\frac{N}{Im(\phi)})=\frac{IN+Im(\phi)}{Im(\phi)}=\frac{N}{Im(\phi)}. Hence$$0=I^n(\frac{N}{Im(\phi)})=\frac{N}{Im(\phi)}.$$So$N=Im(\phi).\$