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?
Do I have to show that $M/I^nM \to N/I^nN$ is surjective?

I am completely stuck here. Need help.

  • $\begingroup$ What is $M^n$ when $M$ is module? $\endgroup$ – 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 $\phi$.

  • $\begingroup$ Alternatively, a nilpotent ideal is contained in the Jacobson radical. One can use Nakayama. $\endgroup$ – Pedro Tamaroff Jan 31 '16 at 18:28
  • $\begingroup$ @Pedro Tamaroff: $\phi(M)$$ is not necessarily a finitely generated module. $\endgroup$ – Bernard Jan 31 '16 at 18:31
  • $\begingroup$ Ah, sure. Pity. $\endgroup$ – Pedro Tamaroff Jan 31 '16 at 18:57
  • $\begingroup$ +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? $\endgroup$ – Ovi Sep 20 '18 at 18:24
  • $\begingroup$ Actually, it's standard reasoning. $\endgroup$ – 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 $N=Im(\phi).$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.