Nilpotent ideal and ring homomorphism

Question

In "Problems and Solutions in Mathematics", 2nd Edition, exercice 1308

Problem statement

Let $I$ be a nilpotent ideal in a ring $R$, let $M$ and $N$ be $R$-modules, and let \begin{equation} f : M \rightarrow N \end{equation} be an $R$-homomorphism. Show that if the induced map \begin{equation} \overline{f} : M / IM \rightarrow N / IN \end{equation} is surjective, then $f$ is surjective.

Beginning of solution

Since $\overline{f}$ is surjective, $f(M) + IN = N$. It follows that \begin{equation} I \cdot N / f(M) = IN + f(M) / f(M) = N / f(M) \end{equation}

My questions

1. Why does $\overline{f}$ being surjective implies $f(M) + IN = N$ ?
2. I understand that $IM$ is a subgroup of $M$ created by applying the scalar product operation between every element of $I$ to every element of $M$. What is the meaning of $I \cdot N / f(M)$ since $N / f(M)$ is a group of cosets ?
3. How do you derive the two equalities: $I \cdot N / f(M) = IN + f(M) / f(M) = N / f(M)$ ? From the first equation you immediately get $I \cdot N / f(M) = I \cdot (f(M) + IN) / f(M)$, but how do you continue from there ?

EDIT: how in particular do you get $I \cdot N / f(M) = IN + f(M) / f(M)$ in question 3, since $IN + f(M) / f(M) = N / f(M)$ is a direct consequence of question 1 ?

Answer 1

1. It is false. It is even meaningless: a priori, $IM$ is not contained in $N$. What is true is that $f(M)+IN=N$.
2. It is what is written just afterwards: $\;(IN+f(M))/f(M)$.
3. It is derived from the correct relation in question $1$.

Once you have $N/f(M)=I\cdot N/f(M)$, you deduce repeatedly: $$N/f(M)=I\cdot N/f(M)=I^2\cdot N/f(M)=\dots=I^k\cdot N/f(M)=\dots$$ If $r$ is the index of nilpotency of $I$ you thus have $$N/f(M)=I^r\cdot N/f(M)=0,\enspace\text{whence}\enspace N=f(M). $$

Note:

If $N$ is a finitely generated $R$-module, it is a simple consequence of Nakayama's lemma.

Answer 2

You have $f(M)+IN=N$. Try showing that $f(M)+I^kN=N$ for all $k$, and then use the fact that $I$ is nilpotent.

For $k=1$, this follows just from surjectivity. Proceeding by induction, suppose $f(M)+I^kN=N$. Then $$ \begin{align*} N&=f(M)+I^kN\\ &=f(M)+I^k(f(M)+IN)\\ &=f(M)+I^kf(M)+I^{k+1}N\\ &=f(M)+I^{k+1}N, \end{align*} $$ the last equality following since $I^kf(M)\subseteq f(M)$. For some sufficiently large $k$, $I^k=0$, so $N=f(M)$.

Answer 3

To see why $f(M) + IN = N$, note that,if $\overline{f}$ is surjective, then for any $x \in N$, there must be $a \in M$, $i \in I$ and $y \in N$ such that $x + iy = f(a)$, so that $x = f(a) - iy \in f(M) + IN$. For the rest see the excellent answers by Bernard and Ben West.