I have the following problem:
Let $R$ be a ring and $a\subset R$ an ideal with $a\subset J(R)$, where $$J(R):=\bigcap_{m\,\in\,\operatorname{mSpec} R} m.$$ Let $M$ be an $R$-module and $N$ a finitely generated $R$-module, $f:M\rightarrow N$ an $R$-module homomorphism and $\overline{f}:M/aM\rightarrow N/aN$ the homomorphism which is induced by $f$.
Show that the following statements are equivalent:
$i)\, f$ is surjective
$ii)\, \overline{f}$ is surjective
I thought about using the universal property of quotient modules
$$ \begin{matrix} M & \xrightarrow{f} & N & \\ \quad \downarrow\pi_M & \nearrow_\tilde{f} \\ M/aM \end{matrix} $$
But I'm not sure if I can use: $\overline{f}=\pi_N\circ\tilde{f}$. Since $\pi_N$ is surjective, the composition of $\pi_N\circ\tilde{f}$ should be surjective, because $\tilde{f}\circ \pi_M$ is surjective. Is that correct? How can I show the other direction? Thanks in advance.
The direction (1) -> (2) is a general fact about induced maps on quotients modules. What does the map $\bar{f}$ do? It takes an element $m + aM$ to an element $n + aN = f(m) + aN$. For surjective $f$, each $n' \in N$ is the image under $f$ of some $m' \in M$, and $\bar{f}(m' + aM) = f(m') + aN = n' + aN$.
For (2) -> (1), first note that if $\bar{f}: M/aM \rightarrow N / aN$ is surjective, then as in the direction we just proved, so too is the induced map $M/J(R)M \rightarrow N/J(R)N$. (By the third isomorphism theorem, $M/J(R)M$ is a quotient of $M/aM$ by $J(R)M / aM$, and sim for $N/J(R)N$).
As mentioned in the comments, the rest is a very standard application of Nakayama's lemma.
From the wiki, statement 4, we have
From above we have a surjective map $\hat{f}: M/J(R)M \rightarrow N/J(R)N$, with $N$ (and hence $N/J(R)N$) finitely generated by assumption. Thus there exists finitely many $m_i \in M$ such that the elements $\hat{f}\big(m_i + J(R)M\big)$ generate $N/J(R)N$. Since $f(m_i) + J(R)M$ generate $N/J(R)N$, we can use Nakayama's lemma to see that $f(m_i)$ generate $N$, and $f$ is indeed surjective.