# Prove that if the induced homomorphism $M/\mathfrak aM \to N/\mathfrak aN$ is surjective, then $f$ it's surjective.

This problem is from Atiyah and Macdonald, Introduction to Commutative Algebra, Exercise 10, Chapter 2.

Let $A$ be a commutative ring with $1 \ne 0$ and let $\mathfrak a$ be an ideal of $A$ contained in the Jacobson radical. Let $M$ be an $A$-module and $N$ a finitely generated $A$-module and let $f: M\to N$ be a homomorphism. If the induced homomorphism $M/\mathfrak aM \to N/\mathfrak aN$ is surjective, then $f$ it's surjective.

I think that I have to consider the isomorphism

$M/aM \cong A/a \otimes_A M$

$N/aN \cong A/a \otimes_A N$

And induce $f$ in this way. But I'm not sure how to proceed.

• Do you have Nakayama's Lemma? – Elchanan Solomon Jul 3 '14 at 10:46
• Yes, I have Nakayama – Miguel Román Jul 3 '14 at 10:50
• When something is contained in the Jacobson radical and things lift from a quotient, Nakayama's lemma is calling to you. – Elchanan Solomon Jul 3 '14 at 11:11

$\DeclareMathOperator{\coker}{coker}$$\DeclareMathOperator{\im}{im}$Let $f : M \to N$. Since tensoring preserves cokernels (by right exactness), $(\coker f) \otimes_A A/a \cong \coker(f \otimes A/a) = \coker(M/aM \to N/aN) = 0$, so $\coker f = a(\coker f)$. Since $\coker f$ is f.g. (being a quotient of $N$ which is f.g.), Nakayama's Lemma gives $\coker f = 0$, i.e. $f$ is surjective.
By the way, the corresponding statement for injectivity is not true - significantly more assumptions are needed (e.g. $M$ Noetherian, $\im(f)$ flat).