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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.

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  • $\begingroup$ Do you have Nakayama's Lemma? $\endgroup$ Jul 3, 2014 at 10:46
  • $\begingroup$ Yes, I have Nakayama $\endgroup$
    – Miguel
    Jul 3, 2014 at 10:50
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    $\begingroup$ When something is contained in the Jacobson radical and things lift from a quotient, Nakayama's lemma is calling to you. $\endgroup$ Jul 3, 2014 at 11:11

1 Answer 1

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$\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).

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