Nakayama's lemma, second version

Question

Let $R$ be a commutative ring with identity, $J$ an ideal that is contained in every maximal ideal of $R$, and $A$ is finitely generated $R-$ module. If $R/J\otimes _R A=0$, then $A=0$.

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Nakayama's lemma says that it is enough to show that $JA=A$.

Hungerford give two hints :

1. use the exact sequence $0\rightarrow J \rightarrow R \rightarrow R/J \rightarrow 0$

2. use the natural isomorphism $R\otimes _R A \cong A$

What I know is that $$ J\otimes _R A \rightarrow R\otimes _R A \rightarrow R/J \otimes _R A \rightarrow 0 $$ is exact sequence. Can you help me?

Answer 1

Since $R/J\otimes_R A=0$, the second exact sequence implies that the map $$J\otimes_R A\to R\otimes_R A\cong A$$ is surjective. $J\otimes_R A$ is generated by elements of the form $j\otimes a$ for some $a\in A$ and $j\in J$. The homomorphism $J\otimes_R A\to A\otimes_R R\cong A$ sends $j\otimes a\mapsto j\otimes a=ja$. The image is thus $JA$ (as $JA$ is the submodule generated by all such elements) and the map is surjective. Hence $JA=A$.