Exercise 2.10

Question

Let $A$ be a ring, $\mathfrak{a}$ an ideal contained in the Jacobson radical of $A$; let $M$ be an $A$-module and $N$ a finitely generated $A$-module, and let $u\colon M 	o N$ be a homomorphism. If the induced homomorphism $M/\mathfrak{a}M 	o N/\mathfrak{a}N$ is surjective, then $u$ is surjective.

Amit Awasthi · Accepted Answer

\begin{proof}
  Letting $L = \operatorname{coker}(u)$, we have the right exact sequence $M 	o N 	o L 	o 0$. Tensoring with $A/\mathfrak{a}$ gives the right exact sequence
  \begin{equation*}
    M/\mathfrak{a}M \overset{\bar{u}}{\longrightarrow} N/\mathfrak{a}N \overset{\bar{\pi}}{\longrightarrow} L/\mathfrak{a}L \longrightarrow 0,
  \end{equation*}
  where we use the isomorphisms $(A/\mathfrak{a}) \otimes_A M \cong M/\mathfrak{a}M$ and $(A/\mathfrak{a}) \otimes_A N \cong N/\mathfrak{a}N$ from Exercise 2.2, where $\bar{u}$ is the induced map of $u$, and $\bar{\pi}$ of $\pi\colon N 	o L$. By assumption, $\bar{u}$ is surjective, hence $L/\mathfrak{a}L = 0$. But by Nakayama's lemma (Prop.~2.6), we then get $L= \operatorname{coker}(u) = 0$, and so $u$ is surjective as claimed.
\end{proof}

Exercise 2.10

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Exercise 2.10

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