Exercise 2.9

Question

Let $0 	o M^{\prime} 	o M 	o M^{\prime\prime} 	o 0$ be an exact sequence of $A$-modules. If $M^{\prime}$ and $M^{\prime\prime}$ are finitely generated, then so is $M$.

Amit Awasthi · Accepted Answer

ewcommand{\dashto}[2]{\smash{\hspace{-0.7em}\begin{array}{ccc} {#1} & \longrightarrow & {#2} \end{array}\hspace{-0.7em}}}

\begin{proof}
  Since $M^{\prime},M^{\prime\prime}$ are finitely generated, by Prop.~2.3 we have the diagram
  \begin{equation*}
    \begin{tikzcd}
      0 ar & A^m ar\dar{\alpha} & A^{m} \oplus A^n ar\dar[dashed]{\beta} & A^n ar\dar{\gamma} & 0\
      0 ar & M' ar{f} & M ar{g} & M'' ar & 0
    \end{tikzcd}
  \end{equation*}
  where $\alpha,\gamma$ are surjective, and each row is exact. Denote the generators of $A^{m} \oplus A^n$ as $x_1,\ldots,x_m,y_1,\ldots,y_n$; identifying $x_i,y_j$ with their preimages and images in $A^m,A^n$ respectively, define $\beta(x_i) = f(\alpha(x_i))$ and $\beta(y_j) = g(\beta(y_j))$. This makes the diagram commute, hence by the snake lemma (Prop.~2.10), we have that $\operatorname{coker} \alpha 	o \operatorname{coker} \beta 	o \operatorname{coker} \gamma$ is exact. But $\operatorname{coker} \alpha = \operatorname{coker} \gamma = 0$, hence $\operatorname{coker} \beta = 0$, and so $\beta$ is surjective, i.e., $M$ is finitely generated by Prop.~2.3.
\end{proof}

Exercise 2.9

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Exercise 2.9

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