Short Exact Sequence and Finitely Generated Modules
Problem 414
Let $R$ be a ring with $1$. Let
\[0\to M\xrightarrow{f} M’ \xrightarrow{g} M^{\prime\prime} \to 0 \tag{*}\]
be an exact sequence of left $R$-modules.
Prove that if $M$ and $M^{\prime\prime}$ are finitely generated, then $M’$ is also finitely generated.
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Proof.
Since $M$ is finitely generated, let $x_1, \dots, x_n$ be generators of $M$.
Similarly, let $z_1, \dots, z_m$ be generators of $M^{\prime\prime}$.
The exactness of the sequence (*) yields that the homomorphism $g:M’\to M^{\prime\prime}$ is surjective.
Thus, there exist $y_1, \dots, y_m\in M’$ such that
\[g(y_i)=z_i\]
for $i=1, \dots, m$.
We claim that the elements
\[f(x_1), \dots, f(x_n), y_1, \dots, y_m\]
generate the module $M$.
Let $w$ be an arbitrary element of $M’$. Then $g(w)\in M^{\prime\prime}$ and we can write
\[g(w)=\sum_{i=1}^m r_iz_i\]
for some $r_i\in R$ as $z_i$ generate $M^{\prime\prime}$.
Then we have
\begin{align*}
g(w)&=\sum_{i=1}^m r_iz_i\\
&=\sum_{i=1}^m r_i g(y_i)\\
&=g\left(\, \sum_{i=1}^m r_iy_i \,\right)
\end{align*}
since $g$ is a module homomorphism.
It follows that we have
\begin{align*}
g\left(\, w- \sum_{i=1}^m r_iy_i \,\right)=g(w)-g\left(\, \sum_{i=1}^m r_iy_i \,\right)=0,
\end{align*}
and thus
\[w- \sum_{i=1}^m r_iy_i \in \ker(g).\]
Since the sequence (*) is exact, we have $\ker(g)=\im(f)$.
Hence there exists $x\in M$ such that
\[f(x)=w- \sum_{i=1}^m r_iy_i.\]
Since $x_i$ generate $M$, we can write
\[x=\sum_{i=1}^n s_i x_i\]
for some $s_i\in R$.
Thus, we have
\begin{align*}
w&=f(x)+\sum_{i=1}^m r_iy_i\\
&=f\left(\, \sum_{i=1}^n s_i x_i \,\right)+\sum_{i=1}^m r_iy_i\\
&=\sum_{i=1}^n s_if(x_i)+\sum_{i=1}^m r_iy_i.
\end{align*}
This proves that any element $w\in M’$ can be written as a linear combination of
\[f(x_1), \dots, f(x_n), y_1, \dots, y_m,\]
and we conclude that $M’$ is generated by these elements and thus finitely generated.
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