Proposition

Let $0 \rightarrow M’ \rightarrow M \rightarrow M’’ \rightarrow 0$ be an exact sequence of $A$-modules. If $M’$ and $M’’$ are finitely generated, then so is $M$.

Solution

Let $i$ denote the map from $M’$ into $M$ and $p$ denote the map from $M$ to $M’’$.

Let $\{ x_1, \cdots, x_n \}$ and $\{ y_1, \cdots, y_m \}$ be generates of $M’$ and $M’’$, respectively. Let $z_i = f(x_i)$ for each $i = 1, \cdots, n$. Since $p$ is surjective, there exists $z_{n + 1}, \cdots, z_{n + m} \in M$ such that $p(z_{n + i}) = y_i$ for each $i$. We claim that $\{ z_1, \cdots, z_{n + m} \}$ generates $M$.

Let $x \in M$. $p$ induces an isomorphism $p’: M / \ker(p) \rightarrow M’’$ because $p$ is surjective. Then $p’(x + \ker(p)) = p(x) = a_{n + 1}y_1 + \cdots + a_{n + m}y_n$. We also have $p’(a_{n + 1}z_{n + 1} + \cdots + a_{n + m}z_{n + m} + \ker(p)) = a_{n + 1}y_1 + \cdots + a_{n + m}y_n$. Since $p’$ is an isomorphism, $x + \ker(p) = a_{n + 1}z_{n + 1} + \cdots + a_{n + m}z_{n + m} + \ker(p)$. In other words, $x - (a_{n + 1}z_{n + 1} + \cdots + a_{n + m}z_{n + m}) \in \ker(p)$.

Similarly, $i$ is an isomorphism between $M’$ and $\Im(i) = \ker(p)$ because $i$ is injective. Then there exists a unique element $a_1x_1 + \cdots + a_nx_n \in M’$ such that $i(a_1x_1 + \cdots + a_nx_n) = x - (a_{n + 1}z_{n + 1} + \cdots + a_{n + m}z_{n + m}))$. This implies that $x = (a_1z_1 + \cdots + a_nz_n) + (a_{n + 1}z_{n + 1} + \cdots + a_{n + m}z_{n + m})$. Therefore, $M$ is indeed generated by the $n + m$ elements.