Finitely generated module in exact sequence

Question

For $A$-modules and homomorphisms $0\to M'\stackrel{u}{\to}M\stackrel{v}{\to}M''\to 0$ is exact. Prove if $M'$ and $M''$ are fintely generated then $M$ is finitely generated.

Answer 1

Suppose $M'$ is generated by $x_1,\dots,x_n$ and $M''$ is generated by $z_1,\dots,z_m$. Let $v(y_i) = z_i$ for $i = 1,\dots,m$. Let $x \in M$. Then there exist $b_1,\dots,b_m \in A$ such that $v(x) = b_1z_1 + \cdots + b_mz_m$. Then $v(x) = v(b_1y_1 + \cdots + b_my_m)$. Hence $x - (b_1y_1 + \cdots + b_my_m) \in Ker(v)$. Since $Ker(v) = Im(u)$, there exist $a_1,\dots,a_n \in A$ such that $x - (b_1y_1 + \cdots + b_my_m) = a_1u(x_1) + \cdots + a_nu(x_n)$. Hence $M$ is generated by $u(x_1),\dots,u(x_n), y_1,\dots,y_m$.

Answer 2

I would consider this as a very special case of the Horseshoe lemma and prove it like that. This is essentially the same prove as in Makoto Kato wrote down, but his is written down more elementary.