# Finitely generated module in exact sequence

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.

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Do this for vector spaces. Then do exactly the same for modules. –  Mariano Suárez-Alvarez Nov 11 '12 at 9:47
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See the proof of the lemma of my answer to this question. math.stackexchange.com/questions/231058/… –  Makoto Kato Nov 11 '12 at 10:01
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 $u(x_1),\dots,u(x_n), y_1,\dots,y_m$.