# Finitely generated modules 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
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 $x - (b_1y_1 + \cdots + b_my_m) \in \operatorname{Ker}v$. Since $\operatorname{Ker}v = \operatorname{Im}u$, 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 $u(x_1),\dots,u(x_n), y_1,\dots,y_m$.

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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.

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This can be a good comment, but is definitely not an answer! – user26857 Dec 30 '15 at 10:53