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.
Finitely generated modules in exact sequence
-
2See the proof of the lemma of my answer to this question. http://math.stackexchange.com/questions/231058/noetherian-rings-and-modules – 2012-11-11
3 Answers
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 $u(x_1),\dots,u(x_n), y_1,\dots,y_m$.
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.
-
1This can be a good comment, but is definitely not an answer! – 2015-12-30