Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
Do this for vector spaces. Then do exactly the same for modules. –  Mariano Suárez-Alvarez Nov 11 '12 at 9:47
    
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. –  Julian Kuelshammer Nov 11 '12 at 9:49
2  
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

2 Answers 2

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 $u(x_1),\dots,u(x_n), y_1,\dots,y_m$.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.