I'm having some trouble understanding the following proof:
We shall only use the assumption that $$ 0 \to M' \xrightarrow{f} M \xrightarrow{g} M'' \to 0 $$ is exact and $M''$ is finitely generated. We see that if $x_1, x_2, . . . , x_n$ are generators for $M''$, then $\bar x_1,\bar x_2, . . . , \bar x_n$ in $\text{coker}(g) = M''/f(M')$ [1] will generate $\text{coker}(g)$. But by the exactness of the sequence, $\text{coker}(g) \simeq M$ [2], hence $M$ will be finitely generated.
My questions are:
- In my algebra class, I remember my teacher saying that if $g: A\to B$ is a homomorphism, then the cokernel of $g$ is defined as the quotient $B/g(A)$. In this case, $g$ is surjective as the sequence of homomorphism is exact, so shouldn't we have that $\text{coker}(g)\simeq (0)$? And why did he write $\text{coker}(g)=M''/f(M')$ instead of $\text{coker}(g)=M''/g(M)$?
- Why don't get why being exact would imply that $\text{coker}(g)\simeq M$. If I'm correct in my previous note, that would imply that $M\simeq 0$, right?
Your questions are valid objections and the proof makes no sense.
Also, the statement in question is false. Let $M'=M$, be any module that is not f.g. and let $M''=0$ and consider the sequence $0 \to M' \to M \to M'' \to 0$, which is just $0 \to M \to M \to 0 \to 0$. Clearly $M''$ is finitely generated, but $M$ is not.
A related, but correct statement is that $M$ is finitely generated if both $M'$ and $M''$ are finitely generated. Another related true statement is that $M''$ is finitely generated if $M$ is.