I recently came across the following statement, however I always fail to prove it. I'm also a bit unsure where the assumption "noetherian" is used:
Let $R$ be a noetherian ring, and $M_i$ be finitely generated $R$-modules. Assume that $M_2 \simeq M_1 \oplus M_3$ and the following sequence is exact:
$$ 0 \rightarrow M_1 \xrightarrow{f} M_2 \xrightarrow{g} M_3 \rightarrow 0$$ Then this sequence splits.
Here is my Idea:
I have to construct an isomorphism $\varphi: M_2 \rightarrow M_1 \oplus M_3$ such that the following diagram commutes:
$$\require{AMScd}
\begin{CD}
M_1 @>{f}>> M_2 @>{g}>> M_3\\
@VVV @VVV @VVV \\
M_1 @>{i}>> M_1 \oplus M_3 @>{p}>> M_3
\end{CD}$$
Left and right maps are the identity, the mittle map $\varphi$ and the lower ones inlucion and projection.
It would also suffice to find a left-inverse to $f$ or a right-inverse to $g$. However I would mainly be interested in the isomorphism for its construction (a.k.a both of the inverses. The rest is taking preimages via inclusion and images via projection). Anyway, on to the proof:
Lets take generators $(\alpha_i)$ of $M_1$, $(\beta_j)$ of $M_2$ and $(\gamma_k)$ of $M_3$. All those sets of generators are of finite cardinality. If i could show, that $f$ maps the $\alpha_i$ uniquely to some of the $\beta_j$, I would be done. I could just map those same $b_j$'s back to the $\alpha_i$'s under $\varphi$.
Since $f$ is injective, there exists a submodule $N \subset M_2$ such that $M_1 \simeq N$ under $f$. (I think here it is that the Noetherian assumption is used) N is also finitely generated, which means that $(f(\alpha_i))$ generates N. Therefore there exists an inverse $f' : N \rightarrow M_1$, such that $f'(f(\alpha_i)) = \alpha_i$. I would now love to say that $n_1$ of the $\beta_j$ generate $N$. But how can I do this? Somehow I need to incorporate exactness...
Since the top row is exact, we know by the first isomorphism Theorem that $M_2/N \simeq M_3$ under $g$. (Again I think here is noetherian used) $M_2/N$ is finitely generated, explicitely by preimages of generators in $M_3$.
But now I am stuck. I just cant manage to show that these things now generate $M_2$. Also I dont know how I now show commutativity. Is this even the right approach?
Edit:
I don't know if I'm allowed to do this, but I have another Idea:
I take a generating set $(\alpha_i)$ of $M_1$. $f$ is injective and an isomorphism to a submodule $N \subset M_2$. Now define $\beta_i := f(\alpha_i)$. So far so good, now comes the shady part:
I map the $\beta_i$'s via $\varphi$ to $(\alpha_i,0)$. I now want to complete a generating set of $M_2$ out of the $\beta_i$'s by adding other elements.
Take a generating set $(\gamma_j)$ of $M_3$ and choose elements $\beta_j \in M_2$ such, that $\beta_j \mapsto \gamma_j$ under $g$. Especially it holds that such $\beta_j \notin N$ by exactness. Now map $\beta_j \mapsto (0, \gamma_j)$ under $\varphi$ and extend linearly to the submodule of $M_2$, generated by $\beta_i$ and $\beta_j$.
Finally, since the Elements $(\alpha_i,0)$ and $(0, \gamma_j)$ generate $M_1 \oplus M_3$ and $M_2 \simeq M_1 \oplus M_3$ is bijective on the generators, $\varphi$ is an isomorphism, $\beta_i,\beta_j$ must generate $M_2$ and the diagram commutes by construction.
Does this kind of work?