Let $A$ be a commutative ring. If $M,N$ are $A$-modules (not necessarily finitely generated), and $\beta:M\longrightarrow N$ is surjective, prove that $$M\simeq\ker(\beta)\oplus\frac{M}{\ker(\beta)}\;\;.$$

($\oplus$ is the direct sum of A-modules).

This is a way to prove the following problem: let $$0\rightarrow L\stackrel{\alpha}{\rightarrow} M\stackrel{\beta}{\rightarrow} N\rightarrow0$$ be a short exact sequence of A-modules. Suppose that there exists $\rho:M\rightarrow L$ A-linear and surjective such that $\rho\circ\alpha=id_{L}$.

Then show that there exist a short exact sequence $$0\rightarrow L\stackrel{\xi}{\rightarrow} L\oplus N\stackrel{\eta}{\rightarrow} N\rightarrow0$$ (in which obiouvsly $L$ and $M$ must be considerated up to isomorphism and $\xi$ and $\eta$ are defined in the obviouvs way) and an isomorphism (and this is the tricky part) $$ \varphi:M\longrightarrow L\oplus N $$ such that the diagram you obtain by drawing the first sequence over the second, is commutative.

Proving the first thing I wrote is enough to do all remaining work. However what I'm interested in, is the solution of the exercise (that is the remaining part), hence, if someone is able to solve at least one of the two questions, I would be a happy guy! Thank you all!

  • 2
    $\begingroup$ It is not true that kernels of modules are always direct factors. So you will need to do something other than this for the exercise. $\endgroup$
    – anon
    Nov 9, 2013 at 20:30
  • $\begingroup$ How do you know that very first short exact sequence exists? $\endgroup$ Nov 9, 2013 at 20:30
  • 3
    $\begingroup$ It's the existence of $\rho$ that guarantees the direct sum decomposition of $M$. Without using $\rho$, you can't show the sequence splits. $\endgroup$ Nov 9, 2013 at 20:32
  • $\begingroup$ Excuse me, how can we construct the map $\rho$? $\endgroup$
    – H. Jackson
    Apr 1, 2017 at 14:37

1 Answer 1


As already remarked in the comments, your first claim is not true. What you need is a map $\rho \colon M \to \text{ker}(\beta)$ with $\rho|_{\text{ker}(\beta)} = 1_{\text{ker}(\beta)}$. With this additional assumption, there is hardly any difference between the first claim (with $\text{ker}(\beta)$) and the second claim (with the short exact sequences).

For the proof, just write down the obvious map $M \to \text{ker}(\beta) \oplus M/\text{ker}(\beta)$ (there now is one, because you have $\rho$) and prove that it is injective and surjective.

Edit (for completeness' sake). The obvious map $\phi \colon M \to \text{ker}(\beta) \oplus M/\text{ker}(\beta)$ is $\phi(m) = (\rho(m), \bar m)$.

Injectivity: Suppose $\phi(m) = 0$, i.e., $\rho(m) = 0$ and $\bar m = 0$. Then $m \in \text{ker}(\beta)$ and because $\rho|_{\text{ker}(\beta)} = 1_{\text{ker}(\beta)}$, $m = \rho(m) = 0$.

Surjectivity: Take $(l, \bar n) \in \text{ker}(\beta) \oplus M/\text{ker}(\beta)$. Now note that $\phi(n) = (\rho(n), \bar n)$. We can now 'fix' the left component by adding $l - \rho(n)$. Using $l - \rho(n) \in \text{ker}(\beta)$, we get $\phi(n + l - \rho(n)) = (l, \bar n)$.

It might be instructive to cast this argument purely in terms of the short exact sequences.


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