If $\pi:M \to A^n$ is a surjection from $A$-module $M$ to free module $A^n$, then $M\cong A^n \oplus \ker\pi$.

Question

Let $A$ be a commutative ring. Let $\pi:M \to A^n$ be a surjective module homomorphism, where $M$ is an $A$-module and $A^n$ is the free module of rank $n$.

Prove that $M\cong A^n \oplus \ker\pi$.

Thoughts:

I know that since $A^n$ is free, it has a basis $\{e_1,…,e_n\}\subset A^n$. Now I can choose some $\{m_1,…,m_n\}\subset M$ where each $m_i$ maps to $e_i$. I can see that the submodule generated by $\{m_1,…,m_n\}$ is isomorphic to $A^n$, but how do I prove something is a an external direct sum?

Answer 1

You only have to prove that it is an internal direct sum, i.e.

1. $\langle m_1,\dots,m_n\rangle+\ker\pi=M$, and
2. $\langle m_1,\dots,m_n\rangle\cap\ker\pi=\{0\}.$