Question 2.12 in Atiyah and MacDonald is similar to this:
Let M be a finitely generated A-module and $φ : M \longrightarrow A^n$
a surjective homomorphism. Show $\ker φ$ is finitely generated.
I like the solution posted here because it is short: https://users.math.msu.edu/users/ruiterj2/Math/Documents/Fall%202018/Commutative%20Algebra/Commutative%20Algebra%20Homework%202.pdf
Proof. Since $A^n$ is a free, and hence projective, A-module, the short exact sequence
$0 \longrightarrow \ker φ \longrightarrow M \longrightarrow A^n \longrightarrow 0$
splits, so $M \cong A^n ⊕ \ker φ$. Since $M$ and $A^n$ are finitely generated, so is $\ker φ$.
I understand the short exact sequence says the second function is 1-1 and the third, $φ$, is onto which we know from the problem statement, and this sequence is similar to the rank-nullity theorem – and the implication that $\ker φ$ is finitely generated, but how can $M \cong A^n ⊕ \ker φ$?
$\ker φ$ is a submodule of M so how can M be isomorphic to a direct sum of the free module $A^n$ and $\ker φ$
Best Answer
Recall that the direct sum of two $A$-modules $P,Q$ is:
$$P\oplus Q=\{(p,q)\mid p\in P, q\in Q\}$$ So $A^n\oplus \ker\varphi$ can be thought of as the set of ordered pairs $(a,\phi)$, with $a\in A^n$ and $\phi\in\ker\varphi$.
Can you now see the isomorphism with $M$?