Let M be a finitely generated A-module and $φ : M → A^n$ a surjective homomorphism. Then $\ker φ$ is finitely generated

Question

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 $\ker φ$

Answer 1

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 $M$?

Answer 2

You can either think of this as the genuine definition of $A^n$ being projective, or an equivalent condition.

As stated in the Wikipedia link provided in the comments, a module $P$ is projective if and only if every short exact sequence of the form

$$0 \rightarrow M \xrightarrow{f} N \xrightarrow{g} P \rightarrow 0$$

splits. That is, there exists a section $$h:P \rightarrow N$$ such that $g \circ h: P \rightarrow P = \text{Id}_P$.

Now, this does not yet give you exactly what you need. The property that $N \cong M \oplus P$ does follow from this, though. You should treat this deduction as an exercise. Try to construct a map $M \oplus P \rightarrow N$ (there is an obvious way to do this, given the existence of the section $h:P \rightarrow N$), and then show that this is an isomorphism.

In general, not all short exact sequences split; that's what is special here (and indeed, what is special about projective modules).