I have a question about an argument from a proof in Hideyuki Matsumura's "Commutative Ring Theory" on page 9, Theorem 2.5:
Let $(A,\mathfrak{m})$ be a local ring; then a finitely generated projective module $M$ over $A$ is free.
The proof works as follows:
Choose a minimal $A$-basis $\omega_1,…,\omega_n$ of
$M$. Take into account that "minimal" means that there cannot exist another system of generators $b_1,…, b_m \in M$ with $m < n$ and $M = \sum_{i=1} ^m A b_i$.
Define a surjective map $\varphi:F \to M$ from the free module $F = Ae_1 \oplus \cdots \oplus Ae_n$ to $M$ by $\varphi(\sum a_i e_i) = \sum a_i\omega_i$. If we set $K = \operatorname{Ker}(\varphi)$ then, from the minimal basis property
$$\sum a_i \omega_i =0 \Rightarrow a_i \in \mathfrak{m} \text{ for all } i. $$
Thus $K \subset \mathfrak{m}F$. Because $M$ is projective, there exists $\psi: M \to F$ such that
$F = \psi(M)\oplus K$,
Why $F = \psi(M)\oplus K$? What I know is this exact sequence splits:
$$0\to \ker\varphi\to F\to M\to 0$$
Where the first non trivial homomorphism is the canonical injection
and the second one $\varphi$, since this sequence splits we have:
$F=\ker\varphi\oplus M$.But why $F = \psi(M)\oplus K$？
Can we show $M＝ \psi(M)$？
Best Answer
It’s actually the same thing up to isomorphism.
You can take $\psi$ as a splitting morphism that goes from M to F.
By definition, a splitting morphism verifies $\phi \circ \psi =id$. So $\psi$ is injective and $\psi(M)$ is isomorphic to M.
The main difference is that we work with internal direct sums (so it’s actually $F=\psi(M)\oplus K$ not just a isomorphism) when you use $\psi(M)$ which lives in F as well as K.