finitely generated projective module over local ring is free

Question

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 $F = \psi(M)\oplus K$, and it follows that $K = \mathfrak{m}K$. (???)

On the other hand, $K$ is a quotient of $F$, therefore finite over $A$, so that $K = 0$ by Nakayama and $F = M$.

Question: why the fact that $F = \psi(M)\oplus K$ implies that $K = \mathfrak{m}K$?

$K \subset \mathfrak{m}F$, then we can deduce $K＝K∩mF$＝$K∩（m \psi(M)\oplus mK）$ But why the last is equal to $mK$?

Thank you for your kind help.

Answer 1

You have $K=K\cap \mathfrak mF=K\cap(\mathfrak m\psi(M)\oplus \mathfrak mK)$. Since $K\cap \psi(M)=0$ ($\because F=\psi(M)\oplus K$), you get $K\cap\mathfrak m\psi(M)=0$. Thus $K=\mathfrak mK$ as required.

Answer 2

Let $k=A/\mathfrak{m}$ be the residue field of $A$. Tensor the exact sequence $G \rightarrow F\rightarrow M\rightarrow 0$ with $k$ (where $G$ is a free module minimally projecting on $K$) to get $F/\mathfrak{m}F\simeq M/\mathfrak{m}M$ ($K\subseteq \mathfrak{m}F$). Now tensor the equality $F=\psi(M)\oplus K$ with $k$ and count vector space dimension.