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 projective module $M$ over $A$ is free.
The understanding problem arises in the verification of the case when $M$ is finite (as $A$-module). 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 $K = \mathfrak{m}K$, more precisely why $K \subset \mathfrak{m}K$? (The other inclusion is trivial.)
Considerations: $F = \psi(M)\oplus K$ is a decomposition as $A$-modules and since $\mathfrak{m} \subset A$ we obtain $\mathfrak{m} F = \mathfrak{m} \psi(M)\oplus \mathfrak{m} K$. since $K \subset \mathfrak{m}F$ it suffices to show that $\mathfrak{m} \psi(M)=0$. Why is it true?
I would like additionally to remark that I found already some other proofs but the intention of this question bases only on the understanding of the explained step in the presented proof.
Best Answer
If $\{\omega_1,...,\omega_n\}$ is a minimal system of generators of $M$, their images $\{\mkern1mu\overline\omega_1,...,\overline\omega_n\}$ in $M/\mathfrak mM$ are a basis of the $A/\mathfrak m$-vector space $M/\mathfrak mM$.
Now tensor the decomposition $F=\psi(M)\oplus K$ with $A/\mathfrak m$: you obtain $$M/\mathfrak mM\simeq F/\mathfrak m F=\bigl(A/\mathfrak m\bigr)^n=\psi(M)/\mathfrak m\psi(M)\oplus K/\mathfrak m K\simeq M/\mathfrak mM\oplus K/\mathfrak mK$$ so $K=\mathfrak mK$. As $K$ is finitely generated, we can appeal to Nakayama again.