# finitely generated projective module over local ring is free

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.

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.
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.
• Sure we can, tensor products are unnecessary, $\mathfrak{m}F=\mathfrak{m}\psi(M)\oplus\mathfrak{m}K$, so just use the definition of a direct sum. Dec 15, 2020 at 7:43
• Every element $f\in F$ has a unique representation $f=a+b$, $a\in \psi(M)$, $b\in K$. If $k\in K$ is arbitrary, then $k$ as an element of $F$ (it is a submodule) can be written in the above form as $0+k$. On the other hand every element of $\mathfrak{m}F$ has a unique decomposition $x+y$, $x\in \mathfrak{m}\psi(M)\subseteq \psi(M)$, $y\in \mathfrak{m}K\subseteq K$. So $k$ as an element of $\mathfrak{m}F$ and hence $F$ can be written uniquely as $x+y$, $x\in \mathfrak{m}\psi(M)$, $y\in \mathfrak{m}K$. Therefore $x=0$, $y=k$, so $K=\mathfrak{m}K$. Dec 15, 2020 at 17:00