Silly thing I cannot wrap my head around. Lem. I.2.2 in Weibel's K-book reads:
If $R$ is a local ring, then every finitely generated projective $R$-module $P$ is free. In fact $P \cong R^p$, where $p = \dim_{R/\mathfrak{m}} (P/\mathfrak{m}P)$.
($\mathfrak{m}$ of course being the maximal ideal of the local ring $R$.)
What troubles me in the proof are the following lines:
Suppose that $P \oplus Q \cong R^n$. As vector spaces over $F=R/\mathfrak{m}$, $P/\mathfrak{m} P \cong F^p$ and $Q/\mathfrak{m} Q \cong F^q$ for some $p$ and $q$. Since $F^p \oplus F^q \cong F^n$, $p+q=n$. Choose elements $\{ e_1 , \dots , e_p \}$ of $P$ and $\{ e'_1 , \dots , e'_q \}$ of $Q$ mapping to bases of $P/\mathfrak{m} P$ and $Q/\mathfrak{m} Q$. The $e_i$ and $e'_j$ determine a homomorphism $R^p \oplus R^q \rightarrow P \oplus Q \cong R^n$.
The thing is plainly this: I don't see how the $e_i$ and $e'_j$ determine a homomorphism $R^p \oplus R^q \rightarrow P \oplus Q \cong R^n$. Can someone explain to me?
Ah! Silly me! The homomorphism in question is $R^p \oplus R^q \rightarrow P \oplus Q$, $(r_1, \dots , r_p , r_{p+1}, \dots , r_{p+q}) \mapsto (r_1 e_1, \dots , r_p e_p , r_{p+1} e'_1 , \dots , r_{p+q} e'_{q})$. Got it.