I have a question about exercise 2.13 in Weibel's K-Book, chapter I.
The goal is to prove that if $P$ is a projective $R$-module of constant rank $1$, then it is finitely generated; the hint suggests to look at the image $\tau_P$ of the natural map $\hom(P,R)\otimes P \to R$, show that it is $R$ (I did that), and then write $1 = \sum_i f_i(x_i)$ (which is of course just a reformulation of $\tau_P = R$)
I assume that the goal would now be to prove that $P$ is generated by the $x_i$, unfortunately I have no clue how to prove that... So my question is : is it what I'm supposed to do, and if so, what could be a further hint ?
If it helps, here's my proof that $\tau_P = R$ : an earlier exercise shows that $\tau_P \subset \mathfrak p \iff P_\mathfrak p = 0$ for a prime ideal $\mathfrak p$. However, $P$ has constant nonzero rank, so $P_\mathfrak p \neq 0$ for all $\mathfrak p$, therefore $\tau_P$ (which is a submodule, hence an ideal of $R$) is contained in no prime ideal, hence is $R$.
Here's what I found in the meantime :
let $R^n \to P$ be associated to the $x_i$.
Then localizing at $\mathfrak p$ we get a map $R_\mathfrak p^n \to P_\mathfrak p$. This map is surjective by Nakayama's lemma (*) and the fact that it surjective mod $\mathfrak p$ (that's the constant rank $=1$ hypothesis and the fact that at least one of the $x_i$'s is nonzero in $P_\mathfrak p/\mathfrak p P_\mathfrak p$ because of the equality $\sum_i f_i(x_i) = 1$)
Thus $R^n\to P$ is locally surjective, hence surjective : $P$ is finitely generated.
(*) To apply Nakayama's lemma I need to show that $P_\mathfrak p$ is a finitely generated $R_\mathfrak p$-module, which is not clear to me... A previous exercise shows that if $P$ has constant rank $=n$ and is finitely generated then $P_\mathfrak p \cong R_\mathfrak p^n$, but there is the assumption of finite generation that I don't have here (and the proof I found for that exercise did use finite generation, precisely to apply Nakayama's lemma)
EDIT : Ah ! But $P_\mathfrak p$ is $R_\mathfrak p$-projective, hence free (projective modules over a local ring are free), hence it must be finitely generated (because its quotient by $\mathfrak p$ is $1$-generated and it is free). Is there a proof that doesn't use "projective over local ring implies free" ?