# Projective module of constant rank $1$ is finitely generated (K-Book)

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" ?

• What is for you the definition of constant rank $1$ ? For me, its $P_\mathfrak{p}\simeq R_\mathfrak{p}$ for all $\mathfrak{p}$. But then$P_\mathfrak{p}$ is a finitely generated $R_\mathfrak{p}$-module, isn't it ? Sep 18 '19 at 17:38
• @GreginGre : Weibel defines the rank as the dimension of $P_\mathfrak p/\mathfrak p P_\mathfrak p$ over $k(\mathfrak p)$ (the residual field). It turns out that these are equivalent for projective modules, but it's not clear that they are for non projective infinitely generated modules Sep 18 '19 at 17:44

Suppose that $$P$$ and $$Q$$ are mutually inverse modules, with $$P \otimes_R Q \cong R$$.

If $$\sum_{i=1}^n p_i \otimes q_i$$ maps to $$1$$ in this isomorphism, then we let $$P' = (p_1, \cdots, p_n)$$ and claim that $$P' = P$$.

To show this, consider that we have an embedding $$P' \otimes_R Q \subseteq P \otimes_R Q \cong R$$ because $$Q$$ is flat, and since this is surjective by construction, we deduce that $$P' \otimes_R Q = P \otimes_R Q$$. Now we just play with tensor products....

Elaborated in spoiler below, if you want

$$P' \cong P' \otimes_R R \cong P' \otimes_R Q \otimes_R P = P \otimes_R Q \otimes_R P \cong P \otimes_R R \cong P$$ The above series of maps just induces the inclusion $$P' \subseteq P$$, so we conclude $$P' = P$$.

• Thanks for your answer ! Is it clear that $P\otimes \hom(P,R) \cong R$ from $\tau_P = R$ ? Sep 19 '19 at 6:10
• No it is not clear! I'm very sorry, I glossed over that part; I was thinking you already knew that $P$ was invertible. Sep 19 '19 at 6:47
• Well I know it if I again use projective over local implies free, but if I use this I know how to solve the question anyway Sep 19 '19 at 8:52