# 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 ? – GreginGre 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 – Maxime Ramzi 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 $\tau_P = R$ ? – Maxime Ramzi 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. – Badam Baplan 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 – Maxime Ramzi Sep 19 '19 at 8:52