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" ?

  • $\begingroup$ 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 ? $\endgroup$
    – GreginGre
    Sep 18 '19 at 17:38
  • $\begingroup$ @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 $\endgroup$ 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 $P' = P$.

  • $\begingroup$ Thanks for your answer ! Is it clear that $P\otimes \hom(P,R) \cong R$ from $\tau_P = R$ ? $\endgroup$ Sep 19 '19 at 6:10
  • $\begingroup$ No it is not clear! I'm very sorry, I glossed over that part; I was thinking you already knew that $P$ was invertible. $\endgroup$ Sep 19 '19 at 6:47
  • $\begingroup$ 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 $\endgroup$ Sep 19 '19 at 8:52

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