# Is a projective module of constant finite rank finitely generated?

If $R$ is a (commutative) ring and $P$ is a projective $R$-module, then every localization of $P$ at a prime of $R$ is free by Kaplansky's theorem, and has a well-defined rank. If these ranks are all finite, must $P$ be finitely generated?

The answer is no: one can take $R=k[x_1,x_2,\dots]/(x_1^2-x_1,x_2^2-x_2,\dots)$ and $P=\bigoplus_{n=1}^\infty R/\bigl(1-x_1x_2\cdots x_{n-1}(1-x_n)\bigr)$; then $P$ is projective, not finitely generated, and at each prime at most one of its summands is nonzero. At each prime $\mathfrak{p}$ of $R$, then, $P_\mathfrak{p}$ is a free rank-$1$ $R_\mathfrak{p}$-module, unless $\mathfrak{p}=(1-x_1,1-x_2,\dots)$, for which $P_\mathfrak{p}=0$.

Or here's another example: let $R$ be the ring of continuous functions $[0,1]\to\mathbb{R}$, and let $P\subseteq R$ be the ideal consisting of those functions that vanish on a neighborhood of zero. Then again $P_\mathfrak{p}$ is free of rank 1 at every prime $\mathfrak{p}$ of $R$, except for those $\mathfrak{p}$ consisting of functions vanishing at $0$; at such $\mathfrak{p}$ we have $P_\mathfrak{p}=0$.

Such examples must be reasonably complicated due to this result of Bass's:

Proposition 4.2. Suppose $R$ has only finitely many primes minimal above $0$... If $P$ is a locally finitely generated projective $R$-module, then $P$ itself is finitely generated.

("Big Projective Modules are Free", 1963)

However, I'm wondering if it's possible to remove the conditions on $R$ if we assume not only that $P$ is projective, but that its rank is a continuous function $\mathrm{Spec}(R)\to \mathbb{N}$. Since continuity of the rank function is equivalent to the rank being locally constant, we can work locally on affine opens where the rank is constant. Thus the question reduces to:

If $P$ is a projective $R$-module of constant finite rank, is $P$ finitely generated?

• Yes, according to Exercises I.2.13 and I.2.14 of Weibel's K-Book. The hints there say: use the determinant to reduce to constant rank one, and use the trace to show finite generation in the rank one case (much like showing quasicompactness of affine schemes). Aug 16 '14 at 9:20
• Thanks for the reference; that's exactly what I'm looking for! But the hint to use the trace is a little fuzzy: in the exercise where he introduces the properties of traces, he also talks about an integer-valued rank function, which makes it sound like $P$ is implicitly assumed to be finitely generated. Aug 18 '14 at 10:14
• I think the rank statement is not so relevant because we are assuming constant rank one anyway. The argument probably should go: the image of $\check{P}\otimes P\to R$ is an ideal, and constant rank one implies that this ideal can not be contained in any prime ideal. Then you write $1=\sum f_i(x_i)$, so that the $x_i$ are the generators. It does not seem that finite generation is used in this argument (but I could be missing something). Aug 18 '14 at 12:52
• I was concerned that Weibel's exercises about the trace were only valid in the finitely-generated case, but the relevant statement (that for each prime $\mathfrak{p}$, we have $\tau_P\subset\mathfrak{p}$ iff $P_\mathfrak{p}=0$) is in fact true for general projective $P$. (That also explains why the supports of the non-constant-rank projective modules in the question were open—their complements are the closed subschemes cut out by their trace ideals!) The rest is just checking that the generators are what you think they are. If you expand your comment into an answer, I'll accept it. Aug 18 '14 at 13:31

For any commutative ring $R$ with 1, any projective $R$-module of constant finite rank is finitely generated. This is the content of Exercises I.2.13 and I.2.14 of Weibel's $K$-book. The argument goes as follows (all the relevant hints are in Weibel's book), I hope I did not introduce any tacit finiteness assumptions.
1) In the constant rank one case, we have the dual $\check{P}=\operatorname{Hom}_R(P,R)$ and the evaluation map $\operatorname{ev}:\check{P}\otimes_R P\to R$ whose image $\tau_P=\operatorname{im}\left(\operatorname{ev}:\check{P}\otimes_RP\to R\right)$ is called the trace. By a result of Kaplansky, $P_{\mathfrak{p}}$ is free for any prime ideal $\mathfrak{p}\subseteq R$, and we can choose a generator $x\in P_{\mathfrak{p}}$. Then $\check{x}\otimes x\mapsto 1$ under $\check{P_{\mathfrak{p}}}\otimes_{R_{\mathfrak{p}}}P_{\mathfrak{p}}\to R_{\mathfrak{p}}$. In particular, $\tau_P\not\subseteq\mathfrak{p}$ since otherwise $P\otimes_R R/\mathfrak{p}=0$ contradicts the constant rank one assumption (as pointed out in the comments of Owen Biesel). Since $\tau_P$ is an ideal, it contains $1$ and we can write $1=\sum f_i(x_i)$. For each $\mathfrak{p}$, some $x_i$ will be a generator of $P_{\mathfrak{p}}$, so the $x_i$ are generators for $P$.
2) If $P$ has constant rank $r$, then $\bigwedge^r P$ has constant rank one and by Step 1 has finitely many generators. These are of the form $\sum a_i y_{1i}\wedge\cdots\wedge y_{ri}$ and the (finitely many) $y_{ij}$ appearing generate $P$.
• There actually is a bit of finiteness I think you've introduced. To conclude that $\tau_P\nsubseteq\mathfrak{p}$, you need to know that $\mathrm{Hom}_R(P,R)_\mathfrak{p} \cong \mathrm{Hom}_{R_\mathfrak{p}}(P_\mathfrak{p},R_\mathfrak{p})$, but generally this requires $P$ to be finitely presented. However... Aug 19 '14 at 9:56
• ... if it were the case that $\tau_P\subseteq\mathfrak{p}$, then we find (e.g. using a dual basis for $P$) that $P=\mathfrak{p}P$, so $P\otimes_R \kappa(\mathfrak{p})=0$. But this is impossible, since $P_\mathfrak{p}$ is assumed to be free of rank 1, so $P\otimes_R \kappa(\mathfrak{p})\cong \kappa(\mathfrak{p})$. Aug 19 '14 at 9:59
• Dear Mathias, your sentence "Since $\tau_P$ is an ideal, it contains $1$ " is a bit strange, to put it mildly... Sep 29 '20 at 18:47
• @GeorgesElencwajg: True. I probably should have written that $\tau_P\subseteq R$ is an $R$-submodule not contained in any prime ideal and therefore contains 1. Oct 1 '20 at 7:41