I thought I had a fairly good understanding of finite projective modules over commutative rings, but I recently asked myself a few questions about them and this exposed how little I actually know. My first question was: can a finite projective module of constant rank $n$ be generated by $n$ elements? The answer is no - a simple example is a non-principal ideal in a Dedekind ring (projective of rank 1 but not generated by one element). (Edit: A finite rank $n$ projective module which can be generated by $n$ elements is necessarily free. There is a surjection from $A^n$ to the module, and a surjection between two finite projective modules of the same constant rank is necessarily an isomorphism). The interesting thing about this example though, is that such an ideal can always be generated by two elements. So, it led me to some other questions:

  1. Let $n<N$ be natural numbers. Is there a commutative ring $A$ and a finite, projective $A$-module $P$ of constant rank $n$ which cannot be generated by fewer than $N$ elements? (Clarification: Can one find such an $A$ and $P$ for every such pair $n, N$?)

  2. If the answer to 1 is no for some $n, N$, then can one find for each $n, N$ a finite module $M$ (not necessarily projective) such that $M_\mathfrak{p}$ can be generated by $n$ elements for each prime ideal $\mathfrak{p}$, but $M$ cannot be generated by fewer than $N$ elements?

  3. Does there exist a commutative ring $A$ and an $A$-module (resp. a projective $A$-module) $M$ which is not finitely generated but such that all localizations $M_\mathfrak{p}, \mathfrak{p} \in Spec(A)$ are finitely generated?

  4. Let $n>0$. Does there exist a commutative ring $A$ and an $A$-module (resp. a projective $A$-module) $M$ which is not finitely generated but such that all localizations $M_\mathfrak{p}, \mathfrak{p} \in Spec(A)$ can be generated by $n$ elements?

Until just a moment ago, I thought I had solved 3 and 4. I thought it was true that: if $M$ is any $A$-module and all localizations $M_\mathfrak{p}$ are finitely generated, then so is $M$.

Incorrect Proof. Let $\mathfrak{p} \in Spec(A)$, and choose a surjective map $A_\mathfrak{p}^n \to M_\mathfrak{p}$. Then since $A^n$ is finitely presented, $\text{Hom}_A(A^n, M)_\mathfrak{p} = \text{Hom}_{A_\mathfrak{p}}(A_\mathfrak{p}^n, M_\mathfrak{p})$ so this surjection can be taken to be the localization of some $f : A^n \to M$. Then since localization is exact,

$U_\mathfrak{p} = \{\mathfrak{q} \in Spec(A) | f_\mathfrak{q} \text{is surjective} \} = \{ \mathfrak{q} | (\text{Coker}(f))_\mathfrak{q} = 0 \} = Spec(A) \setminus Supp(\text{Coker}(f))$

is an open neighborhood of $U_\mathfrak{p}$. Let $D(a_\mathfrak{p})$ be a principal neighborhood of $\mathfrak{p}$ contained in $U_\mathfrak{p}$. Then $M_{a_\mathfrak{p}}$ is a finitely generated $A_{a_\mathfrak{p}}$-module: for every prime $\mathfrak{q}$ of $A_{a_\mathfrak{p}}$, $(f_{a_\mathfrak{p}})_\mathfrak{q}$ is surjective, so $f_{a_\mathfrak{p}}$ is surjective. Also,

$$\bigcup_{\mathfrak{p}} D(a_\mathfrak{p}) = Spec(A)$$

so there is a finite subcover, indexed by $\mathfrak{p}_1, \dots , \mathfrak{p}_n$. Set $a_i = a_{\mathfrak{p}_i}$. Then $a_1, \dots , a_n$ generated the unit ideal of $A$, so $A_{a_1} \times \cdots \times A_{a_n}$ is a faithfully flat $A$-algebra. Further, each $M_{a_i}$ is finitely generated as an $A_{a_i}$ module as we noted earlier. Hence

$$M \otimes (A_{a_1} \times \cdots \times A_{a_n}) = M_{a_1} \times \cdots \times M_{a_n}$$

is a finitely generated $A_{a_1} \times \cdots \times A_{a_n}$-module. By faithful flatness, $M$ is a finitely generated $A$-module.

The error in the proof comes when I show that $U_\mathfrak{p}$ is open: $\text{Coker}(f)$ need not be finitely generated, so $Supp(\text{Coker}(f))$ need not be closed in $Spec(A)$.

So, in the end I am unsure about all of 1,2,3 and 4. I would love to hear anyone's thoughts!

  • $\begingroup$ If $I$ is an ideal which is not finitely generated in a ring whose all localizations are noetherian you are done. For instance, take $R$ a countable direct product of $\mathbb F_2$. All its localizations are fields. (This answers 3 and 4.) $\endgroup$
    – user26857
    Nov 22, 2016 at 9:12
  • $\begingroup$ Great example for 3 and 4, thanks! This also solves the projective case since everything is projective over a field. Also, for 1 and 2, I want to know if it's possible for every pair $n, N$. $\endgroup$ Nov 22, 2016 at 17:52
  • $\begingroup$ Since an example for 1 gives an example fro 2, the `more generally' is a bit confusing. For $N=n+1$, such examples are standard (there are stably free modules of any rank which are not free). I would suspect that a close analysis of the universal examples (which is what works for $N=n+1$ case) might give you such examples, though I have not tried them. $\endgroup$
    – Mohan
    Nov 23, 2016 at 14:35
  • $\begingroup$ @Mohan Do you mean that any stably free module of rank $n$ can be generated by $n+1$ elements? I'm not sure about that... Certainly if $P \oplus A$ is free, but maybe not in general. $\endgroup$ Nov 24, 2016 at 19:54
  • $\begingroup$ I said no such thing. I only said there are stably free modules of rank $n$ which are generated by $n+1$ elements and not free, thus not by less than $n+1$ elements. This can be done for any $n$. $\endgroup$
    – Mohan
    Nov 24, 2016 at 21:08

1 Answer 1


The comment above should also answer 1) and 2): Let $V$ be a $n$-dimensional vector space over $\mathbb F_2$. Take $R=\prod_{i \geq 1} \mathbb F_2$ and $M=\prod\limits_{1 \leq i \leq N} V \times \prod\limits_{i > N} 0$.

  • $\begingroup$ These examples with countable products of fields are a little unfamiliar to me. Do you think you could state more explicitly what happens when you localize $M$? $\endgroup$ Nov 24, 2016 at 19:57

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