# Every finitely generated flat module over a ring with a finite number of minimal primes is projective

Over a commutative ring $R$, a finite type locally free (weak sense) module for which the rank function is locally constant is projective.

If we notice that for each minimal prime $p$ of the ring, the rank function is constant on the adherence of $p$ in the Zariski topology (because if $p\subset q$ the rank at $p$ is equals to the rank at $q$) on finite locally free (weak sense) modules, then if the ring has only a finite number of minimal primes, the rank function is always locally constant on finite flat modules. Am I right ?

Edit 2: Answering a request of Ben the detailed reasoning is the following: The image of the rank function is finite in $\mathbb{N}$. Every reciprocal image of an integer $n$ is a finite union of closed sets like the $V(p)$. So it is a closed set. The disjoint union of these closed sets is the whole spectrum. Each one of them is therefore open.

I am asking this simple question because I read this great answer here where an important paper of Raynaud-Gruson is mentioned. It gives amongst a lot of generalizations the simple criteria that if $R$ has a finite number of associated primes then without any other hypothesis on $R$ every f.g. flat modules is actually projective. My reasoning above seems quite simple and gets a slightly more general result, but perhaps I am wrong ?

Edit: since nobody answered my question I dug around but I did not find any reference to this simple criteria. I only found the criteria about the finiteness of the number of associated primes. Is it equivalent ? Is there a counterexample with an infinite number of embedded primes but a finite number of minimal (isolated) primes ?

Edit 3: the answer is here.

• I may be missing a simple point, but how exactly do you attempt to cover the spectrum by open(!) subsets where the rank is constant?
– Ben
Sep 29 '15 at 20:03
• @Ben Thank you very much for your time. My simple reasonning is the following : the image of the rank function is finite in $\mathbb{N}$. Every reciprocal image of an integer $n$ is a finite union of closed sets like the $V(p)$. So it is a closed set. The disjoint union of these closed sets is the whole spectrum. Each one of them is therefore open right ? Sep 29 '15 at 20:20
• Aah, good point. If the subsets are not disjoint but the rank function is constant on both of them, then it agrees at the intersection and thus is constant on their union; that is what I missed. Seems fine to me now, nice observation!
– Ben
Sep 29 '15 at 20:45
• @Ben +1 for your comment. Thanks ! Unfortunately this question has not received enough attention. But if I am right it is a simple result I never saw anywhere (there is a close but less general one as I quoted in the question, and that is reached through a much more complicated approach ...) Sep 29 '15 at 20:54
• Dear @brunoh, I think I have an example of a ring with unique minimal, but infinitely many associated primes. Would you like me to post it here or on MO? By the way, there, you asked for the etiquette of cross posting: when a question like yours isn't getting enough attention here, then there could be a point of posting it on MO too. But you should've waited for the bounty time here to elapse. Also, always tell people on both sites that and where you cross-posted, so users can check out the other site for useful comments or answers.
– Ben
Oct 1 '15 at 12:21

Upon request, I repost my MathOverflow answer (c.f. MO/218737). There are two questions here, namely, whether the argument given in the question is okay, and whether the criterion really is more general than the one by Raynaud-Gruson. The answer to both questions is yes. The following holds:

Let $A$ be a commutative with unity and $M$ a flat finitely generated $A$-module. Then the rank function $\mathrm{Spec}(A)\to\mathbb{N}_0$, $\mathfrak{p}\mapsto \mathrm{rk}_{A_\mathfrak{p}}(M_{\mathfrak{p}})$, is constant on all irreducible components of $\mathrm{Spec}(A)$. In particular, if $\mathrm{Spec}(A)$ has finitely many maximal irreducible components (i.e., if $A$ has finitely many minimal prime ideals), then the rank function is locally constant, hence $M$ is projective.

The (very nice and short) proof was given by the OP in the question text and in a comment. (With the same arguments, one can further generalise this to schemes; see loc. cit.)

The criterion by Raynaud-Gruson requires finitely many associated primes, so to be sure that the above criterion is more general, we need an example of a ring with finitely many minimal, but infinitely many associated prime ideals. The following example has a single minimal prime ideal, but infinitely many associated prime ideals. Picture it as a line with infinitely many embedded points. (Just as $k[x,t]/(t^2,xt)$ has associated primes $(t)$, the unique minimal one, and $(x,t)$, the "embedded point".)

Let $H$ be the ring of holomorphic functions on the complex line $\mathbb{C}$ (with coordinate $z$) and $f\in H$ a non-trivial holomorphic function with infinitely many zeros (e.g., $f(z) = \sin(z)$). We consider the ring $A:=H[t]/(t^2,tf)$. Since $A/(t)\cong H$ is an integral domain, the ideal $(t)$ is prime. On the other hand, this ideal happens to be the nilradical of $A$, which is the intersection of all prime ideals; thus, $(t)$ is the only minimal prime. However, for each root $\alpha\in\mathbb{C}$ of $f$ we recognise the associated prime ideal $(t,z-\alpha)\subset A$, so there are infinitely many of them. In fact, if $g\in H$ is the unique entire function with $(z-\alpha)g = f$, then it is easy to see that $(t,z-\alpha) = \mathrm{ann}_{A}(tg)$.

• Dear @brunoh, I'm sorry for the late response. I wanted to construct a new example for this post, but what I had in mind didn't work out and then I got busy, so I forgot to repost the known one. I hope it's okay that I didn't copy-paste the proof.
– Ben
Oct 4 '15 at 20:10
• better late than never. You got a well deserved IMO second bounty for your good work ! Oct 4 '15 at 21:33