Equivalent properties of integral domain f.g projective modules

Question

Question :

Let $R$ be an integral domain and $K$ be its fraction field. Let $M$ be an $R$ module. Then the following are equivalent:

(1) $M$ is projective $R$-module such that $[M\otimes_R K : K]$ is finite.

(2) $M$ is finitely generated and $M_{m}$ is free over $R_m$ for all maximal ideals $m$.

(3) $M$ is finitely generated and projective $R$ module.

Attempt :

I have proven the implications $(3)\implies (1)$ and $(3)\implies (2)$.

But I am not able to show any other implication. For example, for showing $(2)\implies (3)$, I don't think it is necessary to invoke the proof of the fact that finitely generated flat modules over integral domains are projective. I am not sure whether a simpler proof exists or not.

For $(1)\implies (3)$, I have no idea how to show M is finitely generated.

Any help would be greatly appreciated.

Answer 1

This is just a long comment about $(2) \implies (3)$.

Being flat is a local property, so, even disregarding finiteness, $(2)$ is immediately equivalent to "$M$ is f.g. and flat".

Over most rings, it's not the case that finitely generated flat modules are projective, so I don't see how you can avoid 'invoking,' or proving for yourself, that f.g. flat modules over domains are projective.

That said, this follows immediately from the following result which everyone should have up their sleeve:

Proposition: Let $A \subseteq B$ an extension of rings. If $M$ is an f.g. flat $A$-module and $M \otimes_A B$ is a projective $B$-module, then $M$ is projective.

This originated in On finitely generated flat modules, S. Jøndrup. For a self-contained exposition, I like Stenström's concise introduction to flat and pure modules in Rings of Quotients, chapter I, sections 10-11, where the result appears as 11.6.

To prove what you want over a domain, this is all you need, because when you extend the module $M$ to $M \otimes_R K$ you get a $K$-vector space.

Your result also follows from the characterization of f.g. projective modules as f.g. flat modules with locally constant rank function, see [Stacks Lemma 10.77.2(8)]. This is a really useful characterization and worth understanding. Whenever a ring has finitely many minimal primes, the rank function is constant.

...Generalizing beyond domains...

More generally, it's not hard to show that finitely generated flat modules are projective over both semi-local rings. For this, you might first cite a companion result to the previous one:

Proposition Let $A$ a ring with Jacobson radical $J$ and $M$ an f.g. flat $A$-module. If $M/JM$ is projective as an $A/J$ module, then $M$ is projective as an $R$-module.

This is due to Vasconcelos in On Finitely Generated Flat Modules, Theorem 2.1. From here the semi-local case is easy, having been reduced to showing that over a finite product of fields an f.g. flat module is projective. Similarly, we get another way to see the result for rings with finitely many minimal primes. If a ring has finitely many minimal primes, then its quotient by the nilradical $N(R)$ has its ring of fractions a finite product of fields, so $M/N(R)M$ is projective over $R/N(R)$. Then since $N(R) \subseteq J$, clearly $M/JM$ is projective over $R/J$, and the cited result implies $M$ is projective.

All in all these two basic results allowed us to deduce the following cute

Conclusion Let $A$ be a subring of a ring $B$ such that either $\operatorname{maxSpec}(B)$ or $\operatorname{minSpec}(B)$ is finite. Then every f.g. flat $A$-module is projective.

Of course this subsumes the results about Noetherian rings and domains.

Bonus remark If $A$ is a ring such that $\operatorname{maxSpec}(A)$ or $\operatorname{minSpec}(A)$ is compact, then every f.g. flat ideal of $A$ is projective. (Here you need to interpret "compact" in the appropriate topology.)

Answer 2

For $(1) \Rightarrow (3)$ choose $N$ such that $M \oplus N$ is free. Let $X$ be a basis so that $M \oplus N \simeq R^{|X|}$, where here it's important to note that $R^{|X|}$ is a direct sum of $|X|$ copies of $R$. Tensoring with $K$ gives $(M \otimes_R K) \oplus (N \otimes_R K) \simeq K^{|X|}$. Now $M \otimes_R K$ is spanned by finitely many vectors which are nonzero in finitely many coordinates. Projecting just to those coordinates we can assume $X$ is finite. Then $M \oplus N$ is finitely generated so its quotient $M$ is finitely generated.

A good takeaway is to remember that a projective module is finitely generated if and only if it's a summand of a finitely generated free module.

For the locally free $\Rightarrow$ projective argument I know of no way to do it other than the standard arguments using flat modules, and those can be looked up in pretty much any commutative algebra reference.