Flatness and projectivity: when is the localization of a ring a projective module?

Projective modules and flat modules are two important concepts in algebra, because they characterize those modules for which a general functorial construction (Hom module and tensor product, respectively) behave better than what is the case for general modules.

This blog post came out of reading a confusion on a student's exam: projective modules are flat, but not all flat modules are projective. Since localization gives flat modules, it is easy to obtain a an example of a flat module which is not projective (see below, $\mathbf Q$ works, as a $\mathbf Z$-module), but my question was to understand when the localization of a commutative ring is a projective module.

$\gdef\Hom{\operatorname{Hom}}\gdef\Spec{\operatorname{Spec}}\gdef\id{\mathrm{id}}$

Let me first recall the definitions. Let $R$ be a ring and let $M$ be a (right)$R$-module.

The $\Hom_R(M,\bullet)$-functor associates with a right $R$-module $X$ the abelian group $\Hom_R(M,X)$. By composition, any linear map $f\colon X\to Y$ induces an additive map $\Hom_R(M,f)\colon \Hom_R(M,X)\to \Hom_R(M,X)$: it maps $u\colon M\to X$ to $\phi\circ u$. When $R$ is commutative, these are even $R$-modules and morphisms of $R$-modules. If $f$ is injective, $\Hom_R(M,f)$ is injective as well, but if $f$ is surjective, it is not always the case that $\Hom_R(M,f)$ is surjective, and one says that the $R$-module $M$ is projective if $\Hom_R(M,f)$ is surjective for all surjective linear maps $f$.

The $\otimes_R$-functor associates with a left $R$-module $X$ the abelian group $M\otimes_R X$, and with any linear map $f\colon X\to Y$, the additive map $M\otimes_R X\to M\otimes_R Y$ that maps a split tensor $m\otimes x$ to $m\otimes f(x)$. When $R$ is commutative, these are even $R$-modules and morphisms of $R$-modules. If $f$ is surjective, then $M\otimes_R f$ is surjective, but if $f$ is injective, it is not always the case that $M\otimes_R f$ is injective. One says that $M$ is flat if $M\otimes_R f$ is injective for all injective linear maps $f$.

These notions are quite abstract, and the development of homological algebra made them prevalent in modern algebra.

Example. — Free modules are projective and flat.

Proposition. — An $R$-module $M$ is projective if and only if there exists an $R$-module $N$ such that $M\oplus N$ is free.
Indeed, taking a generating family of $M$, we construct a free module $L$ and a surjective linear map $u\colon L\to M$. Since $M$ is projective, the map $\Hom_R(M,u)$ is surjective and there exists $v\colon M\to L$ such that $u\circ v=\id_M$. Then $v$ is an isomorphism from $M$ to $u(M)$, and one can check that $L=u(M)\oplus \ker(v)$.

Corollary. — Projective modules are flat.

Theorem (Kaplansky). — If $R$ is a local ring, then a projective $R$-module is free.

The theorem has a reasonably easy proof for a finitely generated $R$-module $M$ over a commutative local ring. Let $J$ be the maximal ideal of $R$ and let $k=R/J$ be the residue field. Then $M/JM$ is a finite dimensional $k$-vector space; let us consider a family $(e_1,\dots,e_n)$ in $M$ whose images form a basis of $M/JM$. Now, one has $\langle e_1,\dots,e_n\rangle + J M = M$, hence Nakayama's lemma implies that $M=\langle e_1,\dots,e_n\rangle$. Let then $u\colon R^n\to M$ be the morphism given by $u(a_1,\dots,a_n)=\sum a_i e_i$; by what precedes, it is surjective, and we let $N$ be its kernel. Since $M$ is projective, the morphism $\Hom_R(M,u)$ is surjective, and there exists $v\colon M\to R^n$ such that $u\circ v=\id_M$. We then have an isomorphism $M\oplus N\simeq R^n$, where $N=\ker(v)$. Moding out by $J$, we get $M/JM \oplus N/JN \simeq k^n$. Necessarily, $N/JN=0$, hence $N=JN$; since $N$ is a direct summand of $R^n$, it is finitely generated, and Nakayama's lemma implies that $N=0$.

Example. — Let $R$ be a commutative ring and let $S$ be a multiplicative subset of $R$. Then the fraction ring $S^{-1}R$ is a flat $R$-module.
Let $u\colon X\to Y$ be an injective morphism of $R$-modules. First of all, one identifies the morphism $S^{-1}R\otimes_R u\colon S^{-1}R\otimes_R X\to S^{-1}R\otimes_R Y$ to the morphism $S^{-1}u\colon S^{-1}X\to S^{-1}Y$ induced by $u$ on fraction modules. Then, it is easy to see that $S^{-1}u$ is injective. Let indeed $x/s\in S^{-1}X$ be an element that maps to $0$; one then has $u(x)/s=0$, hence there exists $t\in S$ such that $tu(x)=0$. Consequently, $u(tx)=0$, hence $tx=0$ because $u$ is injective. This implies $x/s=0$.

Theorem. — Let $R$ be a commutative ring. If $M$ is a finitely presented $R$-module, then $M$ is locally free: there exists a finite family $(f_1,\dots,f_n)$ in $R$ such that $R=\langle f_1,\dots,f_n\rangle$ and such that for every $i$, $M_{f_i}$ is a free $R_{f_i}$-module.
The proof is a variant of the case of local rings. Starting from a point $p\in\Spec(R)$, we know that $M_p$ is a finitely presented flat $R_p$-module. As above, we get a surjective morphism $u\colon R^n\to M$ which induces an isomorphism $\kappa(p)^n\to \kappa(p)\otimes M$, and we let $N$ be its kernel. By flatness of $M$ (and an argument involving the snake lemma), the exact sequence $0\to N\to R_p\to M\to 0$ induces an exact sequence $0\to \kappa(p)\otimes N\to \kappa(p)^n\to \kappa(p)\otimes M\to 0$. And since the last sequence is an isomorphism, we have $\kappa(p)\otimes N$. Since $M$ is finitely presented, the module $N$ is finitely generated, and Nakayama's lemma implies that $N_p=0$; moreover, there exists $f\not\in p$ such that $N_f=0$, so that $u_f\colon R_f^n\to M_f$ is an isomorphism. One concludes by using the quasicompactness of $\Spec(R)$.

However, not all flat modules are projective. The most basic example is the following one.

Example. — The $\mathbf Z$-module $\mathbf Q$ is flat, but is not projective.
It is flat because it is the total fraction ring of $\mathbf Z$. To show that it is not projective, we consider the free module $L={\mathbf Z}^{(\mathbf N)}$ with basis $(e_n)$ and the morphism $u\colon L\to\mathbf Q$ that maps $e_n$ to $1/n$ (if $n>0$, say). This morphism is surjective. If $\mathbf Q$ were projective, there would exist a morphism $v\colon \mathbf Q\to L$ such that $u\circ v=\id_{\mathbf Q}$. Consider a fraction $a/b\in\mathbf Q$; one has $b\cdot 1/b=1$, hence $b v(1/b)=v(1)$. We thus see that all coeffiencients of $v(1)$ are divisible by $b$, for any integer $b$; they must be zero, hence $v(1)=0$ and $1=u(v(1))=0$, a contradiction.
The proof generalizes. For example, if $R$ is a domain and $S$ does not consist of units, and does not contain $0$, then $S^{-1}R$ is not projective. (With analogous notation, take a nonzero coefficient $a$ of $v(1)$ and set $b=as$, where $s\in S$ is not $0$; then $as$ divides $a$, hence $s$ divides $1$ and $s$ is a unit.)

These recollections are meant to motivate the forthcoming question: When is it the case that a localization $S^{-1}R$ is a projective $R$-module?

Example. — Let $e$ be an idempotent of $R$, so that the ring $R$ decomposes as a product ot two rings $R\simeq eR \times (1-e)R$, and both factors are projective submodules of $R$ since their direct sum is the free $R$-module $R$. Now, one can observe that $R_e= eR$. Consequently, $R_e$ is projective. Geometrically, $\Spec(R)$ decomposes as a disjoint union of two closed subsets $\mathrm V(e)$ and $\mathrm V(1-e)$; the first one can be viewed as the open subset $\Spec(R_{1-e})$ and the second one as the open subset $\Spec(R_e)$.

The question was to decide whether this geometric condition furnishes the basic conditions for a localization $S^{-1}R$ to be projective. With the above notation, we recall that $\Spec(S^{-1}R)$ is homeomorphic to a the subset of $\Spec(R)$ consisting of prime ideals $p$ such that $p\cap S=\emptyset$. The preceding example corresponds to the case where $\Spec(S^{-1}R)$ is open and closed in $\Spec(R)$. In this case, we view $S^{-1}R$ as a quasicoherent sheaf on $\Spec(R)$, it is free of rank one on the open subset $\Spec(S^{-1}R)$, and zero on the complementary open subset. It is therefore locally free, hence the $R$-module $S^{-1}R$ is projective.

Observation. — The set $\Spec(S^{-1}R)$ is stable under generization. If $S^{-1}R$ is a projective $R$-module, then it is open.
The first part is obvious: if $p$ and $q$ are prime ideals of $R$ such that $p\subseteq q$ and $q\cap S=\emptyset$, then $p\cap S=\emptyset$. The second part follows from the observation that the support of $S^{-1}R$ is exactly $\Spec(S^{-1}R)$, combined with the following proposition.

Proposition. — The support of a projective module is open.
I learnt this result in the paper by Vasconcelos (1969), “On Projective Modules of Finite Rank” (Proceedings of the American Mathematical Society 22 (2): 430‑33). The proof relies on the trace ideal $\tau_R(M)$ of a module: this is the image of the canonical morphism $t\colon M^\vee \otimes_R M\to R$. (It is called the trace ideal, because when $M$ is free, $M^\vee\otimes_R M$ can also be identified with the module of endomorphisms of finite rank of $M$, a split tensor $\phi\otimes m$ corresponds with the endomorhism $x\mapsto \phi(x)m$, and then $t(\phi \otimes m)=\phi(m)$ is its trace.) Now, if $p$ belongs to the support of $M$, then $\tau_R(M)_p=R_p$, while if $p$ does not belong to the support of $M$, one has $M_p=0$, hence $\tau_R(M)_p=0$. In other words, the support of $M$ is the complement of the closed locus $\mathrm V(\tau_R(M))$ of $\Spec(R)$.

On the other hand, one should remember the following basic property of the support of a module.

Proposition. — The support of a module is stable under specialization. The support of a finitely generated module is closed.
Indeed, for every $m\in M$ and $p\in \Spec(R)$, saying that $m=0$ in $M_p$ means that there exist $s\in R$ such that $s\notin p$ with $sm=0$. In other words, this set is $\mathrm V(\mathrm{ann}_R(m))$. This shows that the support of $M$ is the union of the closed subsets $\mathrm V(\mathrm{ann}_R(m))$; it is in particular stable under specialization. If $M$ is finitely generated, this also shows its support is $\mathrm V(\mathrm{ann}_R(M))$, hence is closed.

At this point, one can go either following Vasconcelos (1969) who shows that a projective module $M$ of the form $S^{-1}R$ is finitely generated if and only if its trace ideal is. In particular, if $R$ is noetherian and $S^{-1}R$ is a projective $R$-module, then $\Spec(S^{-1}R)$ is closed. It is thus open and closed, and we are in the situation of the basic example above.

One can also use a topological argument explained to me by Daniel Ferrand: a minimal prime ideal of $R$ that meets $\Spec(S^{-1}R)$ is disjoint from $S$, hence belongs to $\Spec(S^{-1}R)$. Consequently, $\Spec(S^{-1}R)$ is the union of the irreducible components of $\Spec(R)$ that it meets. If this set of irreducible components is finite (or locally finite), for example if $\Spec(R)$ is noetherian, for example if $R$ is a noetherian ring, then $\Spec(S^{-1}R)$ is closed.

I did not find the time to think more about this question, and it would be nice to have an example of a projective localization which does not come from this situation.