Associated prime ideals and regular elements in polynomial rings

This post is here to record a fact that ought to be better known and for which I ought to have known a proper proof beforehand.

Let's start by defining two thirds of the terms of the title.

By polynomial ring, I simply mean a ring of polynomials $A[T_1,T_2,\dots]$ over a commutative ring $A$, in any number of indeterminates. The most important case here will however be that of a ring of polynomials in one single indeterminate, everything would follow from that one by “standard” arguments. For simplicity, I'll write $A[T]$ for such  a ring of polynomials.

Regular elements in a ring $R$ are those which are not zero divisors. Explicitly, $a\in R$ is regular if $ab=0$ implies $ b=0$ for any $b\in R$. Or $a$ is a zero divisor if there exists a nonzero $b\in R$ such that $ab=0$.

The classic result linking these two notions that I have in mind is the following result.

Proposition. — Let $A$ be a commutative ring. If $P\in A[T]$ is not regular, then there exists a nonzero $a\in A$ such that $aP=0$.

There is a classic, rather elementary, proof of that result which I recall first. Let us first treat the case where is only one indeterminate. By assumption, there exists a nonzero polynomial $Q\in Q[T]$ such that $PQ=0$, and we may choose $Q$ to be of minimal degree. Write $P=\sum_{k=0}^m a_k T^k$ and $Q=\sum_{k=0}^n b_k T^k$, where $m$ and $n$ are the degrees of $P$ and $Q$, so that $a_m\neq 0$ and $b_n\neq 0$.

Lemma. — One has $a_k Q=0$ for all $k$.

The result being obvious for $k>m$, we may argue by decreasing induction on $k$, assuming that $a_\ell Q=0$ for all $\ell>k$. Then, in the product $PQ$, the term of largest degree is $a_k b_n T^{k+n}$, so that $a_k b_n=0$. In particular, the polynomial $a_kQ$ has degree $<n$. Since $P(a_k Q)=a_k PQ=0$, and by the choice of $Q$, we have $a_kQ=0$, as claimed.

Given the lemma, we see that $a_k b_\ell=0$ for all integers $k$ and $\ell$. Choosing any nonzero coefficient $b_\ell$ of $Q$, we obtain $b_\ell P=0$. This concludes the proof of the proposition when there is only one indeterminate.

To treat the general case, it is probably possible to make an induction on the number of indeterminates that appear in an equality $PQ=0$, where $Q$ is a nonzero element of $A[T]$. Another possibility, closer to the above proof, would be to choose a graded monomial ordering, for example the `deglex' order (we first order monomials by total degree, and then by lexicographic order with respect to some linear ordering of the indeterminates). I leave the details to later thought.

The last concept of the title is that of associated prime ideal. There are in fact two different notions for this, and in absence of noetherian hypotheses, it is better to consider the one coined as “weakly associated” by N. Bourbaki (Algèbre commutative) or D. Lazard (“Autour de la platitude”, Bulletin de la S. M. F., tome 97 (1969), p. 81-128). This is also the choice I made in my book, (Mostly) commutative algebra, Universitext, 2021. And so let me skip the adverb “weakly”, defining a prime ideal $P$ of $A$ to be an associated prime to an $A$-module $M$ if there exists $m\in M$ such that $P$ is minimal among the set of prime ideals containing the annihilator $\operatorname{ann}_A(m)$ of $m$. (When $A$ is noetherian, more generally when $P$ is finitely generated, one can prove that there exists $m$ such that $P$ equals the annihilator of $m$, and this is the — sometimes — more classic definition of an associated prime ideal.) The set of associated prime ideals of $M$ is denoted by $\operatorname{Ass}_A(M)$ (or even $\operatorname{Ass}(A)$ when $M=A$), where the abbreviation stands for assassin — In mathematics, forcing something to vanish is often, too often, related to the vocabulary of murder…

This looks like it is a strange notion, but it has some algebraic content, as summaried by the following proposition for which proof I refer to Theorem 6.5.8 of my above mentioned book.

Proposition — Let $A$ be a commutative ring and let $M$ be an $A$-module.

1) An element $a\in A$ belongs to some associated prime ideal of $M$ if and only if there exists a nonzero $m\in M$ such that $am=0$.

2) An element $a$ belongs to every associated prime ideal of $M$ if and only if the localized module $M_a$ vanishes.

Associated primes also have some geometric meaning when $M=A/I$, for some ideal $I$ of $A$. Indeed, the minimal prime ideals of $A$ containing $I$ are associated, and they represent the irreducible components of the locus $\mathrm V(I)$ in $\operatorname{Spec}(A)$. When the ideal $I$ is equal to its radical, this is all there is, but in general there are more associated prime ideals, and the corresponding closed subsets of $\mathrm V(I)$ are called “embedded” components.

It is now a natural question to elucidate associate prime ideal in the context of polynomial rings. The answer is given by a theorem of J. W. Brewer and W. J. Heinzer in their paper “Associated primes of principal ideals”, Duke Math. Journal, 1974.

Theorem (Brewer & Heinzer). — Let $A$ be a commutative ring. The associated prime ideals of the polynomial ring $A[T]$ are the ideals of the form $p[T]$, where $p$ is an associated prime ideal of $A$.

This theorem implies the above proposition. Indeed, if $P\in A[T]$ is a zero divisor, there exists an associated prime ideal $q\in A[T]$ such that $P\in q$, and $q$ is of the form $p[T]$ for some associated prime ideal $p$ of $A$. By definition, there exists an element $a\in A$ such that $p$ is minimal among the prime ideals annihilating $a$. Then $ap=0$, which implies that $aP=0$. Since $p$ is prime, one has $a\neq 0$, hence the proposition.

It doesn't look so clear to me how to immediately deduce the theorem from the proposition, although that might well be possible, and even easy. Anyway, let us describe the proof given by Brewer and Heinzer (this is theorem 7 in their paper), adjusting the notation and terminology. In fact, I cannot really prove the theorem without using the proposition, but I might have overlooked some easy argument.

We consider an associated prime ideal $q$ of $A[T]$ and let $p=A \cap q$ be its trace on $A$. Note that $p$ is a prime ideal of $A$.

Lemma. — One has $q=p[T]$.

The inclusion $p[T]\subseteq q$ is obvious; let us show the other one. Let $f\in q$ and let us prove $f\in p[T]$. Localizing at $p$, we recall that $A_p$ is a local ring with maximal ideal $pA_p$: in that case, elements of $pA_p$ are zero divisors (because $pA_p$ is associated), and elements outside of $pA_p$ are units. In particular, elements of $A_p[T]$ outside of $pA_p[T]$ have at least one coefficient which is a unit, hence can't be zero-divisors in $A_p[T]$ (by the proposition). This implies that $qA_p=pA_p[T]$ after localization, hence there exists $a\notin p$ such that $af\in p[T]$. Using that $p$ is prime, we get $f\in p[T]$, as was to be shown.

Let us now show that $q$ is an associated prime ideal of $A$. We still assume that $A$ is local with maximal ideal $p$. Let $f\in A[T]$ be such that $q$ is a minimal prime ideal among those containing $\operatorname{ann}(f)$.

Lemma. — One has $q=\sqrt{\operatorname{ann}(f)}$.

The radical of an ideal is the intersection of the minimal prime ideals that contain it, and $q$ is among this set of prime ideals, so that it suffices to prove that $q$ is the only minimal prime ideal of $A[T]$ that contains $\operatorname{ann}(f)$. Let then $q'$ be such a prime ideal. By definition, it is an associated prime ideal of $A[T]$ so that, by what has already been proved of the theorem, there exists a prime ideal $p'$ of $A$ such that $q'=p'[T]$. Since $A$ is local with maximal ideal $p$, one has $p'\subseteq p$, hence $q'=p'[T]\subseteq p[T]=q$. Since $q$ was assumed to be a minimal prime ideal containing $\operatorname{ann}(f)$, we get $q'=q$, as claimed.

From this lemma, one deduces that $p=\sqrt{\operatorname{ann}(f)\cap A}$.

Let then $a$ be a nonzero coefficient of $f$. By definition, one has $\operatorname{ann}_{A[T]}(f)\cap A \subseteq \operatorname{ann}_A(a)$, so that $p = \sqrt{\operatorname{ann}_{A[T]}(f)\cap A} \subseteq \sqrt{\operatorname{ann}_A(a)}$. Since Since $A$ is local with maximal ideal $p$, this implies that $p=\sqrt{\operatorname{ann}_A(a)}$. Consequently, $p$ is the only prime ideal containing $\operatorname{ann}_A(a)$, which implies that is is an associated prime ideal of $A$, concluding the proof of the theorem of Brewer and Heinzer.

On the other hand, there is a slightly different approach to the theorem of Brewer and Heinzer, and I want to sketch it now. This will build on proposition II/3.2 of Lazard's paper mentioned above.

Proposition (Lazard). — Let $B$ be a flat $A$-algebra and let $M$ be a $B$-module. Then $\operatorname{Ass}_A(M)$ equals the image of $\operatorname{Ass}_B(M)$ under the natural continuous map from $\operatorname{Spec}(B)$ to $\operatorname{Spec}(A)$.

Now specialize that proposition to the case where $B=A[T]$ and $M=B$. This says that the image of $\operatorname{Ass}_B(B)$ (the associated prime ideals of $B$) in $\operatorname{Spec}(A)$ is exactly $\operatorname{Ass}_A(A[T])$. On the other hand, the freeness of $A[T]$ as an $A$-module implies that $\operatorname{Ass}_A(A[T])=\operatorname{Ass}_A(A)$. Indeed, one reduces to the case of finite free modules, and then apply the classic inclusion $$ \operatorname{Ass}_A(N) \subseteq \operatorname{Ass}_A(M) \subseteq \operatorname{Ass}_A(N) \cup \operatorname{Ass}_A(M/N) $$ for every submodule $N$ of $M$, which will show by induction that $\operatorname{Ass}_A(M)=\operatorname{Ass}_A(A)$ when $M$ is free. On the other hand, we have seen in the first (and easiest) part of the proof of the theorem of Brewer and Heinzer that every associated prime ideal of $A[T]$ is of the form $q=p[T]$ for some prime ideal $p$ of $A$. Since $p=q\cap A$, this shows that $p$ is an associated prime of $A$ and gives the alternative proof.

Lazard's proof is very similar to (but predates by 5 years) the arguments given by Brewer and Heinzer. I don't really know why they didn't use it in their paper.

The Krull dimension of the semiring of natural numbers is equal to 2

Let $R$ be a ring. Its Krull dimension is the supremum of the lengths $n$ of chains $P_0\subsetneq P_1 \subsetneq\dots\subsetneq P_n$ of prime ideals of $R$. When $R$ is a field, the null ideal is the only prime ideal, and it is a maximal ideal so that its Krull dimension is zero. When $R$ is a principal ideal domain which is not a field, there are two kinds of prime ideals: the null ideal is prime (as in any domain), and the other prime ideals are the maximal ideals of $R$, generated by a prime element $p$. In particular, the Krull dimension of the ring of integers is equal to $1$.

It is classic that these concepts can be defined for semirings as well.

A semiring $R$ is a set endowed with a commutative and associative addition with a neutral element $0$, an associative multiplication with a neutral element $1$, such that addition distributes over multiplication: $(a+b)c=ac+bc$ and $c(a+b)=ca+cb$. When its multiplication is commutative, the semiring is said to be commutative.

Semirings $R$ have ideals: these are nonempty subsets $I$ which are stable under addition ($a+b\in I$ for $a,b\in I$), and stable under multiplication by any element of $R$: for general semirings, one has to distinguish between left, right, and two-sided ideals; for commutative semirings, the notions coincide.

An ideal $P$ of a semiring $R$ is said to be prime if $R\setminus P$ is a multiplicative subset; explicitely, $P\neq R$, and if $ab\in P$, then $a\in P$ or $b\in P$.

An ideal $P$ of a semiring $R$ is said to be maximal if $P\neq R$ and if there is no ideal $I$ such that $P\subsetneq I\subsetneq R$. A semiring $R$ is said to be local if it admits exactly one maximal ideal; this means that the set of non-invertible elements of $R$ is an ideal.

The amusing part comes from the classification of prime and maximal ideals of the semiring $\mathbf N$ of natural numbers, which I learned of via a Lean formalization project led by Junyan Xu.

Theorem.

1. The semiring $\mathbf N$ is local; its maximal ideal is the set $\mathbf N\setminus\{1\}$.
2. The null ideal is a prime ideal.
3. The other prime ideals are the sets $p\mathbf N$, for all prime numbers $p$.

In particular, we have chains $\langle 0\rangle \subsetneq \langle p\rangle \subsetneq \mathbf N\setminus\{1\}$ of prime ideals, hence the result stated in the title of this post:

Corollary. — The Krull dimension of the semiring of natural numbers is equal to $2$.

Proof.

1. The element $1$ is the only unit of $\mathbf N$, and $\mathbf N\setminus\{1\}$ is obviously an ideal, necessarily its unique maximal ideal.
2. The null ideal is a prime ideal, as in any semiring which is a domain.
3. Let now $P$ be a nonzero prime ideal of $\mathbf N$ and let $p$ be the smallest nonzero element of $P$. Then $p\neq 1$ (otherwise, $P=\mathbf N$, which is not prime). The hypothesis that $P$ is prime implies that one of the prime factors of $p$ belongs to $p$; by the choice of $p$, this must be $p$ itself, so that $p$ is a prime number. Then $P$ contains the set $p\mathbf N $ of multiples of $p$, which is a prime ideal of $\mathbf N$. Let us assume that $p\mathbf N\subsetneq P$ and let $n\in P\setminus p\mathbf N$. By assumption, $p$ does not divide $n$, so that these two integers are coprime. By the following proposition, $P$ contains every integer at least equal to $(p-1)(n-1)$; in particular, it contains both a power of $2$ and a power of $3$; since $P$ is prime, it contains $2$ and $3$, and it contains any integer at least $(2-1)(3-1)=2$, hence $P=\mathbf N\setminus\{1\}$. This concludes the proof.

Proposition. — Let $a$ and $b$ be nonzero coprime natural numbers. For any integer $n\geq (a-1)(b-1)$, there are natural numbers $u$ and $v$ such that $n=au+bv$.

Proof. — Since $a$ and $b$ are coprime, there are integers $u$ and $v$ such that $n=au+bv$. Replacing $u$ by $u+b$ and $v$ by $v-a$, we may assume that $0\leq u$. Replacing $u$ by $u-b$ and $v$ by $v+a$, we may assume that $u < b$. Then \[ bv = n - au \geq (a-1)(b-1)-a(b-1)= -(b-1). \] This implies $v\geq0$, so that $u$ and $v$ are natural numbers.

On the other hand, not all natural numbers $< (a-1)(b-1)$ can be written as such as sum. For example, $ab-a-b$ can't. Indeed, if $ab-a-b=au+bv$, hence $ab=a(u+1)+b(v+1)$, then $b$ divides $a(u+1)$, hence $b$ divides $u+1$, and similarly $a$ divides $v+1$. Then $ab$ is the sum of two nonzero multiples of $ab$, a contradiction. The precise distribution of the natural numbers $< (a-1)(b-1)$ which can be written as $au+bv$, for some natural numbers $u$ and $v$ is complicated, but at least one result is known : such an integer $n$ can be written in this form if and only if $ab-a-b-n$ cannot! One direction is clear, if $n$ and $ab-a-b-n$ can both be written in this form, then so can their sum, which is $ab-a-b$, a contradiction. On the other hand, let $n$ be an integer that cannot be written in this form, and write it as $n=au+bv$, for some integers $u$ and $v$, with $0\leq u< b$. By assumption, $v<0$, hence $v\leq -1$. Then \[ ab-a-b-n=ab-a-b-au-bv=a(b-1-u)+b(-v-1).\] We see that $b-1-u\geq 0$ and $-v-1\geq 0$, which shows that $ab-a-b-n$ can be written in the desired form.

Another open question of the style of the proposition had been raised by Frobenius: consider mutually coprime integers $a_1,\dots,a_r$; then any large enough integer $n$ can be written as $n=a_1u_1+\dots+a_ru_r$, for some natural numbers $u_1,\dots,u_r$, but when $r\geq 3$, there is no known formula for largest natural number that cannot be written in this form. The case $r=2$ that we discussed here was due to Sylvester (1884).

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.

Associated prime ideals

$\gdef\ann{\mathop{\mathrm{ann}}} \gdef\Ass{\mathop{\mathrm{Ass}}}\gdef\Spec{\mathop{\mathrm{Spec}}}$

I would like to go back to a quite delicate question of commutative algebra, that of associated prime ideals of modules. In most textbooks (Bourbaki, Matsumura…), this concept is considered for modules over a noetherian ring, while it is also necessary to consider it in a greater generality for some applications in algebraic geometry. For my book, (Mostly) commutative algebra (Springer Nature, 2021), I preferred to introduce the general concept (§6.5), because I observed that the initial proofs are in fact easier. In yesterday's class (Cohomology of coherent sheaves, 2nd year of Master course at Université Paris Cité), some remarks of a student, Elias Caeiro, helped me simplify two steps of the treatment I proposed in my book.

Definition. — Let $A$ be a ring and let $M$ be an $A$-module. Say that a prime ideal $P$ of $A$ is associated to $M$ if there exists an element $m\in M$ such that $P$ is minimal among all prime ideals containing $\ann_A(m)$.
We write $\Ass_A(M)$ (sometimes spelt out as “assassin”) for the set of all associated prime ideals of $M$.
(Here, $\ann_A(m)$ is the annihilator of $m$, the ideal of all $a\in A$ such that $am=0$.)

There is a geometric way to intepret this definition: it means that in the spectrum $\Spec(A)$, the irreducible closed set $V(P)$ (of which $P$ is the generic point) is an irreducible component of $V(\ann_A(m))$. Thanks to this remark, associated prime ideals are compatible with localisation: \[ \Ass_{S^{-1}A}(S^{-1}A) = \Ass_A(M) \cap \Spec(S^{-1}A), \] where $\Spec(S^{-1}A)$ is identified as the subset of $\Spec(A)$ consisting of prime ideals $P$ which are disjoint from $S$. In particular, $P$ is associated to $M$ if and only if the maximal ideal $PA_P$ of the local ring $A_P$ is associated to the module $M_P$.

Here is what the associated prime ideals mean, from the point view of module theory.
Proposition. — Let $a\in A$.
a) The multiplication by $a$ is injective in $M$ if and only if $a$ does not belong to any associated prime ideal of $M$.
b) The localized module $M_a$ is zero if and only if $a$ belongs to all associated prime ideals of $M$.
c) In particular, $M=0$ if and only if $\Ass_A(M)=\emptyset$.

Proof. — a) If $a$ belongs to the associated prime ideal $P$, then $a$ belongs to the associated prime ideal $PA_P$ of $M_P$, which means that there exists $m\in M$ such that $PA_P$ is the only prime ideal containing $\ann_{A_P}(m)$. Consequently, $a$ is nilpotent modulo $\ann_{A_P}(m)$ and there exists $n\geq 0$ and $b\in A\setminus P$ such that $a^nb\in\ann_A(m)$. Take a minimal such $n$. Since $b\notin P$, one has $n\geq 1$; then $a^{n-1}b m\neq0$, while $a\cdot a^{n-1}bm=0$ and the homothety $(a)_M$ is not injective. Conversely, if $(a)_M$ is not injective, take $m\neq0$ in $M$ such that $am=0$; the annihilator $\ann_A(m)$ is not equal to $A$, hence $V(\ann_A(m))\neq \emptyset$; take an irreducible component of this closed subset — equivalently a minimal prime ideal $P$ among those containing $\ann_A(m)$; one has $a\in \ann_A(m)$, hence $a\in P$.
b) follows from c), with $a=1$.
c) The module $M$ is zero if and only if the multiplication by $0$ is injective on $M$. By a), this is equivalent to the fact that $\Ass_A(M)$ is empty.

Corollary. — A prime ideal $P$ is in the support of $M$ if and only if it contains some associated prime ideal.
The prime ideal $P$ belongs to the support of $M$ if and only if $M_P\neq0$, if and only if $\Ass_{A_P}(M_P)$ is not empty, if and only if there exists an associated prime ideal of $M$ which belongs to $\Spec(A_P)$, that is, is contained in $P$.

For noetherian rings, one has the following characterization of associated prime ideals, which is usually taken at their definition.

Theorem. — Let $A$ be a noetherian ring and $M$ be an $A$-module. A prime ideal $P$ of $A$ is associated to $M$ if and only if there exists $m\in M$ such that $P=\ann_A(m)$.
If $P=\ann_A(m)$, then $P$ is associated to $M$. Conversely, let $m\in M$ and let $P$ be a minimal prime ideal of $A$ among those containing $\ann_A(m)$. We first assume that $A$ is local with maximal ideal $P$; then $P$ is the only prime ideal of $A$ that contains $\ann_A(m)$, which implies that any element of $P$ is nilpotent modulo $\ann_A(m)$. Since $P$ is finitely generated (because $A$ is noetherian), there exists an integer $n$ such that $P^n\subseteq \ann_A(m)$. Take a minimal such $n$. Since $\ann_A(m)\subseteq P$, one has $n\geq 1$; then $P^{n-1}\not\subseteq\ann_A(m)$ so that there exists $b\in P^{n-1}$ such that $bm\neq0$. Then $ab\in P^n$ for every $a\in P$, so that $P\subseteq \ann_A(bm)$, and $\ann_A(bm)\subseteq P$ because $bm\neq0$. Consequently, $P=\ann_A(bm)$. In the general case, we use the case of a local ring to obtain $m\in M$ such that $\ann_{A_P}(m/1)=PA_P$. Consequently, $\ann_A(m)\subseteq P$, and for every $a\in P$, there exists $b\notin P$ such that $abm=0$. Using that $P$ is finitely generated, one finds $b\notin P$ such that $abm=0$ for every $a\in P$; then $\ann_A(bm)=P$, as was to be shown.

From that point on, both presentations converge. One deduces from the preceding theorem that if $A$ is noetherian and $M$ is finitely generated, there exists a composition series $0=M_0\subseteq M_1 \subseteq \dots \subseteq M_n=M$, with successive quotients $M_k/M_{k-1}$ of the form $A/P_k$, for some prime ideals $P_k$ of $A$, and then $\Ass_A(M)$ is contained in $\{P_1,\dots,P_n\}$, in view of the following lemma. In particular, $\Ass_A(M)$ is finite.

Lemma. — Let $M$ be an $A$-module and let $N$ be a submodule of $M$; then $ \Ass_A(N)\subseteq \Ass_A(M)\subseteq \Ass_A(N)\cup \Ass_A(M/N)$.
The first inclusion $\Ass_A(N)\subseteq \Ass_A(M)$ follows from the definition. Let us prove the second one. Let $P\in\Ass_A(M)$ and let $m\in M$ be such that $P$ is a minimal prime ideal of $A$ among those containing $\ann_A(m)$. Let $m'$ be the image of $M$ in $M/N$. If $P$ contains $\ann_A(m')$, then $P$ is also minimal among such prime ideals, hence $P\in\Ass_A(M/N)$. Otherwise, there exists $b\in \ann_A(m')$ such that $b\notin P$. Let us prove that $P$ is minimal among the prime ideals containing $\ann_A(bm)$. First of all, let $a\in\ann_A(bm)$; then $abm=0$, hence $ab\in P$, hence $a\in P$ since $b\notin P$. Since $\ann_A(m)\subseteq\ann_A(bm)$, it also follows that $P$ is minimal among the prime ideals containing $\ann_A(bm)$. Since $b\in\ann_A(m')$, one has $bm'=0$, hence $bm\in N$ and $P\in\Ass_A(N)$.