Extensions of primes, flatness, and intersection flatness
aa r X i v : . [ m a t h . A C ] A p r Extensions of primes, flatness, andintersection flatness
Melvin Hochster and Jack Jeffries
Abstract.
We study when R → S has the property that prime ideals of R extend to prime ideals or the unit ideal of S , and the situation where thisproperty continues to hold after adjoining the same indeterminates to bothrings. We prove that if R is reduced, every maximal ideal of R contains onlyfinitely many minimal primes of R , and prime ideals of R [ X , . . . , X n ] extendto prime ideals of S [ X , . . . , X n ] for all n , then S is flat over R . We givea counterexample to flatness over a reduced quasilocal ring R with infinitelymany minimal primes by constructing a non-flat R -module M such that M = P M for every minimal prime P of R . We study the notion of intersectionflatness and use it to prove that in certain graded cases it suffices to examinejust one closed fiber to prove the stable prime extension property. Dedicated to Roger and Sylvia Wiegand onthe occasion of their 150th birthday
1. Introduction
All rings in this paper are assumed commutative, associative, with multiplica-tive identity. Although we were originally motivated in studying the Noetheriancase, some of our results hold in much greater generality.We say that an R -algebra S or that the homomorphism R → S has the primeextension property if for every prime P of R , P S is prime in S or the unit idealof S . We say that the R -algebra S or the homomorphism R → S has the stableprime extension property if for every finite set of indeterminates X , . . . , X n overthese rings, R [ X , . . . , X n ] → S [ X , . . . , X n ] has the prime extension property.One of our results, Theorem 3.5, asserts the following. Theorem . Let R be a reduced ring such that every maximal ideal containsonly finitely many minimal primes; in particular, this holds if R is reduced andlocally Noetherian. If R → S has the stable prime extension property, then S is flatover R . Key words and phrases.
Base change, flatness, intersection flatness, prime extension property,prime ideal, ring extension, stable prime extension property.2020
Mathematics Subject Classification . Primary 13B02, 13B10, 13B40, 13C11.The first author was partially supported by National Science Foundation grants DMS–1401384 and DMS–1902116.The second author was partially supported by National Science Foundation grant DMS–1606353.
In the case R is a domain, this follows from a result of Picavet [ , Theorem 3.7],with a different proof. Note that the hypothesis that R be reduced is quite necessary,since R → R/ N R , where N R is the ideal of nilpotent elements of R , also has thestable prime extension property. It is easy to see that the hypothesis of having thestable prime extension property cannot be weakened to having the prime extensionproperty in Theorem 1.1; for example, since the surjection C J X, Y K / ( XY ) ։ C J X K sending Y
0, satisfies the prime extension property, but is not flat. Note thatin this example, the prime extension property is lost if we adjoin an indeterminate Z to both rings, since ( X, Y − Z ) yields a prime in C J X, Y K [ Z ] / ( XY ) but not in C J X K [ Z ]. The hypothesis on the minimal primes of R is also necessary: in Section 4,we construct an inclusion of quasilocal rings in which the source is reduced thatsatisfies the stable prime extension property, but is not flat. A more refined butmore technical version of Theorem 1.1 is given in Theorem 3.6. Our results hereare related to results of the first author on radical ideals in [ ].We note that there are significant examples where the stable prime extensionproperty holds: in particular, the following result from §
5, Theorem 5.12 generalizes[ , Corollary 2.9] (also cf. [ , Theorem 12.1(viii)]). This surprisingly enables oneto deduce that the stable prime extension property holds by examining one closedfiber. Theorem . If K is an algebraically closed field, S is a Z -graded K -algebra(but we are not assuming that S is Noetherian nor that S = K ) and F , . . . , F n are positive degree forms of S with coefficients in K that form a regular sequenceand generate a prime ideal Q of S , then K [ F , . . . , F n ] → S has the stable primeextension property. This and related results in § intersection flatness .Some basic properties of this notion are established in §
5, and applied to givesufficient criteria for the stable prime extension property. The results in § ]. Intersection flatnessis also closely related to the notion of content in the sense of Ohm and Rush [
10, 12 ](see also [
3, 4 ]), which we recall in §
2. Basic properties
We collect some basic properties of the prime extension property and the sta-ble prime extension property. These notions were studied by Picavet [ ] underthe names “prime producing” and “universally prime producing.” Many of theproperties established in this section also appear in [ , § Proposition . Let R → S be a ring homomorphism, and let R λ → S λ bea direct limit system of ring homomorphisms indexed by λ .(a) If R → S and S → T both have the prime extension property (respectively,stable prime extension property), then so does the composite map R → T .(b) R → S has the prime extension property if and only if for every surjection R ։ D , where D is a domain, D ⊗ R S is a domain or zero.(c) R → S has the stable prime extension property if and only if for everymap R → D , where D is a domain finitely generated over R , D ⊗ R S isa domain or zero.(d) If R λ → S λ has the prime extension property (respectively, stable primeextension property) for each λ , so does lim −→ λ ( R λ → S λ ) . XTENSIONS OF PRIMES AND FLATNESS 3 (e) R → S has the stable prime extension property if and only if for everymap R → D , where D is a domain, D ⊗ R S is a domain or zero.(f ) R → S has the stable prime extension property if and only if for everyhomomorphism R → R ′ , R ′ → R ′ ⊗ R S has the prime extension property,and this holds if and only for every homomorphism R → R ′ , R ′ → R ′ ⊗ R S has the stable prime extension property. Proof.
Part (a) is clear, since
T /P T ∼ = T / ( P S ) T , and (b) is a consequence ofthe fact that D ∼ = R/P . For (c), given a domain D finitely generated over R , takea surjection R [ X , . . . , X n ] ։ P . Since R [ X , . . . , X n ] → S [ X , . . . , X n ] has theprime extension property, S ⊗ R D ∼ = S [ X , . . . , X n ] ⊗ R [ X , ..., X n ] D is a domain orzero by (b). Conversely, if P ⊆ R [ X , . . . , X n ] is a prime for which P S [ X , . . . , X n ]is neither prime nor the unit ideal, then setting D = R [ X , . . . , X n ] /P yields D ⊗ R S ∼ = S [ X , . . . , X n ] /P S [ X , . . . , X n ] which is neither a domain nor zero. For(d), note that if P is a prime of lim −→ λ R λ and P λ is the contraction of P to R λ ,then P = lim −→ λ P λ , and the result for the prime extension property follows from thefact that a direct limit of rings each of which is a domain or zero is itself a domainor zero. The result for stable prime extension property is then a consequence ofthe fact that direct limit commutes with adjoining the variables. Part (e) is aconsequence of (c) and (d): given a map from R to a domain D , one may write D = lim −→ λ D λ for a directed system of domains that are finitely generated over R ,and S ⊗ R D ∼ = lim −→ λ ( S ⊗ R D λ ). Part (f) follows at once from the characterizationin (e) and the isomorphisms ( R ′ ⊗ R S ) ⊗ R ′ D ∼ = S ⊗ R D . (cid:3) The name “stable prime extension property” is partly motivated by state-ment (f) above. We will use the characterizations of the prime extension propertyin part (b) and the stable prime extension property in part (e) of the previousproposition repeatedly below.
Proposition . Let R → S be any ring homomorphism.(a) If R → S has the prime extension property or the stable prime extensionproperty and W is a multiplicative system in R (respectively, V is a mul-tiplicative system in S ) then W − R → W − S (respectively, R → V − S )has the same property.(b) If A is an ideal consisting of nilpotent elements of R , then R → S has theprime extension property (respectively, the stable prime extension prop-erty) if and only if R/ A → S/ A S has that property.(c) The map R/P R → S/P S has the prime extension property (respectively,the stable prime extension property) for every minimal prime P of R ifand only if R → S has that property.(d) If S is a polynomial ring in any family of variables over R , then R → S has the stable prime extension property.(e) If R is Noetherian, a formal power series ring in any family of variablesover R has the prime extension property. Proof.
For part (a), suppose that R → S has the prime extension propertyand W − R ։ D for some domain D . Since every prime of W − R is expandedfrom R , we can write D ∼ = W − D ′ for some domain D ′ with R ։ D ′ . Then, byProposition 2.1(b), S ⊗ R D ′ is either a domain or zero, so W − S ⊗ W − R D ∼ = W − S ⊗ W − R W − D ′ ∼ = W − ( S ⊗ R D ′ ) MELVIN HOCHSTER AND JACK JEFFRIES is either a domain or zero, so W − R → W − S has the prime extension property.The argument for stable prime extension property in the same, except replacingsurjections to domains with general maps to domains, and using Proposition 2.1(e).The case with V in place of W is similar.Part (b) follows from the fact all maps from R to a domain D must have A in their kernel, so S ⊗ R D ∼ = S/ A S ⊗ R/ A D ; thus the condition of Proposi-tion 2.1(b) holds or fails simultaneously for the two given maps, and likewise forProposition 2.1(e).For the forward implication of (c), a map R → D to a domain D must containa minimal prime P in the kernel, so S ⊗ R D ∼ = S/P S ⊗ R/P D for that prime P , andthe conclusion follows from Proposition 2.1(b) and (e). The reverse follows fromProposition 2.1(f).Parts (d) and (e) follow from the fact that S/P S may be identified with thecorresponding polynomial or power series ring over
R/P , since for any ideal (re-spectively, finitely generated ideal) I , the expansion IS is the same as the ideal ofpolynomials (respectively power series) all of whose coefficients are in I . Moreover,in the polynomial case, the hypothesis continues to hold after adjoining indetermi-nates to both rings. (cid:3)
3. Flatness
We say that T is a geometrically reduced and irreducible algebra over the field K if for every field extension K ⊆ L , L ⊗ K T is a domain. If D ⊆ L is a domain,then D ⊗ K T ⊆ L ⊗ K T . Thus K → T is geometrically reduced and irreducible ifand only if it has the stable prime extension property.Observe that a direct limit of geometrically reduced and irreducible algebras isagain geometrically reduced and irreducible, since tensor products commute withdirect limits. If T ⊆ T is a K -subalgebra, then L ⊗ K T ⊆ L ⊗ K T , so that T is geometrically reduced and irreducible if and only if all K -subalgebras aregeometrically reduced and irreducible, which happens if and only if all finitelygenerated K -subalgebras are geometrically reduced and irreducible.If P is a prime ideal of R , we let κ P = R P /P R P , which is canonically isomor-phic with the field of fractions of R/P . The fiber of R → S over a prime ideal P of R is κ P ⊗ R S . Proposition . If R → S has stable prime extension property, then for allprimes P of R , the fiber κ P ⊗ R S is a geometrically reduced and irreducible κ P -algebra. If, moreover, R → S is flat, the converse holds, i.e. if R → S is flat,then R → S has the stable prime extension property if and only if κ P ⊗ R S is ageometrically reduced and irreducible κ P -algebra for all primes P of R . Proof.
Assume that R → S has the stable prime extension property. If L isany extension field of κ P , we have a composite map R → κ P → L , and L ⊗ κ P ( κ P ⊗ R S ) ∼ = L ⊗ R S is a domain by Proposition 2.1(e).Now assume that R → S is flat and that all fibers are geometrically reducedand irreducible. Suppose that we have a homomorphism R → D , where D is adomain, and let P be the kernel. Let L be the field of fractions of D . Then D ⊗ R S is flat over D . Hence, the elements of D r { } are nonzerodivisiors, and to showthat D ⊗ R S is a domain it suffices to show that L ⊗ R S is a domain. Since we have XTENSIONS OF PRIMES AND FLATNESS 5 an injection κ P ֒ → L , we have the identification L ⊗ κ P ( κ P ⊗ R S ) ∼ = L ⊗ R S , andthis ring is a domain by the hypothesis that κ P ⊗ R S is geometrically reduced andirreducible. (cid:3) Our next goal is to show that under mild conditions on the reduced ring R , thecondition that R → S has the stable prime extension property forces the flatnessof S over R . See Theorem 3.5 below. We need some preliminary results.The result that intersecting two ideals commutes with extension from R to S is often stated for the case where S is flat over R . We prove that the result holdsunder a much weaker assumption: that for one of the ideals A , S/ A S is flat over R/ A . In fact, we assume even less: Lemma . If R → S is any ring homomorphism and A , B ⊆ R are idealssuch that Tor R/ A ( R/ ( A + B ) , S/ A S ) = 0 , which holds, for example, if S/ A S is flatover R/ A , then A S ∩ B S = ( A ∩ B ) S . Hence, if A , . . . , A h are ideals of R suchthat S/ A i S is flat over R/ A i for ≤ i ≤ h − , then T i ( A i S ) = ( T i A i ) S . Proof.
We use an overline to indicate images of elements in R/ A or S/ A S :context should make it clear which is meant. Let u ∈ A S ∩ B S . Let { g λ | λ ∈ Λ } be a generating set for B , and take a free resolution · · · → ( R/ A ) ⊕ Γ d −→ ( R/ A ) ⊕ Λ d −→ R/ A → R/ ( A + B ) as an ( R/ A )-module, where d maps the generator indexed by λ to g λ . We can write u = k X j =1 g λ j s λ j ∈ A S with s λ , . . . , s λ k ∈ S and λ , . . . , λ k ∈ Λ. Let s be the vector with coordinate s λ j in index λ j , and all other coordinates zero. The relation u = 0 above specifies anelement in ker( S/ A S ⊗ R/ A d ), namely, the vector s with coordinate s λ j in index λ j , and all other coordinates zero.The vanishing of Tor R/ A ( R/ ( A + B ) , S/ A S )implies that s ∈ im( S/ A S ⊗ R/ A d ), i.e., the vector s of coefficents is an ( S/ A S )-linear combination of finitely many, say h , vectors r , . . . , r h with coefficients in R/ A that each give relations on the g λ in R/ A . We lift each r i to a vector r i with coefficients in R by choosing arbitrary preimages for the nonzero coordinates,and zero in the zero coordinates. Note that each r i and hence each r i , has finitesupport, so there is a finite subset Λ = { λ , . . . , λ w } that contains the support ofall of the vectors s , r , . . . , r h . Then, we have(1) s ≡ P hi =1 u i r i modulo A S (as vectors, coordinatewise) for some elements u i ∈ S , and(2) for every i , P λ ∈ Λ r i λ g λ = P λ ∈ Λ r i λ g λ ∈ A , and hence A ∩ B , becauseof the presence of the g λ .From (1) we have(3) for λ ∈ Λ , s λ = P hi =1 u i r i λ + f λ for some f λ ∈ A S . MELVIN HOCHSTER AND JACK JEFFRIES
Then u = X λ ∈ Λ g λ s λ = X λ ∈ Λ g λ h X i =1 u i r i λ + f λ ! = h X i =1 u i X λ ∈ Λ r i λ g λ + X λ ∈ Λ f λ g λ and the result follows because the terms in the first sum on the right are in ( A ∩ B ) S by (2) and each f λ g λ ∈ ( AB ) S . The final statement follows by a straightforwardinduction on h . (cid:3) Proposition . Let R → S have the stable prime extension property.(a) If ( R, m ) → ( S, n ) is a local homomorphism of quasilocal rings, then thekernel is contained in the nilradical of R . Thus, if R is reduced, then themap is injective.(b) For every prime ideal P of R , either P S = S or P S contracts to P in R . Proof.
For part (a), suppose that a ∈ R is in the kernel and not nilpotent.By forming the quotient by a prime of R not containing a and the extension of thisprime to S , we obtain an example where R and S are domains and a is a nonzeroelement in the kernel, using Proposition 2.1(f). Adjoin two indeterminates X, Y toboth rings. Then X − aY is prime in R [ X, Y ]: this is true in F ( Y )[ X ], where F is the fraction field of R , and the contraction of ( X − aY ) F ( Y )[ X ] to ( R [ Y ])[ X ]is generated by X − aY , by the division algorithm for monic polynomials. Theexpansion of the prime ( X − aY ) R [ X, Y ] to S [ X, Y ] is X S [ X, Y ], a contradiction.For part (b), suppose Q ′ = P S = S contracts to Q in R . Then the map( R/P ) Q → ( S/P S ) Q ′ has the stable prime extension property by Proposition 2.2(a),and so must be injective by part (a). But ( Q/P ) R Q is in the kernel, so that Q = P . (cid:3) We need one more small observation for the proof of the main theorem.
Lemma . Let A be an N -graded ring with ( A , m ) quasilocal. Denote by A = m + A > the unique maximal homogeneous ideal. Then every element of A r A is a nonzerodivisor on A and on every Z -graded A -module. Proof. If f ∈ A r A , write f = f + f ′ , with f ∈ A r m and f ′ ∈ A > . Thereis some h ∈ A with hf = 1. If f u = 0 with u = 0, let v be the lowest degree termof u . Then 0 = hf u = v + higher degree terms, and so v = 0, a contradiction. (cid:3) Theorem . Let R be a reduced ring such that every maximal ideal con-tains only finitely many minimal primes. If R → S has the stable prime extensionproperty, then S is flat over R . Proof.
First, flatness is local on the maximal ideals of S and their contractionsto R . Hence, by Proposition 2.2(a), we may assume that ( R, m ) → ( S, n ) is aninjective local homomorphism of quasilocal rings that has the stable prime extensionproperty, and that ( R, m ) is reduced with finitely many minimal primes P , . . . , P h .We proceed by induction on h .We need only show that Tor R ( R/I, S ) = 0 for all ideals I of R , which isequivalent to the injectivity of I ⊗ S → S . Consider α : R [ It ] ⊗ R S ։ S [ It ]. It willsuffice to prove the injectivity of the map α : the fact that α is injective in degreeone is exactly what we need. If h = 1, both rings are domains containing S and,therefore, R , and so the (non)injectivity of the map α is unaffected by localizing at XTENSIONS OF PRIMES AND FLATNESS 7 R r { } . But then the isomorphism is clear, since I becomes either the zero idealor the unit ideal, so α identifies with R ⊗ R S ∼ = −→ S or R [ t ] ⊗ R S ∼ = −→ S [ t ].Now assume that k ≥
2. We shall prove W = R r S j P j consists of nonze-rodivisors on R [ It ] ⊗ R S . Assuming this, by Proposition 2.2(a) we may localizeboth rings at W preserving the stable prime extension property without affecting(non)injecivity of α and so reduce to the case where R becomes a finite product offields, and S becomes a product of algebras over these fields. The injectivity of α is local on the prime ideals of R . But after localization, I becomes either the zeroideal or the unit ideal, and we have injectivity in either case, as above.It remains to show that if r ∈ W , then r is not a zerodivisor on T = R [ It ] ⊗ R S .We consider the N -grading on T induced by the usual grading on R [ It ] (by giving R degree zero and t degree one) and giving S degree zero. Under this grading, T ∼ = S is quasilocal and thus T has a unique maximal homogeneous ideal N . If r is a zerodivisor on T , then we have a nonzero form φ in the annihilator of r in T .Let Q i := P i R [ t ] ∩ R [ It ]. Clearly, the intersection of the Q i is also zero, sinceit is contained in T i ( P i R [ t ]) = ( T i P i ) R [ t ] = 0. Now, R [ It ] → S ⊗ R R [ It ] = T hasthe stable prime extension property by Proposition 2.1(f), and since r is not inany Q i T and these ideals are prime, it follows that φ ∈ T i Q i T , and it sufficesto show that this intersection is zero. To prove this, we may localize at the re-spective homogeneous maximal ideals M , M ′ of R [ It ] and T : by Lemma 3.4, themultiplicative systems that become inverted do not contain any zerodivisors, so T i Q i T injects into its localization at M ′ . But we may then apply Lemma 3.2 tothe homomorphism R [ It ] M → T M ′ , which has the stable prime extension propertyby Proposition 2.2(a), and to the ideals Q i R [ It ] M . We have \ i Q i T ⊆ \ i Q i T M ′ = (cid:16) \ i Q i M (cid:17) T M ′ = (cid:16) \ i Q i (cid:17) M T M ′ = 0 , as required. (cid:3) Theorem . Let h : R → S be a ring homomorphism. If h has the stableprime extension property, then all fibers are geometrically reduced and irreducible.Under the hypothesis that all fibers are geometrically reduced and irreducible, thefollowing are equivalent:(i) R → S has the stable prime extension property.(ii) R/ N R → S/ N R S has the stable prime extension property, where N R isthe nilradical of R .(iii) For every minimal prime P of R , R/P → S/P S is flat.(iv) For every finite intersection J of primes of R , R/J → S/JS is flat.
Proof.
The first statement was already shown in Proposition 3.1. The equiv-alence of (i) and (ii) is Proposition 2.2(b). If R has the stable prime extensionproperty, then for any ideal J ⊆ R that is a finite intersection of primes, R/J hasfinitely many minimal primes, and the map
R/J → S/JS has the stable primeextension property by Proposition 2.1(f), and hence is flat by Theorem 3.5. Thus,(i) implies (iv). The implication (iv) implies (iii) is trivial. If (iii) holds, then
R/P → S/P S has the stable prime extension property for every minimal prime P ,and, hence, (i) follows by Proposition 2.2(c). (cid:3) MELVIN HOCHSTER AND JACK JEFFRIES
4. A counterexample when a quasilocal ring has infinitely manyminimal primes
We construct
R ֒ → S local with ( R, m , K ), R, S quasilocal, R reduced, and S not flat over R such that R ֒ → S has the stable prime extension property. To dothis, we will construct a non-flat module satisfying P M = M for every minimalprime P of R , and take S to be the Nagata idealizer R ⋉ R M of M , which is definedto be R ⊕ M with multiplication ( r ⊕ m )( r ′ ⊕ m ′ ) = rr ′ ⊕ ( rm ′ + r ′ m ), so that M = 0.We first construct an example where R is N -graded over a field K and S is Z -graded. We may then localize. Let (Σ , (cid:22) ) denote a partially ordered set with thefollowing properties:(1) Σ is nonempty.(2) For all σ ∈ Σ, the set { τ ∈ Σ : σ (cid:22) τ } is finite and totally ordered.This implies that for every element τ , there is a unique minimal element τ of Σ with τ (cid:22) τ , and that every element τ ∈ Σ that is not minimalhas a unique immediate predecessor, which we denote τ − . We also write σ ≺ im τ to mean that σ = τ − .(3) For every σ ∈ Σ, there exist incomparable elements τ, τ ′ such that σ = τ − = τ ′− .The height h ( σ ) of σ is the length of a maximal chain of elements descendingfrom σ and is one less than the cardinality of { τ ∈ Σ : σ (cid:22) τ } , since this set issuch a chain. We let Σ + denote the set of nonminimal elements of Σ. Let (Σ , (cid:22) )be a partially ordered set satisfying (1), (2), and (3) above. Let K be a field. Let { X σ : σ ∈ Σ + } be indeterminates over K . Let R = K [ X σ : σ ∈ Σ + ] / I , where I is the ideal generated by the products X σ X τ where σ and τ are incomparable.We let x σ denote the image of X σ in R . Note that the indices of variables not ina given prime ideal Q must be linearly ordered (if two were incomparable, theirproduct is 0 ∈ Q , and so at least one of them is in Q ). It follows at once that theminimal primes of R correspond bijectively to the maximal chains Γ in Σ, where Γcorresponds to P Γ = ( { x γ : γ / ∈ Γ } ) ⊂ R .There is a K -basis for R consisting of products of powers of variables whoseindices form a chain in Σ. Example . If S is a set with two or more elements, the set Σ of finitesequences (including the empty sequence) of elements of S with the relation that σ (cid:22) τ when σ is an initial segment of τ is an example of such a partially ordered set.In this example, the empty sequence is the unique minimal element. If the emptysequence is omitted, the one element sequences are minimal. In this example, σ − is the initial segment of σ that omits the last term of σ , and the height of σ isits length as a sequence. The minimal primes of I are in bijection with N -indexedsequences of elements of S .In fact, if S has two elements, the poset Σ of this example is a subposet (up torelabeling) of any poset satisfying the three conditions above.Let { U σ : σ ∈ Σ } be a free basis for a free R -module M , and let M denote thequotient of M by the submodule spanned by the set of elements { U σ − x τ U τ : τ ∈ Σ and σ = τ − } . XTENSIONS OF PRIMES AND FLATNESS 9
Let u σ denote the image of U σ in M .We shall show that R and M , suitably localized, have the required properties.We first explore what happens in the graded case. We introduce a “multigrading”as follows: the index set will be Z ⊕ Σ , the free abelian group on the elements of Σ.The degree of x σ is σ . The degree of u σ is − P τ (cid:22) σ τ . Since the defining relations X σ X τ for σ, τ incomparable and U τ − − x τ U τ for all τ are multihomogeneous, weobtain compatible gradings on R and M . These gradings yield an N -grading on R ,with R = K , and a Z -grading on S by summimg the components of the respectivemultidegrees.We first give a concrete description of M . Proposition . With the notations introduced above, we have the following:(a) x σ u τ = 0 in M unless σ (cid:22) τ , and x σ u σ = 0 .(b) If τ k − ≺ im τ k ≺ im τ k +1 ≺ im · · · ≺ im τ n − ≺ im τ n is a strict saturated chain of elements of Σ and a i is a positive integer forall i with k ≤ i ≤ n , then x a k τ k x a k +1 τ k +1 · · · x a n τ n u τ n = 0 if any a i ≥ , and is equal to u τ k − if a k = a k +1 = · · · = a n = 1 .(c) The annihilator in R of u τ ∈ M is the monomial ideal generated by all x σ such σ (cid:22) τ , and all products x τ k x τ k +1 · · · x τ n − x τ n with τ k ≺ im τ k +1 ≺ im · · · ≺ im τ n − ≺ im τ n in Σ + . (d) A K -basis for M is given by products µu τ , where µ is a monomial in thevariables x σ such that σ (cid:22) τ . Moreover, the basis elements have mutuallydistinct multidegrees.(e) For all minimal primes P of R , M = P M . Moreover, for any σ ∈ Σ + , (Ann R ( x σ )) M = M .(f ) M is not R -flat. Proof.
For part (a), note that if σ τ , then either σ, τ are incomparableor τ (cid:22) σ . If σ, τ are incomparable, and γ is an immediate successor of τ , then σ, γ are incomparable. Then x σ u τ = x σ ( x γ u γ ) = 0. Similarly, if τ (cid:22) σ , thereis an immediate successor of τ that is not comparable with σ , and the relationfollows in the same way. This justifies the first statement. From this, we have x σ u σ = x σ u σ − = 0.For part (b), we use induction on n − k . If k = n , the result follows at oncefrom the final statement of part (a), and this is also true if a n ≥
2. If a n = 1, wemay replace final part of the product, consisting of x τ n u τ n , by τ n − , and then theresult follows from the induction hypothesis.We prove (c) and (d) simultaneously. Note that the monomials specified in (c)kill u τ by part (b). It is easy to see from part (a) that M is spanned as a K -vectorspace by the terms µu τ such that all variables occurring in µ having subscriptsstrictly less than τ ; we will call the set of such expressions B . For a fixed ω ∈ Σ,we will write B ( ω ) for the elements µu τ ∈ B with τ (cid:22) ω . It is easy to recoveran element of B in M from its multidegree: τ will be the unique largest elementwith a negative coefficient in the multidegree, and the exponent on each variable x σ occurring in µ will be one greater than the coefficient of σ in the multidegree.Thus, each multigraded component of M is at most one-dimensional as a K -vectorspace. However, it remains to see that the elements in B are all nonzero in M .Let V be the “formal” K -vector space spanned by the terms µu τ in B . Weshall give an R -module structure to V by specifying a K -linear endomorphism θ τ of V for every τ ∈ Σ + such that the following two conditions hold:( † ) For all σ, τ ∈ Σ + , θ σ ◦ θ τ = θ τ ◦ θ σ ,( ‡ ) For all σ, τ ∈ Σ + , if σ and τ are incomparable, θ σ ◦ θ τ = θ τ ◦ θ σ = 0.These conditions give V the structure of an R -module. We then verify V ∼ = M as R -modules in such a way that the formal basis element µu τ corresponds to theterm µu τ in V .Let h ( τ ) = n , and let τ ≺ im τ ≺ im · · · ≺ im τ n − ≺ im τ n = τ be the chain ofelements that are (cid:22) τ . For each monomial µ in the x σ for σ (cid:22) τ , let k be one plusthe largest index j such that x τ j does not occur with positive degree in µ , whichwe take to be 0 if all the variables occur; note that k depends on µ , but we omit itfrom the notation for ease of reading. Then µ can be written uniquely in the form ν µ γ µ where γ µ is a monomial in the variables indexed by τ k , τ k +1 , . . . , τ n − and each such variable divides γ µ , and ν µ involves only variables with subscripts τ , . . . , τ k − . Note that if x τ n − does not occur with a positive exponent, then ν µ = µ and γ ν = 1, while if for 1 ≤ j ≤ n each x τ j occurs with positive exponent then ν µ = 1and γ µ = µ . For σ ∈ Σ + , let θ σ be the K -linear endomorphism of V specified onthe basis B by θ σ ( µu τ ) = x σ µu τ if σ (cid:22) τ,ν µ u τ k − if σ = τ and all exponents in γ µ are 1 , σ = τ and some exponent in γ µ is ≥ , σ τ. We need to check that ( † ) and ( ‡ ) hold. Fix τ = τ n of height n . Note that θ σ stabilizes the K -span of B ( τ ), which we denote W ( τ ), and that all x σ kill theseelements unless σ (cid:22) τ . It therefore suffices to consider only the variables x τ i for1 ≤ i ≤ n , and the u τ i for 0 ≤ i ≤ n . To simplify notation, we shall write i insteadof τ i , so that we have 0 ≺ im ≺ im · · · ≺ im n . To verify ( † ) and ( ‡ ) consider the R -module R/ A where A is the monomial ideal generated by all variables whosesubscripts are not (cid:22) n and all monomials of the form x k x k +1 · · · x n . The quotienthas a K -basis consisting of all monomials of the form νγ k for 0 ≤ k ≤ n where γ k = Q nj = k +1 x j and ν is a monomial in x , . . . , x k − .There is a vector space isomorphism between R/ A and W ( τ ) that maps νγ k to ντ k . It is straightforward to verify that for 1 ≤ j ≤ n , x j νγ k corresponds to θ j ( ντ k ). It follows that ( † ) and ( ‡ ) hold for the θ j acting on W ( τ ), and thus ( † )and ( ‡ ) hold on V = lim −→ τ W ( τ ). This gives V the structure of an R -module.The R -linear map from the free module M to V such that U τ · u τ killsthe relations defining M , and so induces an R -linear surjection M ։ V . Since the XTENSIONS OF PRIMES AND FLATNESS 11 elements in M corresponding to B span M , these elements are also a K -basis for M : if they were linearly dependent, their images in V would be as well. It followsthat the map M ։ V is an R -isomorphism. This proves part (d), while part (c)now follows from the fact that Ru τ ∼ = W ( τ ) ∼ = R/ A .Part (e) is clear, since every σ ∈ Σ has at least two incomparable immediatesuccessors τ , ω , and given any minimal prime P = P Γ , at least one of x τ or x ω isin P Γ , say x τ , and then u σ = x τ u τ ∈ P M .It remains only to prove (f). Fix σ ∈ Σ + , and consider the exact sequence0 −−−−→ Ann R ( x σ ) −−−−→ R x σ −−−−→ R. If M were R -flat then tensoring with M would yield thatAnn M ( x σ ) = (Ann R ( x σ )) M. We have Ann R ( x σ ) = (cid:0) { x τ : τ ∈ Σ + and σ, τ are incomparable } (cid:1) . Suppose σ ≺ im σ . Then x σ u σ is a nonzero element of M in Ann M ( x σ ). Hence,if M were R -flat we would have that x σ u σ = P j x τ j µ j u ω j with τ j incomparableto σ and µ j a monomial for each j . Using the Z ⊕ Σ grading, and the fact that eachpiece is a one-dimensional K -vector space, there must be an equality of the form x σ u σ = x τ µu ω , with σ, τ incomparable, and τ (cid:22) ω , for otherwise the right-handside would be zero. Using the module structure as computed above, we can write x τ µu ω = µ u ω with ω (cid:22) ω . Thus we have x σ u σ = µ u ω in M . Since theelements of B form a basis, we must have σ = ω . But then, we have σ (cid:22) σ (cid:22) ω ,and τ (cid:22) ω , so σ and τ are comparable, a contradiction. (cid:3) Theorem . Let R and M be as above. Let m be the homogeneous maximalideal of R . Then R m → ( R ⋉ M ) m ( R ⋉ M ) is a local homomorphism of quasilocalrings, with source R m reduced, that satisfies the stable prime extension property,but is not flat. Proof.
Since R is reduced, R m is as well. Observe that R → R ⋉ M has thestable prime extension property by Proposition 2.2(c), since for any minimal prime P , we have ( R ⋉ M ) /P ( R ⋉ M ) ∼ = R/P . Then, by Proposition 2.2(a), the map R m → ( R ⋉ M ) m ( R ⋉ M ) has the stable prime extension property as well. It remainsonly to show that M m is not flat over R m . We can argue in the same way: if M m were flat, we would have for every σ ∈ Σ + that Ann M m ( x σ ) / (Ann R ( x σ )) M m = 0,and since localization commutes both with taking the annihilator of an elementand with forming quotients, this would imply (cid:0) Ann M ( x σ ) / (Ann R ( x σ )) M (cid:1) m = 0.To see that this does not happen we may apply Lemma 3.4, using the N -grading on R and the Z -grading on M introduced before the statement of Proposition 4.2. (cid:3)
5. Intersection Flatness
Recall (cf. [ , p. 41]) where the terms “intersection-flat” and “ ∩ -flat are used)that an R -module S is intersection flat if S is R -flat and for any finitely generated R -module M ,( { M λ } λ ∈ Λ of M we have S ⊗ R ( T λ M λ )= T λ ( S ⊗ R M λ ) when both are identified with their images in S ⊗ R M . Note that, quite generally, there is an obvious injective map from the firstmodule to the second. Here, S will usually be an R -algebra in which case we alsosay the homomorphism R → S is intersection flat . A flat homomorphism satisfiesthe property ( M whenever Λ is finite.In particular, if S is intersection flat, then for every family of ideals { I λ : λ ∈ Λ } of R the equality ( T λ I λ ) S = T λ ( I λ S ) holds. We shall say that S (or R → S inthe algebra case) is weakly intersection flat for ideals if this condition holds, i.e., if( M = R , and intersection flat for ideals if additionally R → S isflat. We caution the reader that the definition of “intersection flat” given in [ ] isthe notion that we call “intersection flat for ideals” here. Note that when M = R ,we may identify I ⊗ R S with IS .The notion of weak intersection flatness for ideals has been previously studiedin the context of the theory of content. If S is a weakly intersection flat R -module,then, for any s ∈ S , there is a unique smallest ideal I of R such that s ∈ IS , calledthe content of s ; in fact, this characterizes the property of weak intersection flatnessfor ideals by [ , 1.2]. Note that if S is a polynomial ring over R , then this notionof content coincides with the classical notion of content as the ideal generated bythe coefficients of a polynomial. A module S that is weakly intersection flat forideals is called a content module in [
10, 12 ], and called an
Ohm-Rush module in[
3, 4 ].The relationships between the notions in the title of this paper have beenexplored in the works cited in the previous paragraph, and some statements relatedto parts of the lemmas below appear in these sources. We note the followingresult of Rush [ , Theorem 3.2], which should be compared with Theorem 3.5 andProposition 5.6 below. Theorem . Let ϕ : R → S be an injective ring homomorphism thatis weakly intersection flat for ideals. Suppose that R is reduced and that ϕ satisfiesthe prime extension property. Then ϕ is flat. For the most part, it is intersection flatness for ideals that we use for resultson the stable prime extension property (and, as mentioned above, is the definitionused in [ ]). As is shown in Proposition 5.7(b) below, if φ : R → S is module-free(i.e., S is a free R -module) then φ is intersection flat [ , p. 41]. Proposition 5.7below collects some facts about intersection flatness. Note that intersection flatnessfor the Frobenius endomorphism mapping a regular ring to itself is studied in [ ,5.3] and in [ , § Example . Let K be a field and let R = K [ x ] m where m = xK [ x ]. Let f bea formal power series in x that is not in the fraction field K ( x ) of K [ x ]. Then K J x K is intersection flat for ideals over R , since the intersection of any infinite family ofideals is (0) in both rings. However, K J x K is not intersection flat over R . To seethis, let f n denote unique the polynomial of degree at most n that agrees with f modulo x n +1 K J x K . Let M = R ⊕ , and let M n = R (1 , f n ) + R (0 , x n +1 ). When wetensor with b R = K J x K , then b R ⊗ R M n may be identified with the submodule of b R ⊕ spanned by (1 , f ) and (0 , x n +1 ). The intersection of these is b R (1 , f ). But theintersection of the M n in R is 0: if ( g, h ) ∈ M n , then h − f g ∈ m n , so if ( g, h )were a nonzero element of the intersection, we would have that f = g/h would berational over K [ x ]. XTENSIONS OF PRIMES AND FLATNESS 13
Example . K J x, y K is not even intersection flat for ideals over R = K [ x, y ] m ,where m is the maximal ideal ( x, y ) in the polynomial ring K [ x, y ]. Let f ∈ K J x K be transcendental over K [ x ] and let f n be as in Example 5.2. Then the ideals I n = ( y − f n , x n +1 ) R when extended to b R = K J x, y K agree with ( y − f, x n +1 ) b R ,and so their intersection is ( y − f ) b R . But their intersection in R is 0, for if0 = G ( x, y ) ∈ R were in the intersection we may clear the denominator and as-sume that G ∈ K [ x, y ], and then we would have G ( x, f ) = 0, contradicting thetranscendence of f . Discussion . Note that a family of R -submodules of M/N has the form M λ /N where the M λ are submodules of M containing N , and we have \ λ ( M λ /N ) = (cid:0) \ λ M λ (cid:1) /N and so \ λ (cid:0) S ⊗ R ( M λ /N ) (cid:1) = \ λ (cid:0) ( S ⊗ R M λ ) / ( S ⊗ R N ) (cid:1) = (cid:18)\ λ ( S ⊗ R M λ ) (cid:19) / ( S ⊗ R N ) . Under the assumption that R → S is intersection flat, this identifies with S ⊗ R (cid:18)\ λ M λ (cid:19)!, ( S ⊗ R N ) = S ⊗ R (cid:18)\ λ M λ (cid:19)(cid:30) N ! = S ⊗ R (cid:0)\ λ ( M λ /N ) (cid:1) , where the equalities are canonical identifications of submodules of S ⊗ R M or of S ⊗ R ( M/N ).The following result clarifies when properties related to intersection flatnessover R imply flatness. One consequence is that in the definition of intersection flat,it is not necessary to assume flatness separately: the property ( R , implies it. Proposition . Let R be a ring and S an R -module. The following areequivalent.(i) S is R -flat.(ii) For every R -module M , the property ( holds for every finite family ofsubmodules of M .(iii) The property ( holds for every finite family of submodules of M = R (and, hence, for every finite family of ideals in M = R ).(iv) For every element f ∈ R , Ann S ( f ) = (Ann R ( f )) S , and the property ( holds for every finite family of ideals of M = R . Proof.
The implication (i) ⇒ (ii) is well-known, and (ii) ⇒ (iii) is clear. Toshow that (iii) implies (iv), consider the submodules spanned by (1 ,
0) and (1 , f )in R . Their intersection is Ann R ( f ) ×
0. If we expand to S , the intersection isAnn S ( f ) ×
0, while the expansion of the intersection is (Ann R ( f )) S ×
0, from whichthe result follows.It remains to show (iv) ⇒ (i); i.e., that (iv) implies that S is flat. It sufficesto show that for every finitely generated ideal I = ( f , . . . , f n ) R of R , we havethat Tor R ( R/I, S ) = 0. This follows if every relation P ni =1 f i s i = 0 is an S -linearcombination of relations on f , . . . , f n over R . We prove this by induction on n . If n = 1, this follows from the fact that Ann S ( f ) = (Ann R ( f )) S . Now suppose we know the result for f , . . . , f n − , n ≥
2. The relation P ni =1 f i s i = 0 implies that( ⋆ ) − f n s n = n − X i =1 f i s i and this is element is in ( f , . . . , f n − ) S ∩ f n S = (cid:0) ( f , . . . , f n − ) R ∩ f n R (cid:1) S , and sothe equation in ( ⋆ ) arises as an S -linear combination, say with coefficients t j ∈ S ,1 ≤ j ≤ h , of equations ( ⋆ ) − f n r nj = n − X i =1 f i r ij where the r ij ∈ R . These may be rewritten as relations on f , . . . , f n with coeffi-cients r ij in R . After multiplying these relations by the t j , adding, and subtractingthe sum from the original relation, we get a new relation on f , . . . , f n with coeffi-cients in S , say P n − i =1 f i u i + f n u n = 0, in which both P n − i =1 f i u i = 0 and f n u n = 0.Using the induction hypothesis for the former and the case n = 1 for the latter, wesee that this relation is an S -linear combination of relations over R . (cid:3) Proposition . Let R be a ring, and S be an R -module. The following areequivalent.(i) S is an intersection flat R -module.(ii) For every finitely generated R -module M , the property ( ) holds.(iii) For every finitely generated free R -module M , the property ( ) holds.(iv) For every finitely generated R -module M and every family of submodules { M λ : λ ∈ Λ } such that T λ M λ = 0 , we have T λ ( S ⊗ R M λ ) = 0 . Proof.
The implications (i) ⇒ (ii) ⇒ (iii) are clear. The equivalence between(ii) and (iv) is immediate from Discussion 5.4 by replacing M by M/N and every M λ by M λ /N . Similarly, for (iii) ⇒ (ii), given a module M and a family of submodules,we map a free module G ։ M and work with the family consisting of the inverseimages of the M λ in G . We may then apply Discussion 5.4. It remains only toshow (iv) ⇒ (i); i.e., that (iv) implies that S is flat. But this is immediate fromthe implication (ii) ⇒ (i) in Proposition 5.5, (cid:3) Proposition . Let R be a ring.(a) If R → S is intersection flat (respectively, intersection flat for ideals) and T is an S -module that is intersection flat (respectively, intersection flatfor ideals) over S , then T is intersection flat (respectively, intersectionflat for ideals) over R .(b) Direct sums and direct summands of modules that are intersection flat(respectively, intersection flat for ideals) are intersection flat (respectively,intersection flat for ideals).(c) If R is Noetherian, arbitrary products of modules that are intersection flat(respectively, intersection flat for ideals) are intersection flat (respectively,intersection flat for ideals).(d) The formal power series ring in finitely many variables over a Noetherianring is intersection flat.(e) Let R be a complete local Noetherian ring and S an R -flat algebra. If S/IS is m -adically separated for every ideal I of R , then S is intersectionflat for ideals. If S ⊗ R M is m -adically separated for all finitely generated XTENSIONS OF PRIMES AND FLATNESS 15 R -modules M then R → S is intersection flat. In particular, if S isNoetherian and m S is contained in the Jacobson radical of S , then R → S is intersection flat.(f ) If S is intersection flat (respectively, intersection flat for ideals) over R and W is a multiplicative system in S such that no element of W is azerodivisor on S/IS for any ideal I of R , then W − S is intersection flat(respectively, intersection flat for ideals) over R . In particular, if we take S to be a polynomial ring in an arbitrary set of variables over R and W to consist of a set of polynomials, each of which has a set of coefficientsthat generates the unit ideal of R , then W − S is intersection flat over R . Proof.
The arguments dealing with the case with the property “intersectionflat for ideals” are identical with those for the module case, and are not givenseparately, except for a remark in the proof of part (e).Part (a) is immediate from the definition, and part (b) is a completely straight-forward consequence of the fact that for a direct sum S = L µ S µ , we have( \ M λ ) ⊗ R ( M µ S µ ) ∼ = M µ ( \ λ M λ ) ⊗ R S µ , which will be equal to \ λ (cid:0) M λ ⊗ R ( M µ S µ ) (cid:1) ∼ = \ λ (cid:0)M µ ( M λ ⊗ R S µ ) (cid:1) ∼ = M µ (cid:0)\ λ (cid:0) M λ ⊗ R S µ ) (cid:1) . To prove (c), note that over a Noetherian ring, an arbitrary product of flatmodules is flat, by Chase’s theorem [ , Theorem 2.1]. It suffices to prove theproperty ( R -module M (or when M = R in the ideal case). This follows from the observation that for a finitely presented R -module N = M λ , the map N ⊗ R ( Q µ S µ ) → Q µ ( N ⊗ R S µ ) is an isomorphism.(Using the finite presentation of N , we reduce to the case where N = L ki =1 Re i is free. If an element of Q µ ( N ⊗ R S µ ) has µ -coordinate P ki =1 s i,µ e i , we let itcorrespond to P ki =1 σ i e i in N ⊗ R Q µ S i , where σ i has µ -coordinate s i,µ .)Part (d) is immediate from part (c), since these power series rings, as R -modules, are products of copies of R .To prove part (e), we note that since flatness implies that extension commuteswith finite intersection, we may assume the family M λ is closed under finite inter-section. Let N denote the intersection of the family. Chevalley’s lemma impliesthat for every C ∈ N , there exists λ C ∈ Λ such that M λ C ⊆ N + m C M . It followsthat S ⊗ R M λ C ⊆ S ⊗ R N + S ⊗ R m C M ⊆ S ⊗ R N + m C ( S ⊗ R M ) . Since S ⊗ R ( M/N ) ∼ = ( S ⊗ R M ) / ( S ⊗ R N ) is m -adically separated by hypothesis,the intersection of the modules on the right-hand side as C varies is S ⊗ R N , sothe intersection of { S ⊗ R M λ } is S ⊗ R N . In the ideal case, we only need that forevery ideal I of R , S/IS is m -adically separated. For the last statement of ?? , wesimply recall that every finitely generated S -module is separated with respect tothe Jacobson radical of S .For part (f), note that no element of W is a zerodivisor on any module of theform S ⊗ R N , where N is an R -module: it suffices to consider finitely generated R -modules N , and these have finite filtrations in which the factors have the form R/I .Now consider an element of T λ ( W − S ⊗ R M λ ) in W − S ⊗ R M : after multi-plying by a unit from the image of W , we may assume this element is the im-age of an element u ∈ S ⊗ R M . Then u/ W − ( S ⊗ R M λ ), andsince W consists of nonzerodivisors on S ⊗ R ( M/M λ ), we have that u is in ev-ery S ⊗ R M λ . By the hypothesis on S , this implies that u ∈ (cid:0) S ⊗ R ( T λ M λ ) (cid:1) , andso u/ ∈ W − S ⊗ R ( T λ M λ ), as required. (cid:3) Note that part (e) of Proposition 5.7 considerably strengthens the result of [ ,Proposition 5.3]. Proposition . If R → S is intersection flat for ideals and P is the inter-section of a family of primes { Q λ : λ ∈ Λ } in R such that S/ ( Q λ S ) is zero or adomain for each λ , then S/P S a domain.
Proof. If f g ∈ P S , then for each λ , we have either f ∈ Q λ S or g ∈ Q λ S .Set Λ = Λ f ∪ Λ g , where f ∈ Q λ S for λ ∈ Λ f and g ∈ Q λ S for λ ∈ Λ g . Then, if A = T λ ∈ Λ f Q λ and B = T λ ∈ Λ g Q λ we have radical ideals A, B for which A ∩ B = P .We have f ∈ AS and g ∈ BS . But one of A, B must be P , so f ∈ P S or g ∈ P S . (cid:3) We recall that a
Hilbert ring is a ring in which every prime ideal is an intersec-tion of maximal ideals.
Proposition . If R is a Hilbert ring, R → S is intersection flat for ideals,and for every maximal ideal m of R , S/ m S is zero or a domain, then R has theprime extension property. Proof.
This is immediate from Proposition 5.8, since every prime is an inter-section of maximal ideals by definition. (cid:3)
Using this, we can give some versions of Proposition 3.1 that refer only to closedfibers.
Proposition . Let R be a Hilbert ring, R → S be module-free, and supposethat for every maximal ideal m of R , the fiber κ m ⊗ R S is a geometrically reducedand irreducible κ m -algebra. Then R → S has the stable prime extension property. Proof.
Since R → S is module-free, by base change R [ X , . . . , X n ] → S ⊗ R R [ X , . . . , X n ] ∼ = S [ X , . . . , X n ]is module-free as well, and hence intersection flat. By the Hilbert hypothesis, anymaximal ideal M contracts to a maximal ideal m of R . We then have that S [ X , . . . , X n ] M S [ X , . . . , X n ] ∼ = κ M ⊗ R S ∼ = ( κ m ⊗ R S ) ⊗ κ m κ M and by the hypothesis on the fibers, this quotient is either a domain or zero. ByProposition 5.9, R [ X , . . . , X n ] → S [ X , . . . , X n ] then has the prime extensionproperty, as required. (cid:3) Corollary . Let R be a finitely generated algebra over an algebraicallyclosed field, and let S be a nonzero module-free R -algebra. Suppose that for everymaximal ideal m of R , we have S/ m S is a domain. Then R → S has the stableprime extension property. XTENSIONS OF PRIMES AND FLATNESS 17
Proof.
In this case R is a Hilbert ring, and every residue field is algebraicallyclosed. Thus, if the fiber κ m ⊗ R S is a domain or zero, then it is a geometri-cally reduced and irreducible κ m -algebra, by, e.g., [ , Propositions 4.5.1 and 4.6.1].Proposition 5.10 then applies. (cid:3) Theorem . Let K be an algebraically closed field and let S be an N -graded K -algebra that is a domain. Let F , . . . , F n be a regular sequence of forms in S generating a prime ideal Q of S . Then the K -algebra map of the polynomial ring R = K [ X , . . . , X n ] → S such that X i F i , ≤ i ≤ n , has the stable primeextension property. Proof.
The hypothesis is stable under adjoining finitely many indeterminatesto both rings: we may use these to enlarge the sequence F , . . . , F n . Thus, itsuffices to prove the prime extension property under the given hypotheses. Thehypothesis implies that S is free over R . Indeed, fix a homogeneous basis for S/QS over K . These elements will span S over R by the graded version of Nakayama’slemma. They have no relations by induction on n : given a nonzero relation wemay factor out the highest power of F occurring in all coefficients, since F is anonzerodivisor in S , and then we obtain a nonzero relation on the images of thesegenerators working over S/F S and R/X R . Since every maximal ideal of R hasthe form ( X − c , . . . , X n − c n ) for c , . . . , c n ∈ K , we know that the expansion ofany maximal ideal of R to S has the form ( F − c , . . . , F n − c n ). The result nowfollows from Corollary 5.11 and [ , Proposition 2.8]. (cid:3) Acknowledgments
We thank Neil Epstein for bringing to our attention the connection betweenintersection flatness and the theory of content in the sense of Ohm and Rush. Wealso thank Gabriel Picavet for directing us to his earlier work on the stable primeextension property and flatness for domains.
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Department of Mathematics, East Hall, 530 Church St., Ann Arbor, MI 48109–1043,USA
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