Coherence conditions in flat regular pullbacks
aa r X i v : . [ m a t h . A C ] J un COHERENCE CONDITIONS IN FLAT REGULAR PULLBACKS
JASON BOYNTON AND SEAN SATHER-WAGSTAFF
Abstract.
We investigate the behavior of four coherent-like conditions in reg-ular conductor squares. In particular, we find necessary and sufficient condi-tions in order that a pullback ring be a finite conductor ring, a coherent ring,a generalized GCD ring, or quasi-coherent ring. As an application of theseresults, we are able to determine exactly when the ring of integer-valued poly-nomials determined by a finite subset possesses one of the four coherent-likeproperties. Introduction
Throughout this paper, the term “ring” is short for “commutative ring withidentity”, and “module” is short for “unital module”.Over the last half century, there has been an abundance of research dedicatedto the study of the transference of various ring and ideal-theoretic properties inpullback constructions. It is well-known that these pullback constructions providea rich source of (counter)examples in commutative algebra; see [14] for a survey. Inhis book [9], Gilmer popularized a very special case of a pullback called the D + M construction. Gilmer’s construction begins with a valuation domain V containinga retract field K , meaning that V = K + M for some maximal ideal M of V . Let D be a subring of K , and form the subring D + M ⊂ V . In [5], Dobbs and Papickfind necessary and sufficient conditions on K and D (or M ) in order that D + M is a coherent ring. In [3], Brewer and Rutter dropped the valuation condition on T and found similar conditions on the constituent rings so that the ring D + M iscoherent. In [8], Houston and Gabelli offer improved results on the transference ofcoherence and other coherent-like conditions by removing the assumption that thedomain T contains a retract field.The purpose of this article is to investigate the transference of coherent-like con-ditions in a more general setting that is very similar to the pullback constructionof [7]; see Problem 1.1. We are primarily concerned with finding conditions underwhich the (quasi) coherent, finite conductor, and generalized GCD properties as-cend and descend in a pullback. The construction central to our study, is describedhere explicitly. Start with a ring surjection η : T ։ B and an inclusion of rings ι : A ֒ → B with B = 0, hence A = 0. Let R denote the pullback of these maps,that is, the subring of A × T consisting of all elements ( a, t ) such that ι ( a ) = η ( t ). Date : September 18, 2018.2010
Mathematics Subject Classification.
Key words and phrases.
Coherent rings; finite conductor rings, generalized GCD rings; integervalued polynomials; pullbacks; quasi-coherent rings.Sean Sather-Wagstaff was supported in part by a grant from the NSA.
The natural maps η : R ։ A and ι : R ֒ → T yield a commutative diagram of ringhomomorphisms R (cid:31) (cid:127) ι / / η (cid:15) (cid:15) (cid:15) (cid:15) T η (cid:15) (cid:15) (cid:15) (cid:15) A (cid:31) (cid:127) ι / / B ( (cid:3) )such that Ker( η ) = Ker( η ). The common ideal Ker( η i ) is the largest commonideal of R and T ; it is denoted C and called the conductor of T into R . When C contains a T -regular element, we say that the conductor square ( (cid:3) ) is regular .Conductor squares can also be built as follows. Let T be a commutative ringwith subring R , and suppose that R and T have a common, non-zero ideal. Wecall the largest common ideal C the conductor of T into R . Setting A = R/C and B = T /C , we obtain a commutative diagram ( (cid:3) ) which is a conductor square.It is common in the study of pullback constructions to assume that T is anintegral domain and that C is a maximal ideal of T . However, important examplesare obtained by allowing zero-divisors in the pullback square. For example, let D bean integral domain with field of fractions K , and let E = { e , . . . , e r } ⊂ D . Setting T = K [ X ] and C = ( X − e ) · · · ( X − e r ) K [ X ], we have B = T /C ∼ = Q ri =1 K . Using A = Q ri =1 D in the conductor square, we get R = Int( E, D ) = { g ∈ K [ X ] | g ( E ) ⊂ D } the ring of integer-valued polynomials on D determined by the subset E . Observethat the rings A and B are not integral domains. In fact, Chapman and Glaz haveproposed the following open question. Problem 1.1 ([4, Problem 50]) . Study the ring and ideal-theoretic properties thattransfer in a conductor square where the conductor ideal is not maximal (or evenprime) in the extension ring.This paper is organized as follows. In Section 2, we provide the relevant defini-tions and some background results for pullbacks in which the extension ι : R ֒ → T is flat. In Section 3, we show that, in a non-trivial pullback square, the conductor C is never finitely generated over R , and ι : A ֒ → B is never a faithfully flat ex-tension of rings, whenever ι : R ֒ → T is flat. In light of [2, 8], this suggests thatif R is a coherent-like ring defined by a regular conductor square of the type ( (cid:3) ),then either T is R -flat or C (and T ) is finitely generated over R .The remaining sections, we assume that T is R -flat in the regular conductorsquare ( (cid:3) ). In Section 4, we assume that the conductor C is principal in T andthat a unitary type of condition similar to [16] holds in order to find necessary andsufficient conditions on the constituent rings A and T so that R is a finite conductorring. The proofs of these results all extend naturally to the coherent ring case. Weconclude the section by showing that if E is finite, then the ring Int( E, D ) is afinite conductor ring if and only if D is a finite conductor ring. We note that asimilar result holds for Int( E, D ) in the coherent case. In Sections 5 and 6, we provesimilar results for generalized GCD rings and quasi-coherent rings in a flat regularconductor square.In contrast to the ideal-theoretic methods utilized in [8], our proofs rely slightlymore on module theory. We make frequent use of the results found in [10, 12].
OHERENCE CONDITIONS IN FLAT REGULAR PULLBACKS 3 Background
This section focuses on foundational notions and technical lemmas for the sequel.
Remark 2.1.
Consider the regular conductor square ( (cid:3) ). Then T is a domain ifand only if R is a domain. Indeed, since R is a subring of T , one implication iseasy (and does not use the regularity of ( (cid:3) ). The other implication follows fromthe fact that T is an overring of R , by [2, Proposition 2.5(i)]. Remark 2.2.
It is straightforward to show that the following conditions on theconductor square ( (cid:3) ) are equivalent:(i) A = 0;(ii) C = R ;(iii) T = R ;(iv) A = B ;(v) T = C ; and(vi) B = 0.A trivial conductor square is one that satisfies these equivalent conditions. We areonly interested in non-trivial conductor squares. Definition 2.3.
A flat ring homomorphism R → T is a flat epimorphism if thenatural multiplication map µ : T ⊗ R T → T is bijective (i.e., injective).The next lemma shows that many of our results conform to the “general format”. Lemma 2.4.
Consider the regular conductor square ( (cid:3) ) . If T is flat over R , then R → T is a flat epimorphism.Proof. It suffices to show that the natural multiplication map µ : T ⊗ R T → T isinjective. Let c ∈ C be T -regular. The maps T c −→ cT ⊆ −→ C ⊆ −→ R are injective.Since T is flat over R , the explains the monomorphisms in the top row of thefollowing commutative diagram. T ⊗ R T (cid:31) (cid:127) c ⊗ R T / / µ (cid:15) (cid:15) ( cT ) ⊗ R T (cid:31) (cid:127) / / R ⊗ R T ∼ = (cid:15) (cid:15) T c / / T. The unspecified vertical isomorphism is the natural one (tensor-cancellation). Itfollows that µ is injective, as desired. (cid:3) Remark 2.5.
Consider the regular conductor square ( (cid:3) ). Recall that an A -module F is faithfully flat if it is flat and for every A -module N one has N = 0 if and only if F ⊗ A N = 0. (See also [15, Theorem 7.2].) Lemma 2.4 implies that if T is faithfullyflat over R , then T = R ; see [10, p. 15].We use the following conditions in Sections 4–6. Note that (U2) = ⇒ (U1). Definition 2.6.
Consider a regular conductor square ( (cid:3) ) and an ideal I of R . (FP): T is flat over R and C is a principal ideal of T . (U1): There is an ideal I ′ of R isomorphic to I such that I ′ T = T . (U2): There are elements r, s ∈ R and an ideal U ⊆ R such that U T = T = rT and I r −→ ∼ = rI = sU s ←− ∼ = U . JASON BOYNTON AND SEAN SATHER-WAGSTAFF
In conditions (U1) and (U2), the ideals I ′ and U are “unitary”.Next, we discuss important cases where the conditions from Definition 2.6 hold. Remark 2.7.
Assume that T is a PID and a localization of R . Then conditions(FP), (U1), and (U2) from Definition 2.6 hold for all finitely generated ideals I of R ; argue as in [16, Lemma 2.2]. For instance, let K be a field with algebraic closure K , and set T = K [ x ]. Let θ , . . . , θ r ∈ K with minimal polynomials p , . . . , p r ∈ T .Assume that the p i are pairwise relatively prime. For i = 1 , . . . , r , let A i be adomain with field of fractions K [ θ i ].Set A := A × · · · × A r and B := K [ θ ] × · · · × K [ θ r ], let η : T → B bethe natural surjection f ( f ( θ ) , . . . , f ( θ r )), and let ι : A ֒ → B be the naturalinclusion. Consider the conductor square determined by this data: R (cid:31) (cid:127) ι / / η (cid:15) (cid:15) (cid:15) (cid:15) K [ X ] η (cid:15) (cid:15) (cid:15) (cid:15) A × · · · × A r (cid:31) (cid:127) ι / / K [ θ ] × · · · × K [ θ r ] . ( ⊠ )Note that K [ X ] is a localization of R . Also, not that if E is a finite subset of adomain D , then Int( E, D ) is a special case of this construction.For the next result, recall that a ring R is B´ezout if every finitely generated idealof R is principal. Proposition 2.8.
In the regular conductor square ( (cid:3) ) , assume that T is flat over R . Consider the following conditions. (i) Every non-zero finitely generated ideal of R satisfies the condition (U2). (ii) Every non-zero 2-generated ideal of R satisfies the condition (U2). (iii) Every non-zero finitely generated ideal of R satisfies the condition (U1). (iv) Every non-zero 2-generated ideal of R satisfies the condition (U1). (v) The ring T is a B´ezout domain.The implications (i) = ⇒ (ii) = ⇒ (iv) = ⇒ (v) and (i) = ⇒ (iii) = ⇒ (iv) = ⇒ (v) always hold. If T is a localization of R , then the conditions (i) – (v) are equivalent.Proof. The implications (i) = ⇒ (ii) = ⇒ (iv) and (i) = ⇒ (iii) = ⇒ (iv) are routine.(iv) = ⇒ (v). Assume that every non-zero 2-generated ideal of R satisfies thecondition (U1). To show that T is B´ezout, it suffices to show that every 2-generatedideal of T is principal; that is is sufficient follows from an induction argument on thenumber of generators of a given finitely generated ideal. Let t, u ∈ T and considerthe ideal J := ( t, u ) T . Let c ∈ C be T -regular, noting that we have ct, cu ∈ R . Itfollows that J ∼ = cJ = ( ct, cu ) T , so we may replace J with cJ to assume without lossof generality that t, u ∈ R . Our assumption implies that the ideal I is isomorphicto an ideal I ′ of R such that I ′ T = T . Since T is flat over R , it follows that J = IT ∼ = T ⊗ R I ∼ = T ⊗ R I ′ ∼ = I ′ T = T so J is principal. Note that this argument also shows that T is a domain. Indeed,if 0 = t = u , then the argument above shows that tT ∼ = T , so t is a non-zero-divisoron T .It remains to assume that T is a localization of R , and to prove the implica-tion (v) = ⇒ (i). Assume that T = S − R is a B´ezout domain, and let I be anon-zero finitely generated ideal of R , say I = ( r , . . . , r n ) R . We need to show that OHERENCE CONDITIONS IN FLAT REGULAR PULLBACKS 5 I satisfies condition (U2) from Definition 2.6. The extension IT is finitely generatedover T , hence it is principal, say IT = ( r , . . . , r n ) T = tT with t ∈ T . The condition I = 0 implies that tT = IT = 0, since R ⊆ T , so we have t = 0. Write t = r/s forsome r ∈ R r { } and s ∈ S . Since s is a unit in T , it follows that IT = rT . For i = 1 , . . . , n , write r i = rt i for some t i ∈ T = S − R , and write t i = ρ i /σ i for some ρ i ∈ R and σ i ∈ S . Set σ = σ · · · σ n and σ ′ i = σ/σ i , and note that σ ∈ S ⊆ R satisfies σT = T . Also, we have ρ i , σ ′ i ∈ R , so we set U := ( ρ σ ′ i , . . . , ρ n σ ′ n ) R . Itis straightforward to show that we have I σ −→ ∼ = σI = rU r ←− ∼ = U and U T = T , socondition (U1) is satisfied. (cid:3) The next result shows that many of our results are trivially true when the con-ductor square ( (cid:3) ) is trivial, even if it is not regular.
Proposition 2.9.
Consider the trivial conductor square ( (cid:3) ) . If every non-zero2-generated ideal I of R satisfies condition (U1) from Definition 2.6, then R = T is a B´ezout domain.Proof. The condition R = T follows from the triviality assumption.We now show that R is a B´ezout ring. Let I be a non-zero finitely generatedideal. By assumption, there is an ideal I ′ of R such that I ∼ = I ′ and I ′ T = T . Thecondition R = T implies that I ∼ = I ′ = I ′ R = I ′ T = T = R , so I is principal, asdesired.This also shows that R is a domain, since any principal ideal I = rR satisfies rR = I ∼ = R , so r is a non-zero-divisor. (cid:3) The next two lemmas document some technical facts for use in the sequel.
Lemma 2.10.
Consider the regular conductor square ( (cid:3) ) . Assume that T is flatas an R -module and that C is a principal ideal of T . Let I be an ideal of R . (a) If IT = T , then I ⊇ IC = C and IA = I/C = I/CI ∼ = A ⊗ R I . (b) The natural map µ : C ⊗ R I → I is injective and Tor R ( A, I ) = 0 . (c) If IT is a principal regular ideal of T , e.g., if IT = T , then for all i we have Ext iR ( I, C ) ∼ = Ext iR ( I, T ) ∼ = ( T if i = 00 if i = 0 .Proof. (a) If C/IC ∼ = T /IT = 0, then C = IC ⊆ I . Alternately, if T = IT , then C = T C = IT C = IC .(b) By assumption, we have C = cT ∼ = T for some element c ∈ C . Consider thefollowing commutative diagram: T ⊗ R I ∼ = / / ∼ = c · (cid:15) (cid:15) T I (cid:127) _ c · (cid:15) (cid:15) C ⊗ R I µ / / I The horizontal maps are the natural ones: t ⊗ k tk . It follows that µ is injective.Now, consider the exact sequence0 → C → R → A → R ( − , I ):0 → Tor R ( A, I ) → C ⊗ R I µ −→ I → I/CI → . JASON BOYNTON AND SEAN SATHER-WAGSTAFF
Since µ is injective, it follows that Tor R ( A, I ) = 0, as desired.(c) Since C = cT ∼ = T , the isomorphism Ext iR ( I, C ) ∼ = Ext iR ( I, T ) is automatic.For the other isomorphism, let P be a projective resolution of I over R . The iso-morphisms in the next sequence are Hom-tensor adjointness and Hom-cancellation:Hom T ( T ⊗ R P, T ) ∼ = Hom R ( P, Hom T ( T, T )) ∼ = Hom R ( P, T ) . (2.10.1)Since T is flat over R , the complex T ⊗ R P is a projective resolution of T ⊗ R I ∼ = IT ∼ = T ; the second isomorphism is from the assumption that IT is principal andregular over T . This explains the third isomorphism in the next sequence:Ext iR ( I, T ) ∼ = H i (Hom R ( P, T )) ∼ = H i (Hom T ( T ⊗ R P, T )) ∼ = Ext iT ( T, T ) ∼ = ( T if i = 00 if i = 0.The first isomorphism is by definition, the second one is from (2.10.1), and the lastone is standard. (cid:3) Lemma 2.11.
Consider the regular conductor square ( (cid:3) ) . Assume that T is flatas an R -module and that C is a principal ideal of T . Let I be an ideal of R suchthat IT = T . Then the natural maps C → C · Hom R ( I, R ) → C · Hom R ( I, T ) → Hom R ( I, C ) are isomorphisms, and one has Hom R ( I, R ) / [ C · Hom R ( I, R )] ∼ = Hom A ( IA, A ) . Proof.
Assume that IT = T . We first describe the “natural maps” from the state-ment. For each r ∈ R , let λ r : I → R be given by λ r ( i ) = ri . In particular, themap λ : I → R is the inclusion of I in R . Moreover, for all r, s ∈ R we have rλ s = λ rs and so λ r = rλ . In particular, the map Λ : C → C · Hom R ( I, R ) givenby x xλ = λ x is a well-defined R -module homomorphism. Furthermore, thismap is a monomorphism, as follows. Suppose that x ∈ Ker(Λ), that is, that λ x = 0.By definition, it follows that xI = 0, and this implies that 0 = xIT = xT . Since T contains 1, it follows that x = 0.Consider the inclusion R ǫ −→ ⊆ T , and apply the left-exact functor Hom R ( I, − )to obtain the monomorphism Hom R ( I, R ) ǫ ∗ −→ Hom R ( I, T ). Note that ǫ ∗ simplyenlarges the codomain of a homomorphism I → R from R to the larger ring T .The natural map C · Hom R ( I, R ) Φ −→ C · Hom R ( I, T ) is induced by ǫ ∗ : for elements c j ∈ C and f j ∈ Hom R ( i, R ), we have Φ( P j c j f j ) := P j c j ǫ ∗ ( f j ) = P j c j ( ǫ ◦ f j ).Note that this gives a commutative diagram C · Hom R ( I, R ) Φ / / (cid:127) _ (cid:15) (cid:15) C · Hom R ( I, T ) (cid:127) _ (cid:15) (cid:15) Hom R ( I, R ) (cid:31) (cid:127) ǫ ∗ / / Hom R ( I, T )where the vertical monomorphism are the subset inclusions. Since ǫ ∗ is injective,the commutative diagram shows that Φ is also injective. OHERENCE CONDITIONS IN FLAT REGULAR PULLBACKS 7
Next, consider elements c j ∈ C and g j ∈ Hom R ( I, T ). Note that the function g := P j c j g j satisfies g ( i ) = P j c j g j ( i ) ∈ CT = C , that is, we have Im( g ) ⊆ C ;let g ′ : I → C denote the homomorphism obtained by restricting the codomain of g from T to C . We define C · Hom R ( I, T ) Ψ −→ Hom R ( I, C ) by the formula Ψ( g ) := g ′ . Note that if ρ : C → T denotes the inclusion map and ρ ∗ : Hom R ( I, C ) → Hom R ( I, T ) is the induced monomorphism, then our definition implies that wehave ρ ∗ (Ψ( g )) = ρ ∗ ( g ′ ) = ρ ◦ g ′ = g . Also, note that g and g ′ use the same rule ofassignment: for all i ∈ I , we have g ( i ) = g ′ ( i ), by definition. It follows that g = 0if and only if g ′ = 0, that is, if and only if Ψ( g ) = 0; thus, Ψ is injective.Since the maps Λ, Φ, and Ψ are monomorphisms, to prove that they are isomor-phisms, it suffices to show that the composition Ψ ◦ Φ ◦ Λ is surjective. It is straight-forward to show that, for all x ∈ C and all i ∈ I , we have Ψ(Φ(Λ( x )))( i ) = xi .On the other hand, we consider the following sequence: T ∼ = Hom T ( T, T )= Hom T ( IT, T ) ∼ = Hom T ( T ⊗ R I, T ) ∼ = Hom R ( I, Hom T ( T, T )) ∼ = Hom R ( I, T ) . The first and last steps are Hom-cancellation. The second step is from the as-sumption IT = T . The third step is by the flatness of T over R , which impliesthat the map T ⊗ R I → IT , given by t ⊗ i it , is an isomorphism. The fourthstep is Hom-tensor adjointness. It is routine to show that the composition of theseisomorphisms α : T → Hom R ( I, T ) is given by the formula α ( t )( i ) = it .Multiplying by C yields an isomorphism C = CT α ′ −→ ∼ = C · Hom R ( I, T ) which isgiven by α ′ ( x ) = α ′ ( x ·
1) = xα (1) = α ( x ). In other words, we have α ′ ( x )( i ) = xi =Ψ(Φ(Λ( x )))( i ), so α ′ ( x ) = Ψ(Φ(Λ( x ))) and hence α ′ = Ψ ◦ Φ ◦ Λ. Since α ′ is anisomorphism, it follows that Ψ ◦ Φ ◦ Λ is surjective, as claimed. This shows that Λ,Φ, and Ψ are isomorphisms.To complete the proof, consider the exact sequence0 → C → R → A → . Since Ext R ( I, C ) = 1 by Lemma 2.10(c), the induced long exact sequence inExt R ( I, − ) begins as follows:0 → Hom R ( I, C ) → Hom R ( I, R ) → Hom R ( I, A ) → . By what we have already shown, Hom R ( I, C ) is naturally identified with the sub-module C · Hom R ( I, R ) ⊆ Hom R ( I, R ), so this sequence has the form0 → C · Hom R ( I, R ) ⊆ −→ Hom R ( I, R ) → Hom R ( I, A ) → . Thus, we have the first isomorphism in the next sequence:Hom R ( I, R ) / [ C · Hom R ( I, R )] ∼ = Hom R ( I, A ) ∼ = Hom R ( I, Hom A ( A, A )) ∼ = Hom A ( A ⊗ R I, A ) ∼ = Hom A ( IA, A ) . JASON BOYNTON AND SEAN SATHER-WAGSTAFF
The second and third isomorphisms are cancellation and adjointness. And thefourth isomorphism is from Lemma 2.10(a). This sequence of isomorphisms com-pletes the proof. (cid:3) C is not finitely generated over R The point of this section is to prove Theorem 3.2, which says that, given a non-trivial, flat, regular conductor square ( (cid:3) ), the conductor ideal C is never finitelygenerated over R , and B is never faithfully flat over A . Note, however, that C canbe finitely generated over T (indeed, it can be principal), and B is always flat over A in this setting. Lemma 3.1.
Consider the regular conductor square ( (cid:3) ) , and assume that C isfinitely generated over R (e.g., C satisfies ( † ) over R ). (a) For each prime ideal p ⊂ A , there is a prime ideal q ⊂ B lying over p . (b) For each maximal ideal m ⊂ A , there is a maximal ideal n ⊂ B lying over m . (c) If T is flat as an R -module, then B is faithfully flat as an A -module.Proof. (a) Let P ⊂ R be the contraction of p along the surjection R ։ A . Thelocalized square R P (cid:31) (cid:127) / / (cid:15) (cid:15) (cid:15) (cid:15) T P (cid:15) (cid:15) (cid:15) (cid:15) A P (cid:31) (cid:127) / / B P ( (cid:3) P )is a regular conductor square with conductor C P ; see [2, Notation 2.4]. In particular,the ring B P is non-zero, so it has a maximal ideal, necessarily of the form q P by theprime correspondence under localization. The contraction in A P = A p of q P = q p must be maximal in A p by [2, Proposition 2.5(iii)]; this is where we use the finitegeneration of C . Thus, the contraction in A p of q p must be p p . Another applicationof the prime correspondence implies that the contraction of q in A is p , as desired.(b) Part (a) implies that there is a prime ideal q of B that contracts to m in A . If q is not maximal, let n ⊂ B be a maximal ideal containing q . Then the contractionof n in A must be a prime ideal containing m , that is, the contraction must be m .(c) Since T is flat over R , we know that B is flat over A . Thus, part (b) impliesthat B is faithfully flat over A ; see [15, Theorem 7.2]. (cid:3) A special case of the next result is in [8, Lemma 4.1]; see also [3, Lemma 1].
Theorem 3.2.
Consider the non-trivial regular conductor square ( (cid:3) ) . Assumethat T is flat as an R -module. Then B is not faithfully flat over A , and C is notfinitely generated over R Proof.
Suppose by way of contradiction that B is faithfully flat over A . We derivea contradiction by showing that T = R . To accomplish this, by Remark 2.5 itsuffices to show that T is faithfully flat over R . Since T is flat over R , it suffices toshow that for each maximal ideal m ⊂ R there is a maximal ideal n ⊂ T such that m = R ∩ n . If m contains C , then this follows from Lemma 3.1(b), via the primecorrespondence under quotients. On the other hand, if m does not contain C , thenthere is a unique prime ideal q of T contracting to m in R , by [7, Lemma 1.1.4(3)].As in the proof of Lemma 3.1(b), this yields a maximal ideal of T contracting to m . (Of course, the uniqueness of q implies that q itself is maximal.) OHERENCE CONDITIONS IN FLAT REGULAR PULLBACKS 9
Lastly, if C were finitely generated over R , then B would be faithfully flat over A , by Lemma 3.1(c), contradicting the previous paragraph. (cid:3) We conclude this section by documenting some corollaries and examples.
Corollary 3.3.
Consider the non-trivial regular conductor square ( (cid:3) ) . Assumethat C is maximal in R , that is, that A is a field. Then T is not flat over R .Proof. Since A is a field, the extension A → B is faithfully flat. Now apply Theo-rem 3.2. (cid:3) Example 3.4.
Consider the regular pullback R = Q + x R [ x ] (cid:31) (cid:127) ι / / η (cid:15) (cid:15) (cid:15) (cid:15) R [ x ] η (cid:15) (cid:15) (cid:15) (cid:15) Q (cid:31) (cid:127) ι / / R where η maps x
0. Then Corollary 3.3 shows that R [ x ] is not flat over R . Corollary 3.5.
Consider the non-trivial regular conductor square ( (cid:3) ) . Assumethat R is a Pr¨ufer ring. Then C is not finitely generated over R Proof.
Since R is Pr¨ufer, every overring of R is flat over R , including the overring T . Now apply Theorem 3.2. (cid:3) Example 3.6.
Consider the regular pullback ( ⊠ ) with r = 1, A = Z , K = Q ,and θ = 0. It is well known that the pullback R = Z + x Q [ x ] is Pr¨ufer. Thus, theconductor ideal x Q [ x ] is not finitely generated over R . Corollary 3.7.
Consider the non-trivial regular conductor square ( (cid:3) ) . Assumethat R is noetherian. Then T is not flat over R .Proof. Apply Theorem 3.2. (cid:3)
Example 3.8.
Assume that A is a field with B = A , that B is finitely generatedas an A -module, and T is noetherian. Then R is noetherian by [2, Theorem 3.2],and B is automatically flat over A , but each corollary shows that T is never flatover R . A specific example of this is the ring R = Q + ( x + 1) Q [ x ], arising fromthe data A = Q , B = Q [ i ], and T = Q [ x ].4. Finite Conductor Rings and Coherent Rings
This section is devoted to the study of the transfer of the following two propertiesin flat, regular pullbacks. Note that Gabelli and Houston [8] investigate the specialcase where C is maximal in T . Definition 4.1.
We consider the following coherency conditions on a ring R .(1) finite conductor ring : every 2-generated ideal of R is finitely presented.(2) coherent ring : every finitely generated ideal of R is finitely presented.We investigate the stably coherent situation in [1]. Remark 4.2.
Every B´ezout domain is coherent, and every coherent ring is a finiteconductor ring. In particular, in the case of trivial conductor squares, the resultsof this section follow from Proposition 2.9.
The coherent case of the next result is a special case of [13, Proposition 2.2].
Proposition 4.3.
Consider the regular conductor square ( (cid:3) ) . Assume that T isflat as an R -module. If R is a finite conductor ring (respectively, R is coherent),then T is as well.Proof. Assume first that R is a finite conductor ring, and let J be a 2-generatedideal of T . It suffices to show that J is finitely presented over T . By assumption,there is a T -regular element c ∈ C . It follows that the ideal cJ is isomorphic to J (in particular, it is 2-generated) so it suffices to show that cJ is finitely presentedover T . Furthermore, cJ is generated over T by two elements of cJ ⊆ C ⊆ R .Thus, we replace J with cJ to assume that J is generated by two elements r, s ∈ R .Since R is a finite conductor ring, the ideal I = ( r, s ) R is finitely presented over R . By construction, we have IT = J . From right-exactness, the module T ⊗ R I is finitely presented over T . By flatness, we have T ⊗ R I ∼ = IT = J , that is, J isfinitely presented over T , as desired.The coherent case is proved similarly. (cid:3) The next two results show how our properties of interest descend in a flat, regularpullback square.
Theorem 4.4.
Assume that the regular conductor square ( (cid:3) ) satisfies condition(FP), and every non-zero 2-generated ideal I of R satisfies condition (U1) fromDefinition 2.6. If A is a finite conductor ring, then so are R and T .Proof. Assume that A is a finite conductor ring. By Proposition 4.3, it suffices toshow that R is also a finite conductor ring. Let I be a 2-generated ideal of R . Weneed to show that I is finitely presented over R .By assumption, there is an ideal I ′ ⊆ R isomorphic to I such that I ′ T = T .Thus, we may replace I with I ′ to assume that IT = T . Lemma 2.10(a) impliesthat IC = C .To show that I is finitely presented over R , it suffices by [10, Theorem 5.1.2]to show that T ⊗ R I is finitely presented over T and that I/CI = I/C is finitelypresented over
R/C = A . Note that this uses the fact that C ∼ = T is flat over R and that CT = C .Since T is flat over R , we have T ⊗ R I ∼ = IT = T , which is finitely presented over T . Also, the ideal IA is 2-generated over A , so it is finitely presented over A , since A is a finite conductor ring. That is, the A -module I/CI = I/C = IA is finitelypresented, as desired. (cid:3) Theorem 4.5.
Assume that the regular conductor square ( (cid:3) ) satisfies condition(FP), and every finitely generated non-zero ideal I of R satisfies condition (U1)from Definition 2.6. If A is coherent, then so are R and T .Proof. This is similar to the proof of Theorem 4.4. (cid:3)
Remark 4.6.
Note that Theorem 4.5 is similar in spirit to [10, Theorem 5.1.3]:with some extra assumptions, if A and T are coherent, then so is R . However,our result covers some examples that [10, Theorem 5.1.3] does not cover, and viceversa. For instance, in the case ( ⊠ ) of 2.7, if r >
2, and A is non-noetherian withinfinite weak dimension, then Theorem 4.5 applies while [10, Theorem 5.1.3] doesnot; see, e.g., Corollary 4.13. OHERENCE CONDITIONS IN FLAT REGULAR PULLBACKS 11
On the other hand, Theorem 4.5 is somewhat different from [10, Theorem 5.1.3],in that we only assume that A is coherent. This is a byproduct of the assumption(U1) from Definition 2.6, in light of Proposition 2.8.Our next results treat the ascent of our properties of interest. Theorem 4.7.
Assume that the regular conductor square ( (cid:3) ) satisfies condition(FP) and every non-zero 2-generated ideal I of R satisfies condition (U2) fromDefinition 2.6. Assume further that A is locally a domain and has finite Krulldimension. Then R is a finite conductor ring if and only if A is a finite conductorring; each of these conditions implies that T is a finite conductor ring.Proof. In view of Proposition 4.3 and Theorem 4.5, it remains to assume that R isa finite conductor ring and prove that A is so. Let J = ( a, b ) A be a 2-generatedideal of A . We need to prove that J is finitely presented over A . Assume withoutloss of generality that J = 0, and, further, that a = 0. Let f, g ∈ R be such that η ( f ) = a and η ( g ) = b . Set I = ( f, g ) R .Case 1: IT = T . Since I is 2-generated and R is a finite conductor ring, the ideal I is finitely presented over R . It follows (e.g., by right-exactness of tensor-product)that the A -module I/CI is finitely presented. Lemma 2.10(a) implies that CI = C ,so the ideal I/C = IA = ( f, g ) A = ( a, b ) A = J is finitely presented, as desired.Case 2: there is an ideal I ′ ⊆ R such that I ′ T = T and an element h ∈ I such that the multiplication map I ′ h −→ I is an isomorphism. In particular, I ′ is2-generated, by elements f ′ , g ′ ∈ I ′ such that f = f ′ h and g = g ′ h . Set a ′ = η ( f ′ )and b ′ = η ( g ′ ) and x = η ( h ). It follows that we have a = η ( f ) = η ( f ′ ) η ( h ) = a ′ x and similarly b = b ′ x . In particular, the condition a = 0 implies that x = 0.Consider the natural surjection τ ′ : A → J ′ , represented by the row matrix (cid:0) a ′ b ′ (cid:1) . By Case 1, the ideal J ′ = ( a ′ , b ′ ) A is finitely presented over A , so K ′ :=Ker( τ ′ ) is n -generated for some integer n . Assume without loss of generality that n > J is finitely presented, consider the natural surjection τ : A → J ,represented by the row matrix (cid:0) a b (cid:1) . We need to show that K := Ker( τ ) is finitelygenerated. We accomplish this using [17, Theorem 2.1], which says that it sufficesto show that K p is n -generated over A p for each prime p ⊂ A . (Here is where weuse the finiteness of dim( A ). In the language of [17], this allows us to conclude thatthe function b ( p , M ) := ( µ ( A p , M p ) + dim( A/ p ) if p ∈ Supp A ( K )0 otherwisesatisfies b ( p , M ) n + dim( A ) for all p .)Consider the multiplication map J ′ x −→ J , which is surjective (hence, locallysurjective) by construction. Fix a prime p ⊂ A . If the map J ′ p x −→ J p is injective,then it is an isomorphism (since locally surjective). In this situation, given thedefining matrices for τ and τ ′ , it follows that K p ∼ = K ′ p , so K p is also n -generated.Thus, we assume for the remainder of the proof of Case 2 that the map J ′ p x −→ J p is not injective. Claim: J p = 0. Let z ∈ J ′ and s ∈ A r p be such that the element z/t in J ′ p is non-zero and satisfies 0 = ( z/t ) x = ( z/t )( x/ A p is a domain byassumption, it follows that x/ A p . Thus, we have J p = ( x/ J ′ p = 0, asclaimed.From the claim, it follows that K p ∼ = A p , which is 2-generated. Thus, it is n -generated, since n >
2, as desired.Case 3: the general case. By assumption, there are elements r, s ∈ R and anideal U ⊆ R such that U T = T = rT and I r −→ ∼ = rI = sU s ←− ∼ = U . Case 2 shows that rIA = sU A is finitely presented. As in the proof of Case 2, the map J = IA r −→ rIA is surjective. The assumption rT = T says that r is a unit in T , so r representsa unit in T /C = B . In particular, multiplication by r , restricted to the subset IA ⊆ A ⊆ B is injective, so we have J = IA r −→ ∼ = rIA . Since rIA is finitelypresented over A , so is J , as desired. (cid:3) The next result is proved like the previous one. Compare to [10, 5.1.3].
Theorem 4.8.
Assume that the regular conductor square ( (cid:3) ) satisfies condition(FP) and every non-zero finitely generated ideal I of R satisfies condition (U2)from Definition 2.6. Assume further that A is locally a domain and has finite Krulldimension. Then R is coherent if and only if A is coherent; each of these conditionsimplies that T is coherent. Theorem 4.9.
Assume that the regular conductor square ( (cid:3) ) satisfies condition(FP) and every non-zero 2-generated ideal I of R satisfies condition (U2) fromDefinition 2.6. Assume further that A is a (possibly infinite) product Q λ ∈ Λ A λ ofdomains. Then R is a finite conductor ring if and only if A is a finite conductorring; each of these conditions implies that T is a finite conductor ring, and so iseach ring A λ .Proof. As in the proof of Theorem 4.7, we assume that R is a finite conductor ring,and we prove that A and A λ are finite conductor rings. For A , let J = ( a, b ) A with a = ( a λ ) ∈ Q λ ∈ Λ A λ = A and b = ( b λ ). To show that J is finitely presented over A , we consider two cases.Case 1: for each index λ , either a λ = 0 or b λ = 0. Let f, g ∈ R be such that η ( f ) = a and η ( g ) = b , and set I = ( f, g ) R . Since R is a finite conductorring, the ideal I is finitely presented over R . It follows (e.g., by right-exactness oftensor-product) that the quotient I/CI is finitely presented over A .Sub-case 1a: IT = T . Then Lemma 2.10(a) implies that CI = C , so the ideal I/C = IA = J is finitely presented over A , by the previous paragraph.Sub-case 1b: I is isomorphic to an ideal I ′ such that I ′ T = T via a multiplicationmap I ′ h −→ ∼ = I for a fixed h ∈ R . In particular, I ′ is 2-generated, by elements f ′ , g ′ ∈ I ′ such that f = f ′ h and g = g ′ h . Set a ′ = η ( f ′ ) and b ′ = η ( g ) and x = η ( h ).Write a ′ = ( a ′ λ ) and similarly for b ′ and x . As in the proof of Theorem 4.7, we have a = xa ′ , hence a λ = x λ a ′ λ and similarly for b λ , for all λ ∈ Λ. In particular, since a λ = 0 or b λ = 0 for each λ , we have x λ = 0.Since each ring A λ is a domain, and each coordinate of x is non-zero, the map J ′ := ( a ′ , b ′ ) A x −→ ( a, b ) A =: J is injective; it is surjective by construction. Since I ′ T = T , Sub-case 1a implies that J ′ is finitely presented. Hence, the ideal J ∼ = J ′ is finitely presented. This completes Sub-case 1b. OHERENCE CONDITIONS IN FLAT REGULAR PULLBACKS 13
Now, we complete the proof in Case 1. By assumption, there are elements r, s ∈ R and an ideal U ⊆ R such that U T = T = rT and I r −→ ∼ = rI = sU s ←− ∼ = U .Sub-case 1b implies that rIA = sU A is finitely presented over A . As in the proof ofTheorem 4.7, the map J = IA r −→ rIA is an isomorphism, so J is finitely presentedas well. This completes the proof in Case 1.Case 2: the general case. For each λ ∈ Λ, set e a λ := ( a λ if a λ = 0 or b λ = 01 if a λ = 0 = b λ .Set e a = ( e a λ ) and e J := ( e a, b ) A . Note that e J satisfies the hypotheses of Case 1, so itis finitely presented.For each λ ∈ Λ, set J λ := ( a λ , b λ ) A λ and e J λ := ( e a λ , b λ ) A λ . Note that we have e J λ = ( J λ if J λ = 0 A λ if J λ = 0.From this, it is straightforward to show that J is a direct summand of e J . (Specifi-cally, for each λ ∈ Λ, set J λ = ( J λ = 0 A λ if J λ = 0.Then one has e J ∼ = J L J .) Since e J is finitely presented over A , it is straightforwardto show that each summand (in particular, the summand J ) is finitely presentedover A . This completes Case 2, so we conclude that A is a finite conductor ring.Fix an index µ ∈ Λ. To show that A µ is a finite conductor ring, let J µ =( a µ , b µ ) A µ be a 2-generated ideal of A µ . For all λ ∈ Λ r { λ } , set a λ = 0 = b λ and J λ = 0. Also, set a = ( a λ ) ∈ Q λ ∈ Λ A λ = A and b = ( b λ ). Since A is a finiteconductor ring, the ideal ( a, b ) J = Q λ ∈ Λ J λ is finitely presented over A = Q λ ∈ Λ A λ .It follows readily that J µ is finitely presented over A µ , as desired. (cid:3) The next result is proved like the previous one.
Theorem 4.10.
Assume that the regular conductor square ( (cid:3) ) satisfies condition(FP) and every non-zero finitely generated ideal I of R satisfies condition (U2) fromDefinition 2.6. Assume further that A is a (possibly infinite) product Q λ ∈ Λ A λ ofdomains. Then R is coherent if and only if A is coherent; each of these conditionsimplies that T is coherent, and so is each ring A λ . Remark 4.11.
Let A be a (possibly infinite) product Q λ ∈ Λ A λ of non-zero com-mutative rings with identity. The last paragraph of the proof of Theorem 4.9 showsthat, if A is a finite conductor ring, then the same is true of each factor of A . Thisimplies that any sub-product Q λ ∈ Λ ′ A λ with Λ ′ ⊆ Λ is also a finite conductor ring.Conversely, if Λ is finite, and each A λ is a finite conductor ring, then so is thefinite product A . Example 4.12 shows that this converse fails when Λ is infinite.We deduce that, when Λ is infinite, the conclusion of Theorem 4.9 is very strong.(Similar comments hold for the other classes of rings considered below.) Example 4.12.
Let k be a field. For each integer n >
2, consider the polynomialring S n := k [ X , . . . , X n , Y , . . . , Y n ]. Let a n ⊆ S n denote the ideal generated bythe 2 × (cid:0) X ... X n Y ... Y n (cid:1) , and set A n := S n / a n . Consider the I n := ( X , Y ) A n and the natural surjection τ n : A n → I n withkernel K n . It is straightforward to show that K n contains the following vectors: (cid:0) Y − X (cid:1) , (cid:0) Y − X (cid:1) , . . . , (cid:0) Y n − X n (cid:1) . Moreover, since τ n is minimal, and the entries of thesevectors are homogeneous and linear, we conclude that these vectors are minimalgenerators of K n . In particular, since A n and I n are graded, we conclude that each K n requires at least n generators.Now, set A := Q ∞ n =2 A n , and consider the ideal I := Q ∞ n =2 I n ⊆ A . Since each I n is 2-generated over A n , the ideal I is 2-generated over A . The product τ : A → I ofthe maps τ n has kernel K = Q ∞ n =2 K n . Since each K n requires at least n generators,it follows that K is not finitely generated. Thus, even though each A n is noetherian(hence coherent and a finite conductor ring), the product A is neither coherent nora finite conductor ring. Corollary 4.13.
Consider the regular conductor square ( ⊠ ) from Remark 2.7.Then R is a finite conductor ring (resp., coherent) if and only if each A i is so.Proof. Remark 2.7 says that the hypotheses of Theorems 4.9–4.10 are satisfied. (cid:3)
Corollary 4.14.
Let D be a domain, and let E ⊆ D be a finite subset. Then thering of integer-valued polynomials Int(
E, D ) is a finite conductor ring (respectively,coherent) if and only if D is so.Proof. We have R = Int( E, D ) in the following special case of the conductor square( ⊠ ) with r = | E | : R (cid:31) (cid:127) ι / / η (cid:15) (cid:15) (cid:15) (cid:15) K [ X ] η (cid:15) (cid:15) (cid:15) (cid:15) D r (cid:31) (cid:127) ι / / K r . Thus, the desired conclusion follows from Corollary 4.13. (cid:3) GCD Domains and Generalizaed GCD Rings
Next, we turn our attention to transfer of the following two GCD properties.
Definition 5.1.
We consider the following coherency conditions on a ring R .(1) generalized GCD ring : every principal ideal of R is projective, and every inter-section of two finitely generated flat ideals is finitely generated and flat.(2) GCD domain : R is a domain such that every intersection of two principal idealsis principal. Remark 5.2.
Let R be a domain. It is straightforward to show that R is a GCDdomain if for every 2-generated ideal ( r, s ) R with r, s = 0 the kernel of the naturalmap R → ( r, s ) R is cyclic; indeed, the kernel of this map is isomorphic to rR T sR .Note that whenever this kernel is cyclic, it is isomorphic to R , since it is isomorphicto a non-zero principal ideal in the domain R .In particular, this shows that every B´ezout domain is a GCD domain, and everyGCD domain is a finite conductor domain. Remark 5.3.
A result of Glaz [12] says that R is a generalized GCD ring if andonly if it is a finite conductor ring and locally a GCD domain. Thus, every GCDdomain is a generalized GCD ring, by Remark 5.2. In particular, in the case oftrivial conductor squares, the results of this section follow from Proposition 2.9. OHERENCE CONDITIONS IN FLAT REGULAR PULLBACKS 15
Our transfer results for this context begin with the following ascent result.
Proposition 5.4.
Consider the regular conductor square ( (cid:3) ) . Assume that T isflat as an R -module. If R is a GCD domain (respectively, a generalized GCD ring),then T is as well.Proof. Assume that R is a GCD domain. Remark 2.1 implies that T is a domain.Let t, u ∈ T r { } and consider the intersection tT T uT . Let c ∈ C be a T -regular element. It is straightforward to show that ctT T cuT = c ( tT T uT ) ∼ = tT T uT . Thus, to show that tT T uT is principal, it suffices to show that ctT T cuT is principal. Hence, we assume without loss of generality that t, u ∈ C ⊆ R . Since R is a GCD domain, we have tR T uR = rR for some r ∈ R . The flatness of T over R implies that tT \ uT = tRT \ uRT = ( tR \ uR ) T = rRT = rT as desired.Next, assume that R is a generalized GCD ring, i.e., a finite conductor ring andlocally a GCD domain; see Remark 5.3. Consider a prime ideal Q ⊂ T , and set P := R T Q . Then T Q is a localization of the ring T P . Since R P is a GCD domain,the localized square ( (cid:3) P ) from Lemma 3.1 shows that T P is a GCD domain, bythe previous paragraph. It follows that T Q is a GCD domain as well. Also, since R is a finite conductor ring, Proposition (4.3) implies that T is a finite conductorring, so we conclude that T is a generalized GCD ring, again by Remark 5.3. (cid:3) Theorem 5.5.
Assume that the regular conductor square ( (cid:3) ) satisfies condition(FP) and every non-zero 2-generated ideal I of R satisfies condition (U1) fromDefinition 2.6. If C is contained in the Jacobson radical of R (e.g., if R is local),and A is a GCD domain, then R and T are GCD domains as well.Proof. By Proposition 2.9 and Remark 5.2, we assume that ( (cid:3) ) is non-trivial.Assume that C is contained in the Jacobson radical of R , and A is a GCDdomain. Remark 5.2 shows that A is a finite conductor ring. Thus, Theorem 4.4implies that R is a finite conductor ring. Also, by Proposition 2.8, we know that T is a domain, hence R is also a domain by Remark 2.1.Let I = ( r, s ) R such that r, s = 0. As in the proof of Theorem 4.4, we assumethat IT = T , hence I ⊇ IC = C by Lemma 2.10(a). In particular, we have IA ∼ = I/CI ∼ = A ⊗ R I . Moreover, we have I ( C ; otherwise, the ideal C = I wouldbe 2-generated over R , contradicting Theorem 3.2. Thus, we have IA = 0.According to Remark 5.2, we need to show that the kernel K of the naturalmap R → ( r, s ) R is cyclic. Since R is a finite conductor ring, we know that K is finitely generated. Thus, by Nakayama’s Lemma, to show that K is cyclic, itsuffices to show that K/CK is cyclic over A . Also, Lemma 2.10(b) implies thatTor R ( A, I ) = 0.Consider the exact sequence0 → K → R → I → R ( A, − ):0 → A ⊗ R K | {z } ∼ = K/CK → A → A ⊗ R I | {z } ∼ = IA → . (5.5.1) It follows that
K/CK is isomorphic to the kernel of the natural map A → IA . If η ( r ) = 0 or η ( s ) = 0, then it is straightforward to show that this kernel K/CK iscyclic. Otherwise, the fact that A is a GCD domain implies that K/CK is cyclic,again. We conclude that R is a GCD domain. Thus, T is a GCD domain as well,by Proposition 5.4. (cid:3) Theorem 5.6.
Assume that the regular conductor square ( (cid:3) ) satisfies condition(FP) and every non-zero 2-generated ideal I of R satisfies condition (U1) fromDefinition 2.6. If A is a generalized GCD ring, then so are R and T .Proof. Assume that A is a generalized GCD ring. Note that T is a B´ezout domain,hence a generalized GCD domain, by Proposition 2.8 and Remarks 5.2 and 5.3. Inparticular, Remark 2.1 implies that R is a domain.From Theorem 4.4, we conclude that R is a finite conductor ring. Thus, itremains to show that R is locally a GCD domain. Let P ⊂ R be a prime ideal. If R P ∼ = T P , e.g., if C P , then the fact that T is a generalized GCD ring impliesthat the localization T P ∼ = R P is a GCD domain.Assume for the rest of the proof that R P = T P , thus, C ⊆ P . Consider theregular pullback square ( (cid:3) P ) from the proof of Lemma 3.1. It is straightforwardto show that our assumptions on ( (cid:3) ) imply that T P is flat as an R P -module, thatthe conductor ideal C P is a principal ideal of T P , and that every 2-generated ideal I of R P is isomorphic to an ideal I ′ ⊆ R P such that I ′ T P = T P . Furthermore, theideal C P is contained in the Jacobson radical P P of A P , and the ring A P is a GCDdomain. Thus, Theorem 5.5 implies that R P is a GCD domain, as desired. (cid:3) Remark 5.7.
Comparing the previous two results, one might expect us to have aversion of Theorem 5.5 that does not assume that C is contained in the Jacobsonradical of R . However, Example 5.8 below shows that this fails, even in a very nicecase. Note that this comes from [2, Example 6.12], and that the ring R in thisexample is a generalized GCD domain, by Theorem 5.6, that is not a GCD domain. Example 5.8.
Set T := Q [ X ] and B := Q [ i ] with η : T → B the natural surjec-tion. Set A := Z [ i ] with ι : A → B the inclusion map. Note that the pullbackdetermined by this data is of the form ( ⊠ ) from Remark 2.7: R (cid:31) (cid:127) ι / / η (cid:15) (cid:15) (cid:15) (cid:15) Q [ X ] η (cid:15) (cid:15) (cid:15) (cid:15) Z [ i ] (cid:31) (cid:127) ι / / Q [ i ] . ( ⊠ )Moreover, it is straightforward to show that R = Z + Z X + ( X + 1) Q [ X ]; in otherwords, the elements of R are precisely the polynomials in Q [ X ] such that, when onedivides by X + 1, the division algorithm yields remainder bX + c ∈ Z [ X ]. Also,we have C = ( X + 1) Q [ X ].We claim that R is not a GCD domain. By [6, Theorem 4.2(b)], it suffices toshow that the map U ( Q [ X ]) → Q [ i ] ∗ /U ( Z [ i ]) induced by η is not surjective. (Onecan also show directly that the ideal I := ( X + 1) R T ( X + 1) R is not principal.However, it is shorter to use [6].)Since U ( Q [ X ]) = Q ∗ and U ( Z [ i ]) = {± ± i } , the induced map in question is thenatural one Q ∗ → Q [ i ] ∗ / {± ± i } . Using the norm N ( a + bi ) = a + b , one checks OHERENCE CONDITIONS IN FLAT REGULAR PULLBACKS 17 readily that the element ( i + i ) {± , ± i } is not in the image of this map; essentially,this boils down to the fact that √ / ∈ Q .We continue with more transfer results. Theorem 5.9.
Assume that the regular conductor square ( (cid:3) ) satisfies condition(FP) and every non-zero 2-generated ideal I of R satisfies condition (U2) fromDefinition 2.6. Assume also that A is a domain. If R is a GCD domain, then A isa GCD domain; the converse holds if C is contained in the Jacobson radical of R .Furthermore, these conditions imply that T is also a GCD domain.Proof. Again, it suffices to assume that R is a GCD domain and show that A is aGCD domain. Let J = ( a, b ) A with a, b = 0, and consider the natural map A → J .We need to show that the kernel K of this map is cyclic.As in our previous proofs, let f, g ∈ R be such that η ( f ) = a and η ( g ) = b . Set I = ( f, g ) R , and consider the natural map R → I . Since R is a GCD domain, thekernel L of this map is cyclic over R . Lemma 2.10(b) implies that Tor R ( A, I ) = 0.Thus, applying A ⊗ R − to the short exact sequence0 → L → R → I → → L/CL → A → I/CI → . It follows that the kernel of the induced map A → I/CI is isomorphic to
L/CL ,which is cyclic.Case 1: IT = T . Then Lemma 2.10(a) implies that CI = C , so the ideal I/C isequal to J . Thus, the previous paragraph shows that K ∼ = L/CL , which is cyclic.Case 2: I is isomorphic to an ideal I ′ such that I ′ T = T via a multiplication map I ′ h −→ ∼ = I for a fixed h ∈ R . In particular, I ′ is 2-generated, by elements f ′ , g ′ ∈ I ′ such that f = f ′ h and g = g ′ h . Set a ′ = η ( f ′ ) and b ′ = η ( g ) and x = η ( h ). As inthe proof of Theorem 4.7, we have a = xa ′ In particular, as a = 0, we have x = 0.Since the ring A is a domain, and x is non-zero, the map J ′ := ( a ′ , b ′ ) A x −→ ( a, b ) A =: J is injective; it is surjective by construction. Since I ′ T = T , Case 1implies that the kernel K ′ of the natural map A → J ′ is cyclic. Hence, the sameis true for K ∼ = K ′ . This completes Case 2.Now, we complete the proof. By assumption, there are elements r, s ∈ R andan ideal U ⊆ R such that U T = T = rT and I r −→ ∼ = rI = sU s ←− ∼ = U . Case 2 impliesthat the kernel of the natural map A → rIA = sU A is cyclic. As in the proofof Theorem 4.7, the map J = IA r −→ rIA is an isomorphism, so the kernel of thenatural map A → J is cyclic as well, as desired. (cid:3) Theorem 5.10.
Assume that the regular conductor square ( (cid:3) ) satisfies condition(FP) and every non-zero 2-generated ideal I of R satisfies condition (U2) fromDefinition 2.6. Assume further that A is locally a domain and has finite Krulldimension. Then R is a generalized GCD ring if and only if A is a generalizedGCD ring; these conditions imply that T is a generalized GCD ring.Proof. Again, we assume that R is a generalized GCD ring and show that A is ageneralized GCD ring. Theorem 4.7 shows that A is a finite conductor ring. Thus,it remains to let p be a prime ideal of A and show that A p is a GDC domain. Notethat A p is a domain by assumption. Set P := ν − ( p ) and consider the localized square ( (cid:3) P ) from Lemma 3.1. Byassumption, R P is a GDC domain, so Theorem 5.9 implies that the domain A p ∼ = A P is a GDC domain, as desired. (cid:3) Theorem 5.11.
Assume that the regular conductor square ( (cid:3) ) satisfies condition(FP) and every non-zero 2-generated ideal I of R satisfies condition (U2) fromDefinition 2.6. Assume further that A is a (possibly infinite) product Q λ ∈ Λ A λ ofdomains. If R is a generalized GCD ring, then each of the following rings is ageneralized GCD ring: T , A , and A λ . Conversely, if A is a generalized GCD ring,then each of the following rings is a generalized GCD ring: T , R , and A λ .Proof. Note that A is locally a domain. Indeed, since A is a product of domains, itis straightforward to show that every principal ideal of A is a summand of A , henceprojective, hence flat; now apply [10, Theorem 4.2.2].Now, argue as in the proof of Theorem 5.10, using Theorem 4.9 instead of The-orem 4.7. (cid:3) Corollary 5.12.
Consider the regular conductor square ( ⊠ ) from Remark 2.7.Then R is a generalized GCD ring if and only if each A i is so. Corollary 5.13.
Let D be a domain, and let E ⊆ D be a finite subset. Then thering of integer-valued polynomials Int(
E, D ) is a generalized GCD ring if and onlyif D is so. Quasi-coherent Rings
We conclude with an investigation of the following property.
Definition 6.1.
The ring R is quasi-coherent if every ideal of the form (0 : R a ) isfinitely generated, as is every intersection of finitely many principal ideals. Remark 6.2.
B´ezout domain = ⇒ coherent = ⇒ quasi-coherent = ⇒ finiteconductor ring, and GCD domain = ⇒ generalized GCD ring = ⇒ quasi-coherent.Also, if R is a finite product of domains, then every ideal of the form (0 : R a ) isprincipal, hence finitely generated Proposition 6.3.
Consider the regular conductor square ( (cid:3) ) . Assume that T isflat as an R -module. If R is quasi-coherent, then T is as well.Proof. Assume that R is quasi-coherent. As in the proof of Proposition 5.4, onechecks readily that the intersection of finitely many principal ideals of T is principal.Next, let t ∈ T , and consider the ideal (0 : T t ). Let c ∈ C be T -regular. Itis straightforward to show that (0 : T ct ) = (0 : T t ), so we assume without lossof generality that t ∈ R . Since R is quasi-coherent, the ideal (0 : R t ) is finitelygenerated over R . By flatness, the ideal (0 : T t ) = (0 : R t ) T is finitely generatedover T , as desired. (cid:3) Remark 6.4.
By [11, Proposition 2.4], a domain R is quasi-coherent if and only iffor each finitely generated ideal I the module Hom R ( I, R ) ∼ = ( R : Q ( R ) I ) is finitelygenerated. We generalize this next for finite products of domains. Lemma 6.5.
Let D , . . . , D r be domains, and set R = Q ri =1 D r . Then the followingconditions are equivalent. (i) The ring R is quasi-coherent. OHERENCE CONDITIONS IN FLAT REGULAR PULLBACKS 19 (ii)
Each ring D i is quasi-coherent. (iii) For each finitely generated ideal I of R , the module Hom R ( I, R ) is finitelygenerated over R .Proof. Recall that every ideal I of R decomposes uniquely as I = Q ri =1 I i witheach I i an ideal of D i . Moreover, the ideal I is finitely generated over R if andonly each I i is finitely generated over D i . Also, given D i -modules M i and N i , with M = Q ri =1 M i and N = Q ri =1 N i , there is a natural isomorphism Hom R ( M, N ) ∼ = Q ri =1 Hom D i ( M i , N i ) over R .(i) = ⇒ (ii) Assume that R is quasi-coherent, and let i ∈ { , . . . , r } be given. Let { I i,j } nj =1 be a finite set of principal ideals of D i . For all p ∈ { , . . . , r } r { i } and all j = 1 , . . . , n set I p,j = R p , and set I j = Q rp =1 I p,j which is a principal ideal of R .Since R is quasi-coherent, the intersection T nj =1 I j = Q rp =1 (cid:16)T nj =1 I p,j (cid:17) is finitelygenerated over R , and it follows that the factor T nj =1 I i,j is finitely generated over D i , as desired.(ii) = ⇒ (i) Assume that each domain D i is quasi-coherent, and let I , . . . , I n beprincipal ideals of R . Since each ideal I j is of the form I j = Q ri =1 I i,j where each I i,j is a principal ideal of D i , the fact that D i is quasi-coherent implies that eachintersection T nj =1 I i,j is finitely generated over D i , so the intersection T nj =1 I j = Q rp =1 (cid:16)T nj =1 I p,j (cid:17) is finitely generated over R .(ii) = ⇒ (iii) Assume that each domain D i is quasi-coherent, and let I = Q ri =1 I i be a finitely generated ideal of R . Then each I i is finitely generated over the quasi-coherent domain D i , so Remark 6.4 implies that Hom D i ( I i , D i ) is finitely generatedover D i . It follows that the finite product Q ri =1 Hom D i ( I i , D i ) ∼ = Hom R ( I, R ) isfinitely generated over R , as desired.(iii) = ⇒ (ii) Assume that for each finitely generated ideal I of R , the moduleHom R ( I, R ) is finitely generated over R . To show that D i is quasi-coherent, let I i be a finitely generated ideal of D i . For all p ∈ { , . . . , n } r { i } set I p = D p ,and set I = Q np =1 I p . This ideal is finitely generated over R , so the module Q rp =1 Hom D p ( I p , D p ) ∼ = Hom R ( I, R ) is finitely generated over R , by assumption. Itfollows that the factor Hom D i ( I i , D i ) is finitely generated over D i , and Remark 6.4implies that D i is quasi-coherent, as desired. (cid:3) Remark 6.6.
Notice that some implications in the previous result hold more gen-erally than we have stated. For instance, our proof readily shows that the impli-cation (i) = ⇒ (ii) holds for arbitrary products of rings that are not necessarilydomains, and the converse holds for finite products of rings that are not necessarilydomains. However, we are primarily interested in using the equivalence (i) ⇐⇒ (iii)which seems to need R to be a finite product of domains, as we have assumed inthe lemma. Theorem 6.7.
Assume that the regular conductor square ( (cid:3) ) satisfies condition(FP) and every non-zero finitely generated ideal I of R satisfies condition (U1) fromDefinition 2.6. Assume that A is a finite product of domains. If A is quasi-coherent,then R and T are quasi-coherent as well.Proof. As we have noted before, our assumptions imply that R and T are domainsin this setting. Assume that A is quasi-coherent. By Proposition 6.3, it sufficesto show that R is quasi-coherent. Let I be a finitely generated ideal of R . By Remark 6.4, we need to show that Hom R ( I, R ) is finitely generated. Since everynon-zero finitely generated ideal I of R satisfies condition (U1), we assume withoutloss of generality that IT = T .To show that Hom R ( I, R ) is finitely generated over R , it suffices by [10, Theo-rem 5.1.1(3)] to show that Hom R ( I, R ) / [ C · Hom R ( I, R )] is finitely generated over A and that T ⊗ R Hom R ( I, R ) is finitely generated over T . Since A is quasi-coherentand a finite product of domains, and the ideal IA is finitely generated, the A -moduleHom R ( I, R ) / [ C · Hom R ( I, R )] ∼ = Hom A ( IA, A )is finitely generated; see Lemma 2.11 for the isomorphism.Next, consider the exact sequence0 → Hom R ( I, C ) → Hom R ( I, R ) → Hom R ( I, A ) → R ( I, C ) ∼ = C ∼ = T from thislemma implies that this sequence has the following form:0 → T → Hom R ( I, R ) → Hom R ( I, A ) → . As T is flat over R , the preceding sequence yields the next exact sequence over T :0 → T ⊗ R T → T ⊗ R Hom R ( I, R ) → T ⊗ R Hom R ( I, A ) → . Since R → T is a flat epimorphism by Lemma 2.4, this exact sequence has thefollowing form:0 → T → T ⊗ R Hom R ( I, R ) → T ⊗ R Hom R ( I, A ) → . We have already seen that Hom R ( I, A ) ∼ = Hom A ( IA, A ) is finitely generated over A , hence over R ; the isomorphism is from Lemma 2.11. It follows that the module T ⊗ R Hom R ( I, A ) is finitely generated over T . Since T is also finitely generated over T , the preceding exact sequence shows that T ⊗ R Hom R ( I, R ) is finitely generatedover T , as desired. (cid:3) Theorem 6.8.
Assume that the regular conductor square ( (cid:3) ) satisfies condition(FP) and every non-zero finitely generated ideal I of R satisfies condition (U2)from Definition 2.6. Assume that A is a finite product of domains. Then R isquasi-coherent if and only if A is quasi-coherent; these conditions imply that T isalso quasi-coherent.Proof. Assume that R is quasi-coherent. Again, we only need to show that A isquasi-coherent. Let J = ( r , . . . , r n ) A be a non-zero finitely generated ideal of A ; here each r i is in R , and r i is the residue of r i in A . We need to show thatHom A ( J, A ) is finitely generated over A , by Lemma 6.5. We argue as in the proofsof Theorems 4.9 and 5.9. Consider the ideal I = ( r , . . . , r n ) R .Case 1: IT = T . In this case, Lemma 2.11 implies thatHom R ( I, R ) / [ C · Hom R ( I, R )] ∼ = Hom A ( IA, A ) = Hom A ( J, A ) . Since R is a quasi-coherent domain, the R -module Hom R ( I, R ) is finitely generatedover R . It follows that Hom A ( J, A ) is finitely generated over R , hence over A , asdesired. This concludes Case 1.Case 2: The general case. By condition (U2), there are elements r, s ∈ R andan ideal U ⊆ R such that U T = T = rT and I r −→ ∼ = rI = sU s ←− ∼ = U . As in the proofsof Theorems 4.7 and 5.9, these induce isomorphisms J r −→ ∼ = rJ = sU A s ←− ∼ = U A . By
OHERENCE CONDITIONS IN FLAT REGULAR PULLBACKS 21
Case 1, we know that Hom A ( U A, A ) is finitely generated, hence so is Hom A ( J, A ) ∼ =Hom A ( U A, A ). (cid:3) Corollary 6.9.
Consider the regular conductor square ( ⊠ ) from Remark 2.7. Then R is quasi-coherent if and only if each A i is so. Corollary 6.10.
Let D be a domain, and let E ⊆ D be a finite subset. Then thering of integer-valued polynomials Int(
E, D ) is quasi-coherent if and only if D is so. References
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