Prime Ideals in Infinite Products of Commutative Rings
aa r X i v : . [ m a t h . A C ] S e p Prime Ideals in Infinite Products ofCommutative Rings
CARMELO A. FINOCCHIARO , SOPHIE FRISCH and DANIEL WINDISCH A B S T R A C T
In this work we present descriptions of prime ideals and in particular of maximal idealsin products R = Q D λ of families ( D λ ) λ ∈ Λ of commutative rings. We show that everymaximal ideal is induced by an ultrafilter on the Boolean algebra Q P (max( D λ )). If every D λ is in a certain class of rings including finite character domains and one-dimensionaldomains, then this leads to a characterization of the maximal ideals of R . If every D λ is a Prüfer domain, we depict all prime ideals of R . Moreover, we give an example of a(optionally non-local or local) Prüfer domain such that every non-zero prime ideal is ofinfinite height. Let Λ be a set and ( D λ ) λ ∈ Λ a family of commutative rings. Throughout this work, wedenote by R = Q λ ∈ Λ D λ the product of the rings D λ and by B = Q λ ∈ Λ P (max( D λ )) theproduct of the Boolean algebras ( P (max( D λ )) , ∩ , ∪ ), where P ( M ) denotes the powerset of a set M and max( D ) is the set of all maximal ideals of a commutative ring D .Clearly, B is a Boolean algebra with least element 0 B = ( ∅ ) λ ∈ Λ . We denote elements a ∈ R by a = ( a λ ) = ( a λ ) λ ∈ Λ and elements Y ∈ B by Y = ( Y λ ) = ( Y λ ) λ ∈ Λ .In 1991, Levy, Loustaunau and Shapiro [13] showed that if every D λ is the ring of inte-gers Z , then the maximal ideals of R correspond to ultrafilters on B . Moreover, in thissituation, they gave a description of all prime ideals of R and investigated the orderstructure of chains inside spec( R ). O’Donnell [18] generalized some of these results tomaximal ideals in products of commutative rings and characterized certain classes ofprime ideals in products of Dedekind domains. These considerations have been car-ried on by Olberding, Saydam and Shapiro in [19], [20] and [21], where prime idealsin ultraproducts of commutative rings are explored in very broad settings. Our aimis to extend the initial approach by Levy, Loustaunau and Shapiro to more generalsituations, such as products of Prüfer domains. Ultrafilters on Boolean algebras . For an introduction to Boolean algebras, see [12].We tread ultrafilters in two different ways, that nevertheless can be summarized under C.A. Finocchiaro is supported by GNSAGA of Istituto Nazionale di Alta Matematica, FondazioneCariverona (Research project “Reducing complexity in algebra, logic, combinatorics - REDCOM”within the framework of the program “Ricerca Scientifica di Eccellenza 2018”), and by the Departmentof Mathematics and Computer Science of the University of Catania (Research program “Proprietàalgebriche locali e globali di anelli associati a curve e ipersuperfici” PTR 2016-18). S. Frisch is supported by the Austrian Science Fund (FWF): P 30934 D. Windisch is supported by the Austrian Science Fund (FWF): P 30934MSC: primary 13A15; secondary 13C13, 13F05 inocchiaro, Frisch, Windisch: Prime Ideals in Product Ringsone concept:(1) Let ( B, ∧ , ∨ ) be a Boolean algebra. We denote by 0 the minimal element of B , by ¬ the complement operation on B and by ≤ the canonical order relation on B .A non-empty subset U of B is called a filter in B if it satisfies the followingconditions:(i) 0 / ∈ U .(ii) For all X, Y ∈ U , it follows that X ∧ Y ∈ U .(iii) For all Y ∈ U and all Z ∈ B , we have that Y ≤ Z implies Z ∈ U .A filter U in B is called an ultrafilter in B if it satisfies in addition(iv) For all Y ∈ B , we have that either Y ∈ U or ¬ Y ∈ U .(2) If ( B, ∧ , ∨ ) = ( P (Λ) , ∩ , ∪ ), then we have 0 = ∅ and ¬ A = Λ \ A for every A ⊆ Λ,and ≤ equals set-theoretic inclusion. Moreover, we call an ultrafilter U in P (Λ)an ultrafilter on Λ (as it is usual) and the above properties translate as follows:(i) ∅ / ∈ U .(ii) For all A, B ∈ U , it follows that A ∩ B ∈ U .(iii) For all B ∈ U and all C ⊆ Λ, we have that B ⊆ C implies C ∈ U .(iv) For all A ⊆ Λ, we have that either A ∈ U or Λ \ A ∈ U .(3) If B = B = Q λ ∈ Λ P (max( D λ )), then for all Y, Z ∈ B , we have that Y ∧ Z =( Y λ ∩ Z λ ) λ ∈ Λ , Y ∨ Z = ( Y λ ∪ Z λ ) λ ∈ Λ , ¬ Y = (max( D λ ) \ Y λ ) λ ∈ Λ and 0 = 0 B =( ∅ ) λ ∈ Λ . Furthermore, we have Y ≤ Z if and only if Y λ ⊆ Z λ for all λ ∈ Λ.(4) A non-empty subset M ⊆ B is said to have the finite intersection property , iffor all Y , . . . , Y n ∈ M we have that Y ∧ . . . ∧ Y n = 0. If M ⊆ B has the finiteintersection property, then F = { F ∈ B | ∃ Y , . . . , Y n ∈ M Y ∧ . . . ∧ Y n ≤ F } can easily be seen to be a filter in B containing M . Moreover, it holds that everyfilter in B is contained in some ultrafilter in B . This follows from the fact thatultrafilters in B are exactly the maximal elements with respect to set-theoreticalinclusion in the set of all filters on B .(5) It is not hard to see that if U is an ultrafilter in B , then for all X, Y ∈ B , if X ∨ Y ∈ U , then X ∈ U or Y ∈ U .The above facts will be used throughout this work without any additional reference. The Skolem-property.
A subring T of R = Q D λ is said to have the Skolem-property if for all a (1) , . . . , a ( n ) ∈ T such that the ideal ( a (1) λ , . . . , a ( n ) λ ) is equal to D λ for all λ ∈ Λ, it follows that ( a (1) , . . . , a ( n ) ) = T . 2inocchiaro, Frisch, Windisch: Prime Ideals in Product RingsThe Skolem-property introduced here is a generalization of the particular case where T = Int( D ) = { f ∈ K [ x ] | f ( D ) ⊆ D } ⊆ Q D , where D is a domain with quotient field K . For a deeper insight into this circle of ideas, see [1], [2], [3], [4], [11], [14] and [15].In section 2, it is shown that T having the Skolem-property is equivalent to everymaximal ideal of T being induced by an ultrafilter in B . Note that R always has theSkolem-property. Moreover, the ultrafilters in B inducing maximal ideals of R can becharacterized if every D λ is in the class of commutative rings D satisfying the followingproperty, which we call (+):For all r ∈ D and a ∈ D \ { } there exists d ∈ D such that d is in every maximal idealcontaining a but not containing r and d is in no maximal ideal containing r .It is shown that finite character domains and one-dimensional domains satisfy (+).We also investigate the case where every ultrafilter on B induces a maximal ideal of R .It turns out that this property has strong connections to the topological assumption ofproconstructability on the maximal spectra of the component rings D λ .Further considerations in section 2 describe the minimal prime ideals of subrings T ⊆ R .In particular, we present a proof of the fact that every prime ideal of a product of do-mains R contains exactly one minimal prime ideal. First-order sentences and ultraproducts.
In section 3, we make use of some clas-sical terms of model theory including first-order formulas and ultraproducts, whichwe only consider in the special case of the language of rings including +, · , 0 and 1.Roughly speaking, a first order sentence in this language is a formula only using =, +, · , 0, 1, variables and logical symbols such as quantifiers and sentential connectives, butin such a way that variables only range over the elements of the ring.If F is an ultrafilter on Λ, we denote by R ∗ = Q F λ ∈ Λ D λ the ultraproduct of the D λ ,which is the ring that is constructed by identifying elements r, s ∈ R with the propertythat { λ ∈ Λ | r λ = s λ } is in F .Ultrafilters and ultraproducts are playing an increasingly important role in commuta-tive ring theory, for instance, in the work of Olberding (cf. [19], [20], [21]), Fontanaand Loper (cf. [7], [8], [10], [16, section 5], [17]), and Schoutens (cf. [22], [23]).We will extensively use the following fundamental theorem for ultraproducts [5, Theo-rem 4.1.9]: Theorem of Łoś.
A first order sentence ϕ is satisfied by R ∗ if and only if the set ofall λ ∈ Λ such that D λ satisfies ϕ is in F . Using the Theorem of Łoś, it follows in particular that, if every D λ is an integral do-main (respectively a field) with quotient field K λ , then so is R ∗ , and it can be easilyseen that its quotient field K ∗ is isomorphic to the ultraproduct of the K λ . For a moreprecise and general treatment of the introduced concepts, see [5].3inocchiaro, Frisch, Windisch: Prime Ideals in Product RingsIn section 3, we also apply the fact that being a Prüfer domain is preserved by ultra-products [19, Proposition 2.2]. Knowing this, we are able to describe the valuation onthe quotient field K ∗ of an ultraproduct R ∗ of Prüfer domains D λ having as a valuationring the localization R ∗ M at a maximal ideal M ⊆ R ∗ . By a common generalization ofconcepts introduced in [13] and [20], we are then able to describe all prime ideals in R when each D λ is a Prüfer domain. This leads us to the fact that (in the same situation)every non-minimal prime ideal of R contained in a certain type of maximal ideal (thatalways exists) is of infinite height. Finally, we give an example of a Prüfer domain suchthat every non-zero prime ideal is of infinite height, which can be chosen to be eitherlocal (so a valuation domain) or non-local. Describing all maximal ideals
Let D be a commutative ring. For an ideal I ⊆ D , we denote by V ( I ) the set of allmaximal ideals of D containing I and by D ( I ) = max( D ) \ V ( I ). If I = ( a , . . . , a n ) isfinitely generated, we write V ( I ) = V ( a , . . . , a n ).For an element a ∈ R = Q D λ , we set S ( a ) = ( V ( a λ )) λ ∈ Λ ∈ B = Q P (max( D λ )).Moreover, if U is a filter in B and T ⊆ R is a subring, we define( U ) T = { a ∈ T | S ( a ) ∈ U } , where T is omitted whenever the context determines it. Lemma 2.1.
Let T ⊆ R be a subring, a, b ∈ R and U be a filter in B . Then thefollowing assertions hold:(1) S ( a ) ∧ S ( b ) = ( V ( a λ , b λ )) λ ∈ Λ .(2) S ( a ) ∨ S ( b ) = S ( ab ).(3) ( U ) is an ideal of T .(4) If U is an ultrafilter in B , then ( U ) is a prime ideal of T . Proof. (1), (2) and (3) follow immediately from the relevant definitions.For the proof of (4), let U be an ultrafilter in B and note that 1 / ∈ ( U ), because S (1) = 0 B / ∈ U . If now a, b ∈ T such that ab ∈ ( U ), then by (2) we have that S ( a ) ∨ S ( b ) = S ( ab ) ∈ U . Since U is an ultrafilter, it follows that S ( a ) ∈ U or S ( b ) ∈ U and therefore a ∈ ( U ) or b ∈ ( U ). Proposition 2.2.
For a subring T ⊆ R = Q λ ∈ Λ D λ the following assertions are equiv-alent: 4inocchiaro, Frisch, Windisch: Prime Ideals in Product Rings(a) T has the Skolem-property.(b) For every proper ideal A ⊆ T the set { S ( a ) | a ∈ A } ⊆ B satisfies the finiteintersection property.(c) Every proper ideal of T is contained in an ideal of the form ( U ), where U is anultrafilter in B .(d) Every maximal ideal of T is of the form ( U ) for some ultrafilter U in B . Proof. "(a) ⇒ (b)": Let A ⊆ T be a proper ideal and a (1) , . . . , a ( n ) ∈ A . Then, byLemma 2.1(1), it follows that S ( a (1) ) ∧ . . . ∧ S ( a ( n ) ) = ( V ( a (1) λ , . . . , a ( n ) λ )). Assume tothe contrary that V ( a (1) λ , . . . , a ( n ) λ ) = ∅ for all λ ∈ Λ. Then ( a (1) λ , . . . , a ( n ) λ ) = D λ for all λ ∈ Λ and by the Skolem-property we have that A ⊇ ( a (1) , . . . , a ( n ) ) = T , which is acontradiction."(b) ⇒ (c)": Let A ⊆ T be a proper ideal. Then, by (b), we can pick an ultrafilter U in B such that { S ( a ) | a ∈ A } ⊆ U . Now it follows by definition that A ⊆ ( U )."(c) ⇒ (d)": This is clear."(d) ⇒ (a)": Let a (1) , . . . , a ( n ) ∈ T such that A = ( a (1) , . . . , a ( n ) ) is a proper ideal of T .Let U be an ultrafilter in B such that A ⊆ ( U ). We want to show that ( a (1) λ , . . . , a ( n ) λ )is proper for some λ ∈ Λ. Assume to contrary that ( a (1) λ , . . . , a ( n ) λ ) = D λ for all λ ∈ Λ.Then 0 B = ( V ( a (1) λ , . . . , a ( n ) λ )) = S ( a (1) ) ∧ . . . ∧ S ( a ( n ) ) ∈ U , which is a contradiction. Corollary 2.3.
Let ( D λ ) λ ∈ Λ be a family of commutative rings. Then every maximalideal of R = Q D λ is of the form ( U ) for some ultrafilter U in B . Characterizing ultrafilters that induce maximal ideals
Definition 2.4.
A ring D is said to satisfy property (+) if for all r ∈ D and for all non-zero a ∈ D , there exists d ∈ D such that d is in every maximal ideal of D containing a but not containing r and d is in no maximal ideal of D containing r .We will see that property (+) gives us a setting, where we can characterize theultrafilters in B that induce maximal ideals of R .We first give some easy equivalences to property (+), which will help us to give examplesof classes of domains satisfying and not satisfying it. To do this, we need the followingfact, which follows immediately from [6, Corollary 3]: If I is an ideal in a ring D and r ∈ D , then I ⊆ S Q ∈ V ( r ) Q implies that I ⊆ Q for some Q ∈ V ( r ). Lemma 2.5.
Let D be a ring. Then the following assertions are equivalent:(a) D satisfies property (+).(b) For all r ∈ D , a ∈ D \ { } and Q ∈ V ( r ) we have that ( T M ∈ V ( a ) \ V ( r ) M ) \ Q = ∅ .5inocchiaro, Frisch, Windisch: Prime Ideals in Product Rings(c) For all r ∈ D , a ∈ D \ { } and for all maximal ideals Q ⊆ D we have that T M ∈ V ( a ) \ V ( r ) M ⊆ Q implies that there exists some M ∈ V ( a ) \ V ( r ) such that M ⊆ Q .(d) For all r ∈ D , a ∈ D \ { } and for all maximal ideals Q ⊆ D we have that T M ∈ V ( a ) \ V ( r ) M ⊆ Q implies that Q ∈ D ( r ).(e) For all r ∈ D and for all non-zero a ∈ D , there exists d ∈ D such that thecontainment V ( a ) ∩ D ( r ) ⊆ V ( d ) ⊆ D ( r ) holds. Proof.
The equivalence of (b), (c) and (d) is clear. Also (b) follows immediately from(a). Moreover, (a) and (e) are trivially equivalent. It now suffices to prove "(b) ⇒ (a)".So assume that (a) does not hold. Then T M ∈ V ( a ) \ V ( r ) M ⊆ S Q ∈ V ( r ) Q for some r ∈ D and some non-zero a ∈ D . By the prime-avoidance-like statement before the lemma, itfollows that T M ∈ V ( a ) \ V ( r ) M ⊆ Q for some Q ∈ V ( r ), which contradicts (b). Example 2.6. If D is a domain of finite character, i.e. every a ∈ D \ { } is containedin only finitely many maximal ideals of D , then it is immediate by (c) in Lemma 2.5and the fact that every maximal ideal Q ⊆ D is prime that D satisfies (+).In particular, one-dimensional Noetherian domains (and therefore also principal idealdomains) satisfy (+).In the case that D does not have finite character, the situation is much more involved,as we want to illustrate by the next example. Nevertheless, Proposition 2.8 will enlargethe class of rings of which we know that they satisfy (+) into an important direction. Example 2.7. (1) If K is a field and n ≥
2, then the polynomial ring in n indeter-minates over K is a Noetherian factorial domain of Krull dimension n that is notPrüfer and does not satisfy property (+).(2) The polynomial ring Z [ x ] is a two-dimensional Noetherian factorial domain thatis not Prüfer and does not satisfy property (+).(3) The ring of integer-valued polynomials Int( Z ) is a two-dimensional non-NoetherianPrüfer domain not satisfying (+). Proposition 2.8.
Every one-dimension domain satisfies (+).
Proof.
Let D be one-dimensional, r ∈ D and a ∈ D \{ } . Note that Z = T M ∈ V ( a ) ∩ D ( r ) M is an intersection of prime ideals with a ∈ Z . Therefore Z is a non-zero radical ideal of D . If Z = D , the assertion is trivial, so assume that Z is a proper ideal, which impliesthat D/Z is a reduced zero-dimensional ring (i.e. von Neumann regular).Since ( r + Z ) is a principal ideal of D/Z , there exists e ∈ D such that e + Z is idempo-tent in D/Z and ( r + Z ) = ( e + Z ). We define d = 1 − e and claim that d is the rightchoice for property (+). 6inocchiaro, Frisch, Windisch: Prime Ideals in Product RingsLet M ∈ V ( a ) ∩ D ( r ). Then r / ∈ M and this implies r + Z / ∈ M/Z . (For if wehad r + Z ∈ M/Z , then we could pick m ∈ M such that r + Z = m + Z . Butthen r − m ∈ Z ⊆ M , which would imply r ∈ M , a contradiction.) It follows that e + Z / ∈ M/Z and therefore d + Z ∈ M/Z . With the same argument as before, we get d ∈ M , so M ∈ V ( d ).Now let M ∈ V ( d ). Then d ∈ M , so d + Z ∈ M/Z . Therefore e + Z / ∈ M/Z , whichimplies r + Z / ∈ M/Z and hence r / ∈ M . It follows that M ∈ D ( r ).Note that Proposition 2.8 gives also rise to examples of domains satisfying (+) andnot being of finite character. For instance, let ¯ Z be the integral closure of Z in somealgebraic closure of Q . Then ¯ Z is a one-dimensional Prüfer domain but it is not of finitecharacter. Indeed, every prime number p ∈ Z is contained in infinitely many maximalideals of ¯ Z .We now turn back to the investigation of maximal ideals of R = Q D λ and ultrafiltersin B = Q P (max( D λ )). Proposition 2.9.
Let ( D λ ) λ ∈ Λ be a family of rings satisfying (+) and let U be anultrafilter in B containing an element of the form ( V ( a λ )) λ ∈ Λ , where a λ ∈ D λ \ { } forall λ ∈ Λ. Then ( U ) is a maximal ideal of R = Q D λ . Proof.
Let r ∈ R \ ( U ) and let ( a λ ) λ ∈ Λ be a family such that a λ ∈ D λ \ { } for all λ ∈ Λand ( V ( a λ )) λ ∈ Λ ∈ U . Since every D λ satisfies (+), for each λ ∈ Λ we can pick some d λ ∈ D λ such that V ( a λ ) ∩ D ( r λ ) ⊆ V ( d λ ) ⊆ D ( r λ ) and define d = ( d λ ) λ ∈ Λ . Since r / ∈ ( U ), it follows that S ( r ) / ∈ U and therefore ( D ( r λ )) = ¬ S ( r ) ∈ U , because U is anultrafilter. Hence we have S ( d ) ≥ S ( a ) ∧ ( D ( r λ )) ∈ U , which implies that S ( d ) ∈ U and therefore d ∈ ( U ). On the other hand, we have ( d λ , r λ ) = D λ for all λ ∈ Λ. By theSkolem-property of R it follows that ( d, r ) = R and therefore ( U ) is maximal.We now introduce two new kinds of ideals. The first one will also be the prototypeof minimal prime ideals in subrings T ⊆ R . Let F be an ultrafilter on Λ and T ⊆ R bea subring. Then for an element x ∈ T we set z ( x ) = { λ ∈ Λ | x λ = 0 } and we define(0) T F = { x ∈ T | z ( x ) ∈ F } . Moreover, for a family M = ( M λ ) λ ∈ Λ , where M λ ∈ max( D λ ) for every λ ∈ Λ, we set z M ( x ) = { λ ∈ Λ | x λ ∈ M λ } for an element x ∈ T and define M T F = { x ∈ T | z M ( x ) ∈ F } . We write (0) T F = (0) F and M T F = M F if the choice of T is clear from the context. Lemma 2.10. If T ⊆ R is a subring such that there exists c ∈ T where c λ ∈ D λ is a non-zero non-unit for every λ ∈ Λ and F is an ultrafilter on Λ, then (0) F is anon-maximal ideal of T . 7inocchiaro, Frisch, Windisch: Prime Ideals in Product Rings Proof.
It can be easily seen that (0) F is an ideal of T . Now let c ∈ T as in theassumption of the lemma and let M = ( M λ ) be a family such that each M λ is amaximal ideal of D λ containing c λ . Clearly, M F ⊆ T is a proper ideal with (0) F ⊆ M F and c ∈ M F \ (0) F . Therefore (0) F is not maximal. Proposition 2.11.
Let T ⊆ R = Q λ ∈ Λ D λ be a subring with the property that thereexists c ∈ T such that c λ ∈ D λ is a non-zero non-unit for every λ ∈ Λ, where every D λ is an integral domain. Moreover, let U be an ultrafilter in B such that ( U ) ⊆ T is amaximal ideal. Then U contains an element of the form ( V ( a λ )) λ ∈ Λ , where a λ ∈ D λ \{ } for all λ ∈ Λ. Proof.
First, note that { z ( x ) | x ∈ ( U ) } does not have the finite intersection property.For otherwise there would exist an ultrafilter F on Λ such that ( U ) ⊆ (0) F , whichwould imply that (0) F is maximal. A contradiction to Lemma 2.10.So we can pick x (1) , . . . , x ( n ) ∈ ( U ) such that z ( x (1) ) ∩ . . . ∩ z ( x ( n ) ) = ∅ . Therefore for all λ ∈ Λ we can choose i λ ∈ { , . . . , n } such that x ( i λ ) λ = 0 and therefore a λ := c λ · x ( i λ ) λ isa non-zero non-unit of D λ . If we now set a = ( a λ ) λ ∈ Λ , then ( V ( a λ )) = ( V ( c λ · x ( i λ ) λ )) ≥ ( V ( c λ · x (1) λ , . . . , c λ · x ( n ) λ )) = S ( c · x (1) ) ∧ . . . ∧ S ( c · x ( n ) ) ∈ U . Therefore ( V ( a λ )) ∈ U ,which we wanted to show. Corollary 2.12.
Let ( D λ ) λ ∈ Λ be a family of domains not being fields satisfying property(+) and let R = Q λ ∈ Λ D λ . Then the maximal ideals of R are exactly the ideals of theform ( U ) , where U is an ultrafilter in the Boolean algebra B = Q λ ∈ Λ P (max( D λ )) containing an element of the form ( V ( a λ )) λ ∈ Λ such that a λ ∈ D λ \ { } for all λ ∈ Λ . The finite character case
If every D λ is a domain of finite character (i.e. every non-zero element is only containedin finitely many maximal ideals) and T ⊆ R is a subring such that there exists some c ∈ T such that every c λ ∈ D λ is a non-zero non-unit, then it follows immediately fromProposition 2.11 that if ( U ) ⊆ T is a maximal ideal, then the ultrafilter U must containan element Y = ( Y λ ) such that every Y λ is finite.The next result gives us a statement analogous to Proposition 2.9 in the finite charactercase. Proposition 2.13.
Let ( D λ ) λ ∈ Λ be a family of rings such that for every λ ∈ Λ and forevery r λ ∈ D λ we have that r λ is contained either in all maximal ideals of D λ or in onlyfinitely many of them. Let U be an ultrafilter in B containing an element Y = ( Y λ ) λ ∈ Λ such that Y λ is finite for every λ ∈ Λ. Then ( U ) is a maximal ideal of R = Q λ ∈ Λ D λ . Proof.
Let r ∈ R \ ( U ). Define a = ( a λ ) such that(1) a λ ∈ P for all P ∈ D ( r λ ) ∩ Y λ and 8inocchiaro, Frisch, Windisch: Prime Ideals in Product Rings(2) a λ / ∈ Q for all Q ∈ V ( r λ ).If, for λ ∈ Λ, we have that V ( r λ ) is finite, then this is possible by the Chinese RemainderTheorem. If V ( r λ ) = max( D λ ), then this works by setting a λ = 1. By (2) and theSkolem-property of R , it follows that ( a, r ) = R .To see that a ∈ ( U ), note that S ( r ) / ∈ U and therefore ( D ( r λ )) = ¬ S ( r ) ∈ U . Therefore S ( a ) ≥ ( D ( r λ )) ∧ Y ∈ U and hence a ∈ ( U ). Corollary 2.14.
Let ( D λ ) λ ∈ Λ be a family of domains of finite character not being fieldsand let R = Q λ ∈ Λ D λ . Then the maximal ideals of R are exactly the ideals of the form ( U ) , where U is an ultrafilter in the Boolean algebra B = Q λ ∈ Λ P (max( D λ )) containingan element Y = ( Y λ ) λ ∈ Λ such that Y λ is finite for all λ ∈ Λ . Proconstructability of the maximal spectra
We now want to investigate the connection between a certain topological property ofthe max( D λ ) called proconstructability and the situation that for every ultrafilter U in B the ideal ( U ) ⊆ R = Q D λ is maximal.If D is a commutative ring, then the constructible topology on spec( D ) is a topologyfiner than the Zariski topology on spec( D ) making it a compact Haussdorf space andpreserving certain important properties. The easiest way to describe the closed setsin the constructible topology (which are called proconstructible ) uses the fact thatit is equal to the so-called ultrafilter topology on spec( D ): A subset X ⊆ spec( D )is proconstructible if and only if for each ultrafilter F on X the prime ideal X F = { r ∈ D | V ( r ) ∩ X ∈ F } of D is in X , where V ( r ) = { P ∈ spec( D ) | r ∈ P } . Ifwe consider the subspace X = max( D ) of spec( D ), then this property translates asfollows: X = max( D ) is proconstructible if and only if X F = { r ∈ D | V ( r ) ∈ F } ismaximal for each ultrafilter F on max( D ). Proposition 2.15.
If ( U ) is a maximal ideal of R = Q λ ∈ Λ D λ for every ultrafilter U in B , then max( D λ ) is proconstructible in spec( D λ ) for every λ ∈ Λ. Proof.
Fix λ ∈ Λ and set X = max( D λ ). As noted before the proposition, it suffices toshow that X F = { r ∈ D λ | V ( r ) ∈ F } is in X for every ultrafilter F on X . So let F bean ultrafilter on X . For every r ∈ D λ consider the element Y ( r ) ∈ B defined by setting Y ( r ) µ = D ( r ) if µ = λY ( r ) µ = ∅ if µ = λ for µ ∈ Λ.Now consider the subset G = { Y ( r ) | r ∈ D λ \ X F } of B . Since F is an ultrafilter on X = max( D λ ) and V ( r ) / ∈ F for every r ∈ D λ \ X F , it follows that for all r , . . . , r n ∈ D λ \ X F we have that D ( r ) ∩ . . . ∩ D ( r n ) ∈ F . Hence G has the finite intersection9inocchiaro, Frisch, Windisch: Prime Ideals in Product Ringsproperty as a subset of the Boolean algebra B . Let U be an ultrafilter in B such that G ⊆ U .By assumption ( U ) ⊆ R is a maximal ideal and it can easily be seen that it containsthe kernel of the projection map p : R → D λ . Indeed, if r ∈ R such that r λ = 0, then S ( r ) ≥ Y (1) ∈ U . It follows that p (( U )) ⊆ D λ is a maximal ideal.Now we claim that D λ \ X F ⊆ D λ \ p (( U )). If we know this, it follows that p (( U )) ⊆ X F and therefore X F is maximal, which is what we wanted to show.To prove the claim, assume to the contrary that there exists α ∈ D λ \ X F such that α = p ( f ) for some f ∈ ( U ), i.e. α = f λ . Since S ( f ) and Y ( α ) are in U , it follows that0 B = S ( f ) ∧ Y ( α ) ∈ U , which is a contradiction. Definition 2.16.
A commutative ring D is said to satisfy property (++), if for all r ∈ D there exists some d ∈ D such that D ( r ) = V ( d ).Note that if a ring D satisfies (++), then it also satisfies (+). Indeed, given r ∈ D and a ∈ D \ { } , let d ∈ D such that D ( r ) = V ( d ). Then V ( a ) ∩ D ( r ) ⊆ D ( r ) = V ( d ) ⊆ D ( r ). So D satisfies (+) by Lemma 2.5.Before we will see examples of rings with property (++), we want to illustrate how wecan apply it to our description of maximal ideals of the product ring R . Lemma 2.17.
If ( D λ ) λ ∈ Λ is a family of commutative rings satisfying (++), then ( U )is a maximal ideal of R = Q λ ∈ Λ D λ for every ultrafilter U in B . Proof.
Let U be an ultrafilter in B and choose r ∈ R \ ( U ). Using property (++), let d ∈ R such that D ( r λ ) = V ( d λ ) for every λ ∈ Λ. Then, by the Skolem-property of R ,we have that ( r, d ) = R . Moreover, since S ( r ) / ∈ U , we have that S ( d ) = ( V ( d λ )) =( D ( r λ )) = ¬ S ( r ) ∈ U , hence d ∈ ( U ). This shows that ( U ) is maximal.For a subset X ⊆ spec( D ), where D is a commutative ring, we denote by Cl zar ( X )the closure of X with respect to the Zariski topology, by Cl cons ( X ) the closure of X with respect to the constructible topology and by X sp = { P ∈ spec( D ) | P ⊇ Q for some Q ∈ X } the specialization of X .It is shown in [9, Lemma 1.1] that Cl zar ( X ) = ( Cl cons ( X )) sp for every X ⊆ spec( D ).From this it follow easily that max( D ) is proconstructible in spec( D ) if and only if itis closed with respect to the Zariski topology on spec( D ). Proposition 2.18.
Let D be a commutative ring such that max( D ) is proconstructiblein spec( D ). Then D satisfies property (++).10inocchiaro, Frisch, Windisch: Prime Ideals in Product Rings Proof.
Let J denote the Jacobson radical of D . Since max( D ) is proconstructible, itfollows by the remarks before the proposition that max( D ) is closed with respect tothe Zariski topology. In this case we have that { P ∈ spec( D ) | J ⊆ P } = max( D ) andtherefore D ′ := D/J is a zero-dimensional reduced ring.Let r ∈ D . Since D ′ is zero-dimensional reduced, it follows that there exists some e ∈ D such that e + J ∈ D ′ is idempotent and the principal ideals ( r + J ) D ′ and ( e + J ) D ′ coincide. Let d := 1 − e . Then it can be easily seen that D ( r + J ) = D ( e + J ) = V ( d + J ).From this it is clear that D ( r ) = V ( d ).Note that, if D is zero-dimensional, then max( D ) = spec( D ) is proconstructible.Also, if D is a one-dimensional domain with non-zero Jacobson radical J , then max( D ) = V ( J ) is proconstructible. Hence both zero-dimensional rings and one-dimensional do-mains with non-zero Jacobson radical satisfy (++).The next result is now an immediate consequence of Proposition 2.15, Lemma 2.17 andProposition 2.18. Corollary 2.19.
Let ( D λ ) λ ∈ Λ be a family of commutative rings and R = Q λ ∈ Λ D λ .Then the following assertions are equivalent:(a) ( U ) is a maximal ideal of R for every ultrafilter U in the Boolean algebra B = Q λ ∈ Λ P (max( D λ )) .(b) The subspace max( D λ ) is proconstructible in spec ( D λ ) for every λ ∈ Λ .(c) D λ satisfies property (++) for every λ ∈ Λ , i.e. for every r ∈ D λ there exists d ∈ D λ such that D ( r ) = V ( d ) . In the particular case where | Λ | = 1, we get the following statement: Corollary 2.20.
Let D be a commutative ring. Then max( D ) is proconstructible inspec ( D ) if and only if D satisfies property (++), i.e. for every r ∈ D there exists d ∈ D such that D ( r ) = V ( d ) . Minimal prime ideals
For the rest of this section, let ( D λ ) λ ∈ Λ be a family of integral domains and R = Q D λ .Recall that for an element x ∈ R we set z ( x ) = { λ ∈ Λ | x λ = 0 } and define n ( x ) =Λ \ z ( x ). Moreover, recall the definition of the proper ideal (0) T F = { x ∈ T | z ( x ) ∈ F } of a subring T ⊆ R for an ultrafilter F on Λ. Proposition 2.21.
Let F be an ultrafilter on Λ and T ⊆ R = Q λ ∈ Λ D λ be a subring.(1) The ultraproduct R ∗ = Q F λ ∈ Λ D λ is isomorphic to R/ (0) R F .(2) (0) T F is a prime ideal of T . 11inocchiaro, Frisch, Windisch: Prime Ideals in Product Rings(3) Every minimal prime ideal of T is of the form (0) T F for some ultrafilter F on T . Proof.
To prove (1), note that ϕ : R → R ∗ mapping an element r ∈ R to its equivalenceclass r ∗ ∈ R ∗ is a surjective homomorphism. Its kernel can be easily seen to coincidewith (0) R F .Now, to prove (2), consider the map ι : T / (0) T F → R/ (0) R F with ι ( x + (0) T F ) := x + (0) R F .It clearly is an injective homomorphism. Moreover, by (1) and the Theorem of Łoś, R/ (0) R F is an integral domain, hence so is T / (0) T F . It follows that (0) T F is a prime idealof T .Finally, for the proof of (3), let P ⊆ T be a minimal prime ideal and let M = { n ( x ) | x ∈ T \ P } . We claim that M has the finite intersection property. Assume to thecontrary that there are x , . . . , x n ∈ T \ P such that n ( x ) ∩ . . . ∩ n ( x n ) = ∅ . Then x · . . . · x n = 0 ∈ P and therefore there exists some i ∈ { , . . . , n } such that x i ∈ P ,which is a contradiction. Let F be an ultrafilter on Λ such that M ⊆ F . Clearly, T \ P ⊆ T \ (0) T F and therefore (0) T F ⊆ P . By the minimality of P it follows that P = (0) T F .In the next lemma, we have to restrict our scope to subrings T ⊆ R such that forevery Z ⊆ Λ there exists some x ∈ T such that Z = z ( x ). Note, that there are examplesof such rings T not being equal to a product of commutative rings, e.g. let T be thering generated (in R ) by all elements x ∈ R such that x λ ∈ { , } for all λ ∈ Λ. Lemma 2.22.
Let T ⊆ R = Q D λ be a subring such that for every Z ⊆ Λ thereexists some x ∈ T such that Z = z ( x ). Then (0) T F is a minimal prime ideal of T forevery ultrafilter F on Λ. Moreover, if F and G are two different ultrafilters on Λ, then(0) T F = (0) T G . Proof.
Let F be an ultrafilter on Λ and P ⊆ T be a prime ideal such that P ⊆ (0) T F .Let x ∈ (0) T F and let y ∈ T such that z ( y ) = Λ \ z ( x ). Then x · y = 0 ∈ P . Since P is prime, either x ∈ P or y ∈ P . But y cannot be an element of P ⊆ (0) T F , becauseotherwise x + y ∈ (0) T F and therefore ∅ = z ( x + y ) ∈ F , which is a contradiction. Henceit must hold that x ∈ P .Now, let G be an ultrafilter on Λ different from F . Let Z ∈ G \ F and x ∈ T such that z ( x ) = Z . Then x ∈ (0) T G \ (0) T F . Corollary 2.23.
Let ( D λ ) λ ∈ Λ be a family of commutative integral domains and let T ⊆ R = Q λ ∈ Λ D λ be a subring such that for every Z ⊆ Λ there exists some x ∈ T suchthat Z = { λ ∈ Λ | x λ = 0 } . Then the map F 7→ (0) T F is a bijection between ultrafilterson Λ and minimal prime ideals of T . Let U be an ultrafilter in B and for every Y ∈ U set F Y = { λ ∈ Λ | Y λ = ∅} . Thenwe can define the collection F U = { F Y | Y ∈ U } F U is an ultrafilter on Λ. Proposition 2.24.
Let U be an ultrafilter in B and F be an ultrafilter on Λ. Thenthe containment (0) F ⊆ ( U ) of ideals of R = Q λ ∈ Λ D λ holds if and only if F = F U .In particular, every prime ideal of R contains a unique minimal prime ideal. Proof.
Assume that (0) F ⊆ ( U ) and let F ∈ F . For a subset M ⊆ Λ we denote by χ M the element of R for which the entry at λ ∈ Λ is 1 if λ ∈ M and is 0 if λ / ∈ M . If weset M = Λ \ F , then χ M ∈ (0) F ⊆ ( U ). Therefore Y := S ( χ M ) ∈ U , where Y λ = ∅ if λ / ∈ F and Y λ = max( D λ ) if λ ∈ F . So F = { λ ∈ Λ | Y λ = ∅} and therefore F ∈ F U .Whence F ⊆ F U , which implies F = F U , because F is an ultrafilter.Conversely, let F = F U and let r ∈ (0) F . Then for M = { λ ∈ Λ | r λ = 0 } we have that χ M ∈ (0) F . Therefore Λ \ M ∈ F , which implies that Λ \ M = { λ ∈ Λ | Y λ = ∅} for some Y ∈ U . Clearly, we have S ( χ M ) ≥ Y , so S ( χ M ) ∈ U . Consequently, r = rχ M ∈ ( U ).For the last statement, let P ⊆ R be a prime ideal. Then P contains a minimal primeideal. If Q ⊆ P is a minimal prime ideal, then by Proposition 2.2 there exists anultrafilter F on Λ such that Q = (0) F . In the same way, if M is a maximal idealcontaining P , then by Proposition 2.21(3) we can pick some ultrafilter U on B suchthat M = ( U ). Since (0) F ⊆ P ⊆ ( U ), it follows by the considerations before that F = F U , so (0) F = (0) F U , which therefore is the unique minimal prime ideal containedin P .Proposition 2.24 gives a better understanding of the order structure of the set ofprime ideals in the product R of integral domains in the sense that spec( R ) is a disjointunion of partially ordered sets O , where each O has a unique minimal element. This isalso a starting point for our considerations in the next section. By Proposition 2.24, if we want to characterize all prime ideals of R = Q λ ∈ Λ D λ itis sufficient to describe for every ultrafilter U on B the prime ideals P ⊆ R with(0) F U ⊆ P ⊆ ( U ). So from now on, fix an ultrafilter U in the Boolean algebra B = Q λ ∈ Λ P (max( D λ )) and let F = F U be the corresponding ultrafilter on Λ.Let R ∗ = Q F λ ∈ Λ D λ be the ultraproduct of the D λ with respect to F . We have seenin Proposition 2.21 that R ∗ is isomorphic to R/ (0) F . Let moreover R ∗U denote thelocalization of the integral domain R ∗ at the maximal ideal ( U ) ∗ of R ∗ correspondingto ( U ). Then the prime ideals P ⊆ R with (0) F ⊆ P ⊆ ( U ) are in inclusion preservingone-to-one correspondence with the prime ideals of R ∗U .For r ∈ R , we denote by r ∗ the image of r in R ∗ or the image of r in R ∗U , dependingon in which ring we are working in at the moment.In the following, we want to characterize all prime ideals of R , where every D λ is aPrüfer domain. In [13] this was done for the special case where each D λ is the ring13inocchiaro, Frisch, Windisch: Prime Ideals in Product Ringsof integers. In [20] and [21], special types and chains of prime ideals in ultraproductsof certain commutative rings are described, including for instance all prime ideals inultraproducts of Dedekind domains. Our investigation of prime ideals in products ofgeneral Prüfer domains is new and is different from the one in [21] in the special caseof Dedekind domains. Therefore it also gives a new viewpoint in this situation.From now on, let D λ be a Prüfer domain for every λ ∈ Λ.It is shown in [19, Proposition 2.2] that "Prüfer domain" is preserved by ultraproducts.Therefore R ∗ is a Prüfer domain and R ∗U is a valuation domain. Let K ∗ be the quotientfield of R ∗ and note that it is isomorphic to the ultraproduct with respect to F ofthe quotient fields K λ of D λ . Moreover we extend the notation V ( r λ ) and D ( r λ ) toelements k λ ∈ K λ , namely set V ( k λ ) = { M ∈ max( D λ ) | v M ( k λ ) > } , where v M isthe valuation on K λ corresponding to M , and let D ( k λ ) = max( D λ ) \ V ( k λ ).In the following proposition we are able to partially describe the valuation v on K ∗ that has R ∗U as its valuation ring. Valuations and prime ideals
Proposition 3.1.
Let v be the valuation on K ∗ having R ∗U as valuation ring. Thenfor a, b ∈ R , the following assertions are equivalent:(1) v ( a ∗ ) ≥ v ( b ∗ ).(2) There exists Y ∈ U such that for all λ ∈ Λ and for all P ∈ Y λ it holds that v P ( a λ ) ≥ v P ( b λ ). Proof.
If either a ∗ or b ∗ is equal to 0, then the statement is trivial. So let a ∗ = 0 = b ∗ .Since both (1) and (2) only depend on entries a λ and b λ of a and b for λ in an ultrafilterset of F (and a , b are both non-zero on such a set), we can assume without loss ofgenerality that a λ = 0 = b λ for all λ ∈ Λ.Now assume that (2) holds and let Y ∈ U such that for all λ ∈ Λ and for all P ∈ Y λ we have v P ( a λ ) ≥ v P ( b λ ). Assume to the contrary that v ( a ∗ ) < v ( b ∗ ). Then 0 < v ( b ∗ a ∗ )and therefore b ∗ a ∗ ∈ ( U ) ∗ ⊆ R ∗U . Since localization commutes with forming the quotientmodulo some ideal, we can write b ∗ a ∗ = ( b λ a λ ) ∗ λ . We set Z = ( V ( b λ a λ )) λ . Claim: Z ∈ U .If the claim holds, we know that Y ∧ Z ∈ U , so in particular Y ∧ Z = 0 B . So wecan pick λ ∈ Λ such that Y λ ∩ Z λ = ∅ . Let P ∈ Y λ ∩ Z λ . Since P ∈ Y λ , we have v P ( a λ ) ≥ v P ( b λ ). On the other hand, since P ∈ Z λ = V ( b λ a λ ), we have v P ( b λ a λ ) > v P ( b λ ) > v P ( a λ ). This is a contradiction.To prove the claim, first note that, since b ∗ a ∗ ∈ ( U ) ∗ , we can pick d ∈ ( U ) and c ∈ R \ ( U )such that b ∗ a ∗ = d ∗ c ∗ . By the same argument as in the beginning of the proof, we canchoose c such that c λ = 0 for all λ ∈ Λ and we are then able to write d ∗ c ∗ = ( d λ c λ ) ∗ λ . Since b ∗ a ∗ = d ∗ c ∗ , it follows that ba and dc coincide on a set of F and again, since our considerations14inocchiaro, Frisch, Windisch: Prime Ideals in Product Ringsonly are influenced by entries for λ in some element of F , we may assume that ba = dc .Therefore it follows that Z = ( V ( d λ c λ )) λ = ( { M ∈ max( D λ ) | v M ( d λ c λ ) > } ) λ ≥ ( { M ∈ max( D λ ) | d λ ∈ M ∧ c λ / ∈ M } ) λ = ( V ( d λ )) λ ∧ ( D ( c λ )) λ ∈ U by the choice of c and d .So Z ∈ U and the proof for the implication from (2) to (1) is complete.Now assume that (1) holds and assume to the contrary that for all Y ∈ U there existssome λ ∈ Λ and some P ∈ Y λ such that v P ( a λ ) < v P ( b λ ). By (1), we have that v ( a ∗ b ∗ ) ≥ a ∗ b ∗ ∈ R ∗U . Similar to the proof of the other direction, we can now pick c ∈ R and d ∈ R \ ( U ) such that d λ = 0 for all λ ∈ Λ and ( c λ d λ ) ∗ λ = c ∗ d ∗ = a ∗ b ∗ = ( a λ b λ ) ∗ λ . Asbefore, it is no restriction of generality to assume that a λ b λ = c λ d λ for all λ ∈ Λ. It followsby the choice of d that Y := ( D ( d λ )) λ ∈ U . So by assumption, we can pick λ ∈ Λand P ∈ Y λ such that v P ( a λ ) < v P ( b λ ). Whence 0 ≤ v P ( c λ d λ ) = v P ( a λ b λ ) <
0, which is acontradiction.For every λ ∈ Λ and P ∈ max( D λ ), we denote by S P the totally ordered submonoidof non-negative elements (including ∞ ) of the value group associated to the valuation v P of P . We define S = Q λ ∈ Λ Q P ∈ max( D λ ) S P to be the product of all these monoids.We write elements g ∈ S as g = ( g λ,P ) λ ∈ Λ ,P ∈ max( D λ ) . For g ∈ S , define( U ) g = { x ∈ R | ∃ Y ∈ U ∃ n ∈ N ∀ λ ∈ Λ ∀ P ∈ Y λ v P ( x nλ ) ≥ g λ,P } . It will turn out that the sets ( U ) g are prime ideals of R and that they can be used todescribe all prime ideals of R contained in ( U ) and containing (0) F . Proposition 3.2.
For any g ∈ S with g λ,P > λ ∈ Λ and P ∈ max( D λ ), wehave that ( U ) g is a prime ideal of R contained in ( U ). Proof.
Clearly ( U ) g is an ideal. To see that it is contained in ( U ), let x ∈ ( U ) g andchoose Y ∈ U and n ∈ N such that for all λ ∈ Λ and for all P ∈ Y λ we have v P ( x nλ ) ≥ g λ,P >
0. It follows that S ( x n ) ≥ Y ∈ U , so x n ∈ ( U ), which is a prime ideal. Therefore x ∈ ( U ).Finally, let a, b ∈ R such that ab ∈ ( U ) g and let Y ∈ U and n ∈ N such that for all λ ∈ Λ and all P ∈ Y λ we have v P ( a nλ b nλ ) ≥ g λ,P . Given λ ∈ Λ and P ∈ Y λ , it followsthat g λ,P + g λ,P ≤ v P ( a nλ b nλ ) + v P ( a nλ b nλ ) = v P ( a nλ b nλ ) = v P ( a nλ ) + v P ( b nλ ). Hence v P ( a nλ ) ≥ g λ,P or v P ( b nλ ) ≥ g λ,P . If we define Y a = ( { P ∈ max( D λ ) | v P ( a n ) ≥ g λ,P } ) λ and Y b = ( { P ∈ max( D λ ) | v P ( b n ) ≥ g λ,P } ) λ , then it follows that Y a ∨ Y b ≥ Y ,which implies that Y a ∨ Y b ∈ U . Since U is an ultrafilter, it follows that Y a ∈ U or Y b ∈ U . Say Y a ∈ U . Then there exist Y ′ ∈ U (namely Y ′ = Y a ) and n ′ ∈ N (namely n ′ = 2 n ) such that for all λ ∈ Λ and for all P ∈ Y ′ λ it holds that v P ( a n ′ λ ) ≥ g λ,P , whichmeans per definition that a ∈ ( U ) g . Proposition 3.3.
Let x ∈ ( U ) and g λ,P = v P ( x λ ) if P ∈ V ( x λ ) and g λ,P = ∞ otherwise. Then ( U ) g is the smallest prime ideal P with (0) F ⊆ P ⊆ ( U ) containing x .15inocchiaro, Frisch, Windisch: Prime Ideals in Product Rings Proof.
We already know that ( U ) g ⊆ ( U ). To see that x ∈ ( U ) g , set Y = S ( x ) ∈ U , n = 1. Then for all λ ∈ Λ and for all P ∈ Y λ , we have by definition that v P ( x nλ ) = v P ( x λ ) = g λ,P . So x ∈ ( U ) g .Now let (0) F ⊆ P ⊆ ( U ) be a prime ideal containing x . Since the prime ideals of R containing (0) F and being contained in ( U ) are in inclusion preserving bijectionwith the prime ideals of R ∗U , it suffices to prove the inclusion (( U ) g ) ∗ ⊆ P ∗ of thecorresponding prime ideals in R ∗U . So let r ∈ ( U ) g . We show that r ∗ ∈ P ∗ . Let Y ∈ U and n ∈ N such that for all λ ∈ Λ and P ∈ Y λ we have v P ( r nλ ) ≥ g λ,P andwithout loss of generality choose Y ≤ S ( x ), so that for all λ ∈ Λ and P ∈ Y λ we have v P ( r nλ ) ≥ g λ,P = v P ( x λ ) (this is possible, since S ( x ) ∈ U and therefore Y ∧ S ( x ) ≤ S ( x )is in U ). By Proposition 3.1, it follows that v (( r ∗ ) n ) ≥ v ( x ∗ ), where v is the valuationon K ∗ having R ∗U as valuation ring. Therefore, x ∗ divides ( r ∗ ) n in R ∗U , which impliesthat ( r ∗ ) n ∈ P ∗ , which is a prime ideal and therefore contains r ∗ . This is what wewanted to show. Theorem 3.4.
Let R = Q λ ∈ Λ D λ where every D λ is a Prüfer domain. The primeideals of R are exactly the unions of prime ideals of the form ( U ) g with g ∈ S such that g λ,P > for all λ ∈ Λ and P ∈ max( D λ ) .Proof. Since R ∗U is a valuation domain, the prime ideals of R contained in ( U ) form achain. Therefore every union of ( U ) g is a union of a chain of prime ideals, hence it isprime.Conversely, let (0) F ⊆ P ⊆ ( U ) be a prime ideal of R . For x ∈ P we define an element g ( x ) ∈ S such that for all λ ∈ Λ and for all P ∈ max( D λ ) we have g ( x ) λ,P >
0. Namely,set g ( x ) λ,P = v P ( x λ ) if P ∈ V ( x λ ) and g ( x ) λ,P = ∞ otherwise. We claim that P = [ x ∈ P ( U ) g ( x ) . By Proposition 3.3, ( U ) g ( x ) is the smallest prime ideal contained in ( U ) and containing x .So S x ∈ P ( U ) g ( x ) ⊆ P . On the other hand, if y ∈ P , then by Proposition 3.3 we havethat y ∈ ( U ) g ( y ) and therefore y ∈ S x ∈ P ( U ) g ( x ) . Heights of prime ideals
Recall that for every λ ∈ Λ and P ∈ max( D λ ), we denote by S P the totally orderedsubmonoid of non-negative elements (including ∞ ) of the value group associated to thevaluation v P of P and we defined S = Q λ ∈ Λ Q P ∈ max D λ S P to be the product of allthese monoids.We now define a relation ≪ on S , where g ≪ h : ⇔ ∀ Y ∈ U ∀ n ∈ N ∃ λ ∈ Λ ∃ P ∈ Y λ n · g λ,P < h λ,P g, h ∈ S . Lemma 3.5.
Let g, h ∈ S .(1) If ( U ) h $ ( U ) g , then g ≪ h .(2) Let it hold in addition that there is some Y ∈ U and some N ∈ N such that forall λ ∈ Λ we have | Y λ | ≤ N (e.g. let every D λ be semilocal with a uniform boundon the cardinality of max( D λ )). Then g ≪ h implies ( U ) h $ ( U ) g . Proof.
To see (1), let x ∈ ( U ) g \ ( U ) h . Then there exists some Y ∈ U and some n ∈ N such that for all λ ∈ Λ and P ∈ Y λ we have v P ( x nλ ) ≥ g λ,P . On the other hand, for all Y ′ ∈ U and n ′ ∈ N there exists some λ ∈ Λ and some P ∈ Y ′ λ such that v P ( x n ′ λ ) < h λ,P .It follows immediately that g ≪ h .By the additional assumption in statement (2), we can find Y ∈ U and Y , . . . , Y N ∈ B such that Y = Y ∨ . . . ∨ Y N and for each i ∈ { , . . . , N } and λ ∈ Λ we have | ( Y i ) λ | = 1.Moreover, since Y ∈ U and U is an ultrafilter there exists some i ∈ { , . . . , N } suchthat Y ′ := Y i ∈ U . For each λ ∈ Λ, let P λ be the unique maximal ideal of D λ containedin Y ′ λ and let x λ ∈ D λ such that v P λ ( x λ ) = g λ,P . Clearly, ( x λ ) λ ∈ Λ ∈ ( U ) g \ ( U ) h .We now introduce a special type of ultrafilter that will be helpful to force certainprime ideals in R to have infinite height. Let B be a Boolean algebra that admitscountable joins, i.e. for every countable family ( B n ) n ∈ N the join W n ∈ N B n ∈ B is defined.In words of the partial order on B , every countable subset of B should have a supremum.An ultrafilter G in B is called countably incomplete if there exists a countable family( P n ) n ∈ N of elements of B such that P n / ∈ G for every n ∈ N , W n ∈ N P n = 1 B equals thetop element of B and for all m, n ∈ N we have that m = n implies P m ∧ P n = 0 B .This translates in the following way to our main examples of Boolean algebras: Anultrafilter F on a set Λ is countably incomplete if and only if there exists a countablepartition ( P n ) n ∈ N of Λ such that P n / ∈ F for every n ∈ N . It is shown in [5, Theorem6.1.4] that for every infinite set Λ there exists a countably incomplete ultrafilter on Λ.An ultrafilter U in the Boolean algebra B = Q λ ∈ Λ P (max( D λ )) is countably incompleteif and only if there exists a family ( P n ) n ∈ N of elements of B such that P n / ∈ U for every n ∈ N , W n ∈ N P n = 1 B = (max( D λ )) λ ∈ Λ and for all m, n ∈ N we have that m = n implies P m ∧ P n = 0 B . The family ( P n ) n ∈ N is called a partition of 1 B . Lemma 3.6. If F is a countably incomplete ultrafilter on Λ, then every ultrafilter U in B with F = F U is countably incomplete. Proof.
Let ( P n ) n ∈ N be a partition of Λ such that P n / ∈ F for all n ∈ N . Define Q ( n ) := ( Q ( n ) λ ) λ ∈ Λ ∈ B for each n ∈ N , where Q ( n ) λ = max( D λ ) if λ ∈ P n and Q ( n ) λ = ∅ else. Clearly, W n ∈ N Q ( n ) = 1 B and Q ( m ) ∧ Q ( n ) = 0 B for all m, n ∈ N with m = n .Assume that Q ( n ) λ ∈ U for some n ∈ N . Then P n = { λ ∈ Λ | Q ( n ) λ = ∅} ∈ F U = F ,which is a contradiction. 17inocchiaro, Frisch, Windisch: Prime Ideals in Product RingsFrom now on, we again fix an ultrafilter U in B and set F = F U the induced ultrafilteron Λ. Lemma 3.7.
Let F be countably incomplete and g, h ∈ S such that for all λ ∈ Λ andfor all P ∈ max( D λ ) we have g λ,P > h λ,P > g ≪ h , then there exists some k ∈ S such that g ≪ k ≪ h .(2) If g ≪ ∞ = ( ∞ ) λ ∈ Λ , then there exists some k ∈ S such that g ≪ k ≪ ∞ . Proof. (2) follows immediately by setting h = ∞ in (1).To show (1), we can assume without loss of generality that g λ,P < h λ,P for all λ ∈ Λ and P ∈ max( D λ ). We define the following two complementary elements of the Booleanalgebra B : V = ( V λ ), where V λ = { P ∈ max( D λ ) | ∀ n ∈ N n · g λ,P < h λ,P } and W = ( W λ ), where W λ = { P ∈ max( D λ ) | ∃ N ∈ N N · g λ,P ≥ h λ,P } . Since U is an ultrafilter in B , we either have V ∈ U or W ∈ U . Assume that V ∈ U .Since U is countably incomplete by Lemma 3.6, we can choose a partition ( P n ) n ∈ N of1 B such that P n / ∈ U for all n ∈ N . By setting V ( n ) = P n ∧ V for each n ∈ N , we get apartition ( V ( n ) ) n ∈ N of V such that V ( n ) / ∈ U for every n ∈ N (in the sense that it is apartition of the top element V in the subalgebra B V := { Y ∈ B | Y ≤ V } ). For λ ∈ Λand P ∈ max( D λ ), we define k λ,P = n · g λ,P if P ∈ V ( n ) λ and k λ,P = g λ,P if P / ∈ V λ .Then clearly k ≪ h .To see that g ≪ k , let U ∈ U and n ∈ N . Then U ∧ V ∈ U and therefore there existssome N > n such that U ∧ V ( N ) = 0 B . (Indeed, if for all N > n we would have that U ∧ V ( N ) = ∅ , then ( U ∧ V (1) ) ∨ . . . ∨ ( U ∧ V ( n ) ) = U ∧ V ∈ U . Therefore there wouldexist some i ∈ { , . . . , n } such that U ∧ V ( i ) ∈ U and therefore V ( i ) ∈ U , which is acontradiction.) Pick some λ ∈ Λ and P ∈ U λ ∩ V ( N ) λ . Then k λ,P = N · g λ,P > n · g λ,P ,so g ≪ k .Now consider the case where W ∈ U . For each λ ∈ Λ and P ∈ W λ , there exists some N > h λ,P ≤ N · g λ,P , so we can pick N λ,P ≥ N λ,P · g λ,P 0] := ∞ . For P / ∈ W λ let k λ,P = g λ,P .Let Y ∈ U and n ∈ N . We have to show the following two assertions:(i) There exist λ ∈ Λ and P ∈ Y λ such that n · g λ,P < k λ,P .(ii) There exist λ ∈ Λ and P ∈ Y λ such that n · k λ,P < h λ,P .We can assume without loss of generality that Y ≤ W . First of all, note that { N λ,P | λ ∈ Λ , P ∈ Y λ } is unbounded, because g ≪ h . It follows that the sets S ( i ) := { [ N λ,P / log( N λ,P )] | λ ∈ Λ , P ∈ Y λ } and S ( ii ) := { log( N λ,P ) | λ ∈ Λ , P ∈ Y λ } are also unbounded. To show (i), we use that S ( i ) is unbounded and pick λ ∈ Λ, P ∈ Y λ such that n < [ N λ,P / log( N λ,P )]. It follows that n · g λ,P < [ N λ,P / log( N λ,P )] · g λ,P = k λ,P .For the proof of (ii), we can pick λ ∈ Λ, P ∈ Y λ such that n < log( N λ,P ). It followsthat n · [ N λ,P / log( N λ,P )] ≤ n · N λ,P / log( N λ,P ) < n · N λ,P /n = N λ,P . Hence n · k λ,P = n · [ N λ,P / log( N λ,P )] · g λ,P < N λ,P · g λ,P < h λ,P . Theorem 3.8. Let ( D λ ) λ ∈ Λ be a family of Prüfer domains and set R = Q λ ∈ Λ D λ .Let F be a countably incomplete ultrafilter on Λ and U be an ultrafilter in the Booleanalgebra B = Q λ ∈ Λ P (max( D λ )) such that F equals the unique induced ultrafilter F U on Λ and such that there exist Y ∈ U and N ∈ N with | Y λ | ≤ N for all λ ∈ Λ (e.g. let all D λ be semilocal with a uniform bound on the cardinalities of max( D λ ) ).Then for every prime ideal P ⊆ R with (0) F $ P there exists some prime ideal Q ⊆ R such that (0) F $ Q $ P .In particular, every prime ideal of R strictly containing (0) F is of infinite height.Proof. Let I ⊆ S such that P = S g ∈ I ( U ) g , which exists by Theorem 3.4. Since (0) F ⊆ ( U ) g for all g ∈ I , there must exist some g ∈ I such that ( U ) ∞ = (0) F $ ( U ) g . Itfollows by Lemma 3.5(1) that g ≪ ∞ . So by Lemma 3.7(2) we have that there existssome h ∈ S with g ≪ h ≪ ∞ . Lemma 3.5(2) implies that (0) F $ ( U ) h $ ( U ) g ⊆ P .The assertion follows by setting Q = ( U ) h .We close with an example of a Prüfer domain in which every non-zero prime ideal isof infinite height: Let Λ be an infinite set and ( D λ ) λ ∈ Λ be a family of semilocal Prüferdomains with the property that there exists N ∈ N such that | max( D λ ) | ≤ N for all λ ∈ Λ. Moreover, let F be a countably incomplete ultrafilter on Λ (which exists by [5,Theorem 1.6.4]). Then by Theorem 3.8, every non-minimal prime ideal of R = Q λ ∈ Λ D λ containing (0) F is of infinite height. As noted before, the ring theoretical property ofbeing a Prüfer domain is preserved by ultraproducts.So the ultraproduct R ∗ = Q F λ ∈ Λ D λ ∼ = R/ (0) F is a Prüfer domain in which every non-zero prime ideal is of infinite height. If every D λ is local, then R ∗ is in addition avaluation domain. If every D λ is non-local, then so is R ∗ . 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Finocchiaro, Dipartimento di Matematica e Informatica, Uni-versità degli Studi di Catania, 95125 Catania, Italy E-mail address : [email protected] Sophie Frisch, Department of Analysis and Number Theory (5010), Tech-nische Universitat Graz, Kopernikusgasse 24, 8010 Graz, Austria E-mail address : [email protected] Daniel Windisch, Department of Analysis and Number Theory (5010),Technische Universitat Graz, Kopernikusgasse 24, 8010 Graz, Austria E-mail address : [email protected]@math.tugraz.at