Topology and its Applications

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1 Topology and ts Applcatons 58 (20) Contents lsts avalable at ScenceDrect Topology and ts Applcatons Fntely -convex f -rngs Suzanne Larson Loyola Marymount Unversty, Los Angeles, CA 90045, Unted States artcle nfo abstract MSC: 06F25 54C40 3A5 Keywords: f -Rngs Fntely -convex -Convex Fntely an F-space F-space SV-space SV f -rng Ths paper nvestgates f -rngs that can be constructed n a fnte number of steps where every step conssts of tang the fbre product of two f -rngs, both beng ether a -convex f -rng or a fbre product obtaned n an earler step of the constructon. These are the f -rngs that satsfy the algebrac property that rngs of contnuous functons possess when the underlyng topologcal space s fntely an F-space (.e. has a Stone Čech compactfcaton that s a fnte unon of compact F-spaces). These f -rngs are shown to be SV f -rngs wth bounded nverson and fnte ran and, when constructed from semsmple f -rngs, ther maxmal deal space under the hull-ernel topology contans a dense open set of maxmal deals contanng a unque mnmal prme deal. For a large class of these rngs, the sum of prme, semprme, prmary and z-deals are shown to be prme, semprme, prmary and z-deals respectvely. 20 Elsever B.V. All rghts reserved.. Introducton A commutatve f -rng s -convex f for any u, v A such that 0 u v, theresaw A such that u = wv.gven f -rngs A, A 2, B and surectve l-homomorphsms φ : A B and φ 2 : A 2 B, thefbre product of A and A 2,denoted A B A 2,sthesub-f -rng of A A 2 gven by A B A 2 ={(a, a 2 ): φ (a ) = φ 2 (a 2 )}. Wesayan f -rng (a rng) s a fnte fbre product of the f -rngs (rngs) A, A 2,...,A n f t can be constructed n a fnte number of steps where every step conssts of tang the fbre product of two f -rngs (rngs), both of these f -rngs (rngs) satsfyng ether the property that t s one of the A not used n a prevous step, or t s a fbre product obtaned n an earler step of the constructon. An f -rng A s fntely -convex f t s ether a -convex f -rng or can be wrtten as a fnte fbre product of -convex f -rngs. These are the f -rngs that satsfy the algebrac property that rngs of contnuous functons possess when the underlyng topologcal space s fntely an F-space. A topologcal space X s fntelyanf-space f ts Stone Čech compactfcaton s a unon of fntely many closed F-spaces. An often used constructon of a space that s fntely an F-space but not an F-space begns wth n copes of a compact F-space X and a certan type of closed nowhere dense set A X, and then for each a A, alln copes of a are dentfed as a sngle pont. In [2], t s shown that for a normal space X, C(X) s fntely -convex f and only f X s fntely an F-space. An f -rng A s an SV f -rng f for every mnmal prme deal P of A, A/P s a valuaton doman. A topologcal space s an SV space f C(X) s an SV f -rng. Mel Henrsen and Rchard Wlson ntated the study of SV rngs and spaces wth ther 992 papers (see [6,7]). A dozen or more papers have been wrtten that study SV rngs and spaces and related matters. Spaces that are fntely an F-space were ntroduced n [6] and are of nterest because they are relatvely easy to construct, ther correspondng rng of contnuous functons s an SV f -rng, and the correspondng space of prme deals s relatvely smple. In 2.9, [6] t s shown that f the space X s fntely an F-space then C(X) s an SV f -rng and n [], t s shown that the converse does not hold. Fntely -convex f -rngs were later ntroduced n [2] largely because of ther connecton to spaces that are fntely an F-space and to SV f -rngs. If the space X s compact and fntely an F-space, then, as shown E-mal address: suzanne.larson@lmu.edu /$ see front matter 20 Elsever B.V. All rghts reserved. do:0.06/.topol

2 S. Larson / Topology and ts Applcatons 58 (20) n 5.6 of [5], X contans a dense open set of ponts of ran (.e. X contans a dense open set of ponts x for whch the correspondng maxmal deal M x ={f C(X): f (x) = 0} contans a unque mnmal prme deal). In ths paper, we nvestgate commutatve semprme f -rngs wth dentty element that are fntely -convex. We wll loo at the relatonshp of a fntely -convex f -rng to that of an SV f -rng and an f -rng wth fnte ran and at the bounded rng of elements of a fntely -convex f -rng. We then wll nvestgate propertes of maxmal, mnmal, prme, semprme, prmary and z-deals n fntely -convex f -rngs. We loo at the maxmal and mnmal prme deals n a fntely -convex f -rng and wll show that for a commutatve semprme fntely -convex f -rng wth dentty element that s constructed from semsmple f -rngs, the space of maxmal deals contans a dense open set of maxmal deals of ran under the hull-ernel topology. Ths extends the nown result that a compact space that s fntely an F-space has a dense open set of ponts for whch the correspondng maxmal deal has ran. In the last secton, we show that there s a large class of fntely -convex f -rngs n whch the sum of two prme, semprme, prmary and z-deals s a prme, semprme, prmary and z-deal respectvely. 2. Prelmnares Throughout ths paper, all rngs wll be assumed to be commutatve semprme rngs wth dentty element, and wth the excepton of Secton 4, all rngs wll be f -rngs as well. An f -rng s a lattce ordered rng that s a subdrect product of totally ordered rngs. For general nformaton on f -rngs see []. Gven an f -rng A, weleta + ={a A: a 0}, and for an element a A, weleta + = a 0, a = ( a) 0, and a =a ( a). IfA s an f -rng wth dentty element, let A ={a A: a n for some postve nteger n}. ThenA s a sub- f -rng of A, and s called the subrng of bounded elements. IfA s an f -rng wth dentty element n whch every element a s nvertble, then A s sad to be closed under bounded nverson or to have bounded nverson. An l-homomorphsm φ : A B mappng an f -rng A to an f -rng B s a rng homomorphsm such that for all a, b A, φ(a b) = φ(a) φ(b) and φ(a b) = φ(a) φ(b). ArngdealI of an f -rng s an l-deal f a b and b I mples a I, or equvalently, f t s the ernel of a lattce-preservng homomorphsm (l-homomorphsm). Gven any element a of an f -rng A, there s a smallest l-deal contanng a, and we denote ths by a. Gvenan f -rng A and an l-deal I of A, the quotent rng A/I s n fact an f -rng under the usual rng operatons on A/I and an order gven by a + I b + I f there exsts, 2 I such that a + b + 2 n A. Suppose A s an f -rng and I s an deal of A. ThedealI s semprme (resp. prme) f J 2 I (resp. JK I) mples J I (resp. J I or K I) fordeals J, K.Anl-deal I of an f -rng s a semprme (resp. prme) deal f and only f a 2 I mples a I (resp. ab I mples a I or b I). The f -rng A s called semprme (resp. prme) f {0} s a semprme (resp. prme) deal. It s well nown that n an f -rng, an l-deal I s a semprme deal f and only f t s an ntersecton of prme l-deals whch are mnmal wth respect to contanng I. IfP s a prme l-deal of the f -rng A, thena/p s a totally ordered prme rng and all l-deals of A contanng P form a chan. An deal I of a commutatve rng wth dentty element s pseudoprme f ab = 0mplesa I or b I and s prmary f ab I mples a I or b n I for some natural number n. In a commutatve and semprme rng, a pseudoprme deal contans a prme deal. For an element a of a rng A, weletm A (a) denote the set of all maxmal deals of A contanng a. AndealI of a commutatve rng wth dentty element s called a z-deal f whenever a, b A wth M A (a) = M A (b) and a I, thenb I. Equvalently, I s a z-deal f whenever a, b A wth M A (a) M A (b) and a I, thenb I. It s easly seen that every z-deal s a semprme deal. As s shown n Theorem. of [3], every mnmal prme deal of an f -rng s a z-deal. For any f -rng A, weletmax(a) denote the set of all maxmal deals of A. Ifa A, leth c (a) ={M Max(A): a / M}. The hull-ernel topology on Max(A) s the topology generated by {h c (a): a A}. IfA has an dentty element and satsfes the bounded nverson property then Max(A), under the hull-ernel topology, wll be a compact Hausdorff space (see [4]). A commutatve rng s a valuaton rng f gven any two elements, one dvdes the other. An f -rng A s an SV f -rng f for every mnmal prme deal P of A, A/P s a valuaton doman. A commutatve f -rng A s sad to satsfy the st-convexty condton, ortobe-convex f for any u, v A such that 0 u v, theresaw A such that u = wv. In a commutatve f -rng A wth dentty element and satsfyng the st-convexty condton, u, v A wth 0 u v mples that there s a w A such that 0 w and u = wv. Every commutatve f -rng wth the st-convexty condton s an SV f -rng and the followng lemma shows a further connecton between SV f -rngs and f -rngs satsfyng the st-convexty condton. Lemma. ([2, Lemma 5.8]) Suppose A s a commutatve f -rng wth dentty element and bounded nverson. Then A s an SV f -rng f and only f for every mnmal prme deal P of A, A/P s-convex. Suppose M s a maxmal l-deal of an f -rng A. Theran of M s the number of mnmal prme deals contaned n M f the set of all such mnmal prme deals s fnte, and the ran of M s nfnte otherwse. We let ran A (M) denote the ran of the maxmal l-deal M n the f -rng A. IfA s an f -rng, then the ran of A s the supremum of the rans of the maxmal l-deals of A. The f -rng A s sad to have fnte ran f the ran of A s fnte. A commutatve semprme -convex f -rng wth dentty element has ran as was shown n Theorem 5.6 of [2] and also has the property that all deals are l-deals (see [8]). As a result, the set of all prme deals contaned n a gven maxmal deal of a commutatve semprme -convex f -rng wth dentty element form a chan.

3 890 S. Larson / Topology and ts Applcatons 58 (20) An F-space s a (completely regular) topologcal space X such that n C(X), the rng of all real-valued contnuous functons defned on X, every fntely generated deal s prncpal. A number of condtons, both topologcal condtons on X, and algebrac condtons on C(X), areequvalenttox beng an F-space and appear n 4.25 of [2], of [5], and 2.4 of [8]. One partcular equvalence we wll mae use of s that a topologcal space X s an F-space f and only f C(X) s -convex. For a gven functon f C(X), thezeroset of f s Z( f ) ={x X: f (x) = 0}. A topologcal space X s fntely an F-space f ts Stone Čech compactfcaton, β X, s a unon of fntely many closed F-spaces. See [2] for more nformaton on the Stone Čech compactfcaton of a space X. 3. Basc propertes of fntely -convex f -rngs Gven f -rngs (rngs) A, A 2, B and surectve l-homomorphsms (homomorphsms) φ : A B and φ 2 : A 2 B, recall that the fbre product of A and A 2,denotedA B A 2,sthesub-f-rng (subrng) of A A 2 gven by A B A 2 ={(a, a 2 ) A A 2 : φ (a ) = φ 2 (a 2 )}. It s worth notng that f A = A 2 = B s an f -rng (rng) and the dentty l-homomorphsms (homomorphsms) are used, then the fbre product A B A 2 ={(a, a): a A } = A. On the other hand, f A, A 2 are f - rngs (rngs) and B ={0}, then the fbre product A B A 2 ={(a, a 2 ): a A, a 2 A 2 } = A A 2 (the drect product of A, A 2 ). We say an f -rng (a rng) s a fnte fbre product of the f -rngs (rngs) A, A 2,...,A n f t can be constructed n a fnte number of steps where every step conssts of tang the fbre product of two f -rngs (rngs), both of these f -rngs (rngs) satsfyng ether the property that t s one of the A not used n a prevous step, or t s a fbre product obtaned n an earler step of the constructon. Note that by ncludng the requrement that the f -rngs (rngs) A not be used n more than one step of the constructon, we smply requre every tme a rng s used that s not a fbre product obtaned n an earler step of the constructon, that rng be ncluded as an entry n the lstng of the A s, even f t causes a repetton of rngs n the lstng. For example, we say (A B A ) B2 A s a fnte fbre product of the rngs A, A, A, and we say A A A s a fnte fbre product of A, A. Our defnton of a fnte fbre product of the f -rngs (rngs) A, A 2,...,A n allows for varatons n the steps taen when constructng such a rng. We may assume that the steps n the constructon mae use of the f -rngs (rngs) A, A 2,...,A n n the order lsted and that the frst step of the constructon yelds an f -rng (rng) of the form A B A 2. Stll, later steps could nvolve tang the fbre product of an f -rng (rng) resultng from an earler step and the next A that has not been used n an earler step, could nvolve tang the fbre product of two f -rngs (rngs) resultng from earler steps, or could nvolve tang the fbre product of the next two A that have not been used n an earler step. For example, the constructon of a fnte fbre product of the f -rngs A, A 2, A 3, A 4 could result n an f -rng of the form ((A B A 2 ) B2 A 3 ) B3 A 4 or of the form (A B A 2 ) B3 (A 3 B2 A 4 ). As the next theorem wll ndcate, every fnte fbre product constructed from the f -rngs (rngs) A, A 2,...,A n s somorphc to a fnte fbre product constructed frst by tang the fbre product of A, A 2, then at each stage tang the fbre product of the f -rng (rng) that resulted from the prevous step and the next f -rng (rng) n the lst of the A s. Frst, however, we need a verson of Goursat s lemma for rngs. A smlar lemma, gven n a dfferent context, appears n [6]. Recall also that a subdrect product of the rngs A, A 2,...,A n s a subrng of A A 2 A n for whch each proecton mappng onto A s surectve. Lemma 2. Suppose A, A, A 2 are commutatve semprme f -rngs (rngs) wth dentty element and A A A 2 s a subdrect product of A, A 2. Then there s a commutatve f -rng (rng) B such that A = A B A 2. Proof. We prove the result for f -rngs. Defne I ={a A : (a, 0) A} and I 2 ={a A 2 : (0, a) A}. ThenI, I 2 are l-deals of A, A 2 respectvely. We wll show that A /I = A2 /I 2.Todoso,defneψ: A /I A 2 /I 2 by ψ(a + I ) = a 2 + I 2 where a 2 A 2 s chosen such that (a, a 2 ) A. Frst we wll show that ψ s a well defned mappng. So suppose that a 2, a 3 A 2 such that (a, a 2 ), (a, a 3 ) A. Then(a, a 2 ) (a, a 3 ) = (0, a 2 a 3 ) A and so a 2 a 3 I 2. Ths mples that a 2 + I 2 = a 3 + I 2 and hence ψ s well defned. It s straghtforward to see that ψ preserves the operatons of addton, multplcaton, and tang supremum. To see that ψ s nectve, suppose that ψ(a + I ) = ψ(a + I ).Thenb + I 2 = b + I 2 for some b, b A 2 wth (a, b), (a, b ) A. Sob b I 2, and (0, b b ) A. Then(a a, 0) = (a, b) (a, b ) (0, b b ) A. It follows that a a I and a + I = a + I. That ψ s surectve follows from the fact that A s a subdrect product of A, A 2.Thusψ s an l-homomorphsm and A /I = A2 /I 2. Now let B = A 2 /I 2,letφ be ψ composed wth the natural l-homomorphsm mappng A to A /I, and let φ 2 be the natural l-homomorphsm mappng A 2 to A 2 /I 2. Then t s straghtforward to show that A = A B A 2. Note that any fbre product of the f -rngs (rngs) A, A 2 s a subdrect product of A, A 2. Theorem 3. Suppose A, A, A 2,...,A n are commutatve semprme f -rngs (rngs) wth dentty element and A s a fnte fbre product of the f -rngs (rngs) A, A 2,...,A n. Then there are f -rngs (rngs) B, B 2,...,B n such that A s l-somorphc (somorphc) to (( (A B A 2 ) B2 A 3 ) Bn 2 A n ) Bn A n.

4 S. Larson / Topology and ts Applcatons 58 (20) Proof. We wll prove the result for f -rngs. Suppose A s a fnte fbre product of the f -rngs A, A 2,...,A n. We may suppose that the steps n the constructon of A mae use of the f -rngs A, A 2,...,A n n the order lsted and that the frst step of the constructon yelds A B A 2 for some f -rng B. We proceed by nducton on the number of f -rngs used n the constructon of A. Ifn = 2, A = A B A 2, and there s nothng we need prove. Note that f n = 3, then A s ether (A B A 2 ) B2 A 3 or A 3 B2 (A B A 2 ) for some f -rng B 2. Snce the second of these s somorphc to the frst, the result holds when n = 3. Now suppose that the desred result holds for any fnte fbre product constructed from f-rngs, where 2 < < n. The fnal step of the constructon of A s to tae a fbre product of two f -rngs, say K 0, K, where ether () K 0 s a fnte fbre product nvolvng A, A 2,...,A n obtaned n an earler step of the constructon and K = A n, or () K 0 = A n and K s a fnte fbre product nvolvng A, A 2,...,A n obtaned n an earler step of the constructon, or () K 0 s a fnte fbre product nvolvng A, A 2,...,A t obtaned n an earler step of the constructon for some t < n and K s a fnte fbre product nvolvng A t+, A t+2,...,a n obtaned n an earler step of the constructon. If () holds, by our nducton hypothess K 0 = ((A B A 2 ) B2 A 3 ) Bn 2 A n for some f -rngs B, B 2,...,B n 2. Then A = (((A B A 2 ) B2 A 3 ) Bn 2 A n ) Bn A n for some f -rng B n and we are done. Smlarly, f () holds, K = ((A B A 2 ) B2 A 3 ) Bn 2 A n for some f -rngs B, B 2,...,B n 2, and A = A n Bn (((A B A 2 ) B2 A 3 ) Bn 2 A n ) = (((A B A 2 ) B2 A 3 ) Bn 2 A n ) Bn A n for some f -rng B n. If () holds, then by our nducton hypothess K 0 = ((A B A 2 ) B2 A 3 ) Bt A t for some f -rngs B, B 2,...,B t and K = (((A t+ Ct+ A t+2 ) Ct+2 A t+3 ) Cn A n ) for some f -rngs C t+, C t+2,...,c n.letk t = K 0 and defne recursvely, for = t +,...,n, K = {((( ) ) ) (( ) ) (a, a 2 ), a 3,...,a, a : (a, a 2 ), a 3,...,a K, and there exsts a +, (( ) ) } a +2,...,a n such that (a t+, a t+2 ), a t+3,...,an K. Then K t+ K t A t+ s a subdrect product of K t and A t+, and so by the prevous lemma, there s an f -rng B t such that K t+ = Kt Bt A t+ = (((A B A 2 ) B2 A 3 ) Bt A t ) Bt A t+. Repeatng ths argument for K t+2,...,k n results n K n = ((A B A 2 ) B2 A 3 ) Bn 2 A n. Fnally, A s l-somorphc to a subdrect product of K n and A n, and so by the prevous lemma, A = Kn Bn A n for some f -rng B n.hencea = Kn Bn A n = (((A B A 2 ) B2 A 3 ) Bn 2 A n ) Bn A n for f -rngs B, B 2,...,B n. Every fnte fbre product of the f -rngs A, A 2,...,A n s l-somorphc to a sub- f -rng of A A 2 A n.welet ψ : A A A 2 A n denote an l-embeddng. To ad n our nvestgaton, we adopt the followng notatonal conventon: for =, 2,...,n, π : A A wll denote the proecton mappng of ψ(a) onto A composed wth ψ. Note that for each, π s surectve. Defnton 4. An f -rng A s fntely -convex f t s ether a -convex f -rng or can be wrtten as a fnte fbre product of -convex f -rngs. Next we gve an example of a fntely -convex f -rng that we wll mae use of several tmes. Example 5. Let R[x] denote the rng of polynomals over the reals n one ndetermnate. Totally order R[x] lexcographcally, so that x x 2.NowletA ={ p : p, q R[x], q } under the usual addton and multplcaton of quotents of q polynomals and under the order nduced by the order on R[x]. Thats, f and only f p q 2 p 2 q.thena s a p q p 2 q 2 totally ordered -convex f -rng. Let A 2 ={f C(N): n 0 N, r R such that f (n) = r n n 0 } under the usual addton, multplcaton, and partal order of functons. Then A 2 s also a -convex f -rng. Defne φ : A R by φ ( p q ) = p(0) q(0) and φ 2 : A 2 R by φ 2 ( f ) = r where there exsts n 0 N such that f (n) = r for all n n 0.Bothφ,φ 2 are surectve l- homomorphsms. Then the f -rng A R A 2 ={( p q, f ) A A 2 : -convex. p(0) q(0) = f (n 0), where f (n) = f (n 0 ) for all n n 0 } s fntely A topologcal space X s fntely an F-space f ts Stone Čech compactfcaton s a unon of fntely many closed F-spaces. Suppose X s a compact space that s fntely an F-space. Then, as shown n Theorem 5.3 of [2], C(X) s a fntely -convex f -rng. In fact, f X = X X 2 for some compact F-spaces X, X 2,then C(X) = C(X ) C(X X 2 ) C(X 2 ) where the requred l-homomorphsms are the restrcton mappngs of the form f f X X 2. An nductve argument can be employed to show that f X = n = X for compact F-spaces X, X 2,...,X n then C(X) s fntely -convex. Conversely, f C(X) s fntely -convex (and X s compact), then X s fntely an F-space as s also shown n Theorem 5.3 of [2]. Suppose B s a -convex f -rng and Q asemprmel-deal of B. One type of fntely -convex f -rng that s partcularly nce to wor wth can be constructed as the sub- f -rng of n = B gven by A ={(b, b 2,...,b n ): b b Q for all, }.

5 892 S. Larson / Topology and ts Applcatons 58 (20) Indeed, the f -rng A could be wrtten as a fnte fbre product of n copes of B n the form ([(B B/Q B) B/Q B] B/Q B). We wll say that an f -rng (rng) A s a homogeneously fnte fbre product f there s an f -rng (rng) B and a semprme l-deal (deal) Q of B such that A s l-somorphc to the sub- f -rng (subrng) of B B B gven by {(b, b 2,...,b n ): b b Q for all, =, 2,...,n}. Defnton 6. An f -rng A s homogeneously fntely -convex f there s a -convex f -rng B and semprme l-deal Q of B such that A s l-somorphc to the sub- f -rng of B B B gven by {(b, b 2,...,b n ): b b Q for all, =, 2,...,n}. Because fntely -convex f -rngs were frst ntroduced n connecton wth the study of SV f -rngs and f -rngs of fnte ran, t s approprate to begn wth a theorem that helps to show the relatonshp of a fntely -convex f -rng to that of an SV f -rng and an f -rng wth fnte ran. The proof of ths theorem wll mae use of the fact that a fntely -convex f -rng necessarly satsfes the bounded nverson property as was noted n [2], and the followng well-nown result that follows drectly from Proposton 3 of [3]. Theorem 7. Suppose A s a commutatve semprme f -rng wth dentty element. If M s a maxmal deal of A and a s an element n the ntersecton of all the mnmal prme deals contaned wthn M, there s an element b / Msuchthatab= 0. Theorem 8. Suppose A s a commutatve semprme f -rng wth dentty element. Then () (2) (3) (4). () Asafntely-convex f -rng. (2) A s an SV f -rng wth fnte ran and bounded nverson. (3) For every u, v such that 0 u v, there are fntely many elements w, w 2,...,w n Asuchthat0 = (u w v)(u w 2 v) (u w n v). (4) A s an SV f -rng. Proof. () (2) appears n Theorems 5.5 and 5.7 of [2]. (2) (3): Suppose 0 u v. We frst show that for any maxmal deal M, there exst fntely many elements w M,, w M,2,...,w M,tM and an element a M M + such that (u w M, v)(u w M,2 v) (u w M,tM v)a M = 0. Let M be a maxmal deal and let t M = ran A (M). Suppose the mnmal prme deals contaned n M are P M,, P M,2,...,P M,tM. Snce A s an SV f -rng, each factor f -rng A/P M, s -convex by Lemma. So for each =, 2,...,t M, there s a w M, A such that u + P M, = (w M, + P M, )(v + P M, ) n A/P M,. Then u w M, v P M, for each. Ths mples (u w M, v)(u w M,2 v) (u w M,tM v) t M = P M,. So by the prevous theorem there s an a M 0 such that a M / M and (u w M, v)(u w M,2 v) (u w M,tM v)a M = 0. Now the collecton {h c (a M ): M Max(A)} s an open cover of Max(A). BecauseA has bounded nverson, Max(A) s compact and so there s a fnte subcover of Max(A). Wewlldenotethesubcoverby{h c (a M ), h c (a M2 ),...,h c (a Mn )}. Then n a M = (u w M,v)(u w M,2v) (u w M,t M v) = 0foreach =, 2,...,n. So [ n ] (u w M,v)(u w M,2v) (u w M,t M v) ][ n a M = = [ ] n n = (u w M,v)(u w M,2v) (u w M,t M v) = = 0. a M = Now n = a M s not contaned n any maxmal deal snce n = a M a M 0foreach, and snce A has bounded nverson, every maxmal deal s an l-deal and does not contan (at least) one of the a M.BecauseA has bounded nverson, then n = a M s a unt of A. Ths mples that n = (u w M,v)(u w M,2v) (u w M,t M v) = 0. (3) (4): Suppose u, v A, P s a mnmal prme deal of A, and 0 u + P v + P n A/P. Then there are p, p 2 such that 0 u + p v + p 2 n A. Letu = u + p, v = v + p 2. By (3), there s a fnte number of w A such that (u w v )(u w 2 v ) (u w n v ) = 0 P.SnceP s a prme deal, there s an such that u w v P.Thenu + p w (v + p 2 ) P and t follows that u w v P.Sou + P = (w + P)(v + P). Ths shows A/P s -convex and hence by Lemma, A s an SV f -rng. Property (2) does not mply property () n the prevous theorem. In [], an example of a normal topologcal space X s constructed such that C(X) s an SV f -rng of fnte ran, whle X s not fntely an F-space. Snce a normal topologcal space s fntely an F-space f and only f ts correspondng rng of contnuous functons s fntely -convex, ths C(X) provdes an example of a commutatve semprme SV f -rng wth dentty element and bounded nverson that has fnte ran and bounded nverson and yet s not fntely -convex.

6 S. Larson / Topology and ts Applcatons 58 (20) Note that for a C(X) for X a compact space, propertes (2), (3), and (4) of the prevous theorem are equvalent snce every SV C(X) has fnte ran and bounded nverson. (See 4. of [5].) However, n general f -rngs, property (2) s nether equvalent to property (3) nor to property (4) and we do not now f propertes (3) and (4) are equvalent. The next example demonstrates that nether property (3) nor property (4) of the prevous theorem mples property (2). Example 9. Let βn denote the Stone Čech compactfcaton of the natural numbers N. Letα βn N be a pont for whch there s a G δ set contanng α that fals to be a neghborhood of α (.e. let α be a non-p-pont). Let Y denote the topologcal space N {α} under the subspace topology relatve to βn. Note that Y s an F-space and so C(Y ) s -convex. For each n N, letx n denote the topologcal space obtaned by tang n copes of Y and pastng them together at α. Wewllcall the dentfed pont α n. It s straghtforward to show that for each n, C(X n ) s an SV f -rng of ran n. Then by the prevous theorem, property (3) holds n each C(X n ).NowletX = n= X n. We defne the f -rng A as follows. Let A ={f C(X): f = m + g, where g C(X), g Xn = 0forallbutfntelymanyn and m C(X) s a constant functon }. ItsnotdffculttoseethatA s a sub- f -rng of C(X). We wll show that property (3) of the prevous theorem holds n A. Sosuppose0 u v n A, where u = m + g, v = m 2 + g 2, m,m 2 are constant functons on X and g, g 2 C(X) wth g Xn, g 2 Xn = 0 for all but fntely many n. Lett N be such that g Xn, g 2 Xn = 0 for all n > t. Frst assume that m 2 = 0. Then m = 0 and u Xn = v Xn = 0foralln > t. Puttng ths together wth the fact that for n =, 2,...,t, C(X n ) satsfes property (3), t follows easly that A satsfes property (3). Now assume m 2 0. Then u Xn = m, v Xn = m 2 for all n > t. So(u m m 2 v) Xn = 0foralln > t. Ths, together wth the fact that for n =, 2,...,t, C(X n ) satsfes property (3), mples that A satsfes property (3). By the prevous theorem, then A also satsfes property (4). Now for each n N, a maxmal deal of A s M n ={f A: f (α n ) = 0}. Foreachn, lety n, denote the th copy of Y used n the constructon of X n.thenletp n, ={f A: f (U Y n, ) = 0 for some neghborhood U of α n }.For =, 2,...,n, P n, s a mnmal prme deal contaned n M n. Ths shows that for each n N, M n has ran at least n. SoA has nfnte ran, and property (2) does not hold n A. Also,noteA satsfes the bounded nverson property snce f f and f = m + g where g C(X), g Xn = 0 for all but fntely many n and m C(X) s a constant functon, then f m C(X) wth ( f m ) X n = 0 for all but fntely many n and f = m + ( f m ) A. If A s a commutatve semprme f -rng wth dentty element and bounded nverson, then A has fnte ran f and only f A has fnte ran as shown n Proposton 3.2 of [5]. A commutatve semprme f -rng A wth dentty element and bounded nverson s SV f and only f A s SV. For a commutatve semprme fntely -convex f -rng wth dentty element, we can show that the sub- f -rng of bounded elements s also fntely -convex. Theorem 0. Suppose A s a commutatve semprme f -rng wth dentty element. If A s fntely -convex then A s also fntely -convex. Proof. It wll be suffcent to establsh that () f A s -convex, then A s a -convex f -rng and () f A = A B A 2 where A, A 2 are fntely -convex f -rngs, A, A 2 are fntely -convex f -rngs, and φ : A B, φ 2 : A 2 B are surectve l-homomorphsms then A s fntely -convex. An nducton argument would then show the result holds for all fntely -convex f -rngs. So, frst assume that A s -convex and that 0 u v n A.SnceA s -convex, there s a w A such that u = wv.thenw A and u = (w )v. HenceA s -convex. Next suppose that A = A B A 2 where A, A 2 are fntely -convex f -rngs, A, A 2 are fntely -convex f -rngs, and φ : A B, φ 2 : A 2 B are surectve l-homomorphsms. For =, 2, defne φ : A B to be the restrcton mappng φ A. It s not hard to see that φ preserves the rng and lattce operatons. So to show that φ s a surectve l-homomorphsm mappng A onto B we need only show that t s surectve. Suppose that b B. Then there s an m N such that b m. Snce φ s surectve, there exsts an a A such that φ (a) = b. Thena m A and φ (a m ) = φ (a m ) = φ (a) φ (m ) = b m = b. Henceφ s surectve. It s now straghtforward to show that A = A B A 2.HenceA s fntely -convex. When the f -rng A does not have the bounded nverson property, t s possble for A to be fntely -convex, whle A s not fntely -convex. For example, the f -rng R[x] of polynomals wth real coeffcents under the total orderng n whch x x 2 x 3 has the property that A = R s a -convex f -rng, whle A s not fntely -convex (or even SV). 4. Ideals n fbre products of (unordered) rngs Some of the basc propertes of deals n fntely -convex f -rngs that we wll use do not depend on the exstence of a partal order on the rng. These propertes hold n fnte fbre products of commutatve, but not necessarly partally ordered rngs. The purpose of ths secton s to gather together these basc propertes that do not depend on the exstence of a partal order. The followng lemma provdes us wth a means for constructng deals of varous types n a fnte fbre product of commutatve rngs. Its proof s straghtforward, and omtted.

7 894 S. Larson / Topology and ts Applcatons 58 (20) Lemma. Let A be a commutatve semprme rng wth dentty element. Suppose that A s a fnte fbre product constructed from the rngs A, A 2,...,A n.let {, 2,...,n}. () If M s a maxmal deal of A then (M ) s a maxmal deal of A. (2) If P s a prme deal of A then (P ) s a prme deal of A. (3) If P s a pseudoprme deal of A then (P ) s a pseudoprme deal of A. (4) If P s a semprme deal of A then (P ) s a semprme deal of A. (5) If P s prmary and pseudoprme deal of A then (P ) s a prmary and pseudoprme deal of A. (6) If P s z-deal of A then (P ) s a z-deal of A. In fact, for several types of deals, every deal of that type s of the form gven n the prevous lemma, as our next theorem ndcates. Theorem 2. Let A be a commutatve semprme rng wth dentty element. Suppose that A s a fnte fbre product constructed from the rngs A, A 2,...,A n. () If I s an deal of A and {, 2,...,n} then ({0}) I f and only f I = (π (I)). (2) Every prme deal of A has the form (P ) for some {, 2,...,n} and prme deal P of A. (3) Every mnmal prme deal of A has the form (P ) for some {, 2,...,n} and mnmal prme deal P of A. (4) Every maxmal deal of A has the form (M ) for some {, 2,...,n} and maxmal deal M of A. (5) Every pseudoprme deal of A has the form (I ) for some {, 2,...,n} and pseudoprme deal I of A. (6) Every semprme deal of A s an ntersecton of fntely many semprme deals of the form (I ) for some {, 2,...,n} and semprme deal I of A. (7) Every prmary and pseudoprme deal of A has the form (I ) for some {, 2,...,n} and prmary and pseudoprme deal I of A. (8) Every z-deal of A s an ntersecton of fntely many z-deals of the form (I ) for some {, 2,...,n} and z-deal I of A. Proof. (): Ths follows from the fourth somorphsm theorem. (2), (3): Let P be a prme deal of A. Note that 2 ({0}) πn ({0}) ={0} P.SnceP s prme, π show s prme. Suppose a, b A and a b π (P). Thereexstsa, b, c A such that π (a) = a, π (b) = b, π (c) = a b, ({0}) s an deal for =, 2,...,n, and that ({0}) ({0}) P for some. Nowπ (P) s an deal of A that we wll and c P.Thenc ab ({0}) P.Sncec P,thenab P.SnceP s prme, a P or b P.Thsmplesa π (P) or b π (P). Henceπ (P) s prme. By (), P = (π (P)). Next, we note that f P s a mnmal prme deal, then π (P) must also be a mnmal prme deal; for f not, there s a prme deal Q of A that s a proper subset of π (P), whch then would mply (Q ) s a prme deal properly contaned n (π (P)) = P, contrary to P beng a mnmal prme deal of A. (4): Let M be a maxmal deal of A. Frst suppose for each that π (M) = A.Thenforeach, theresap M + such that π (p ) = π (). Snce n a commutatve rng wth dentty element, every maxmal deal s a prme deal, M s a prme deal. Then ( p )( p 2 ) ( p n ) = 0mples p M for some. Butthenp M mples M, a contradcton. Hence there must exst a such that π (M) s a proper subset of A.DefneM = π (M). ThenM s an deal of A.Now M (M ) and snce M s a maxmal deal of A, M = (M ). (5): Suppose I s a pseudoprme deal of A. Then by (2), there s a prme deal of the form deal P of A ) contaned n I. Because deal of A snce P π (I). ({0}) (P ) I, () mples I = (P ) (for some and prme (π (I)). Nowπ (I) s a pseudoprme (6): Let I be a semprme deal of A. ThenI s an ntersecton of prme deals of A, and by (2), each of these prme deals has the form (P) for some and prme deal P of A.For =, 2,...,n, let{p α : α Λ } denote the collecton of prme deals of A used n formng ths ntersecton. For =, 2,...,n let I = α P α.theneachi s a semprme deal of A and I = n = (I ). (7): Suppose I s a prmary and pseudoprme deal of A. Snce I s pseudoprme, (5) mples I = (I ) for some pseudoprme deal I of A. We wll show that I s also prmary. Suppose that a b I.Leta, b A such that π (a) = a and π (b) = b.thenab (I ) = I. SnceI s prmary, a I or b m I for some m. Thenπ (a) = a I or (π (b)) m = b m I. Hence I s pseudoprme and prmary n A.

8 S. Larson / Topology and ts Applcatons 58 (20) (8): Let I be a z-deal of A. Snceanyz-deal s semprme, I s the ntersecton of the prme deals mnmal wth respect to contanng I, and by (2), each of these prme deals has the form (P) for some and prme deal P of A.Noweach of these prme deals s a z-deal by Theorem. of [3], whch states that n a commutatve rng every mnmal deal n the class of prme deals contanng a z-deal s a z-deal. For =, 2,...,n, let{p α : α Λ } denote the collecton of prme deals of A used n formng ths ntersecton. For =, 2,...,n let I = α P α. Then each I s an ntersecton of z-deals and hence s a z-deal of A.ThenI = n = (I ). The next example demonstrates that we must tae care when usng these methods to construct a mnmal prme deal of a fnte fbre product rng. In fact, when A s a fnte fbre product constructed from the rngs A, A 2,...,A n and P s a mnmal prme deal of A, the prme deal (P ) s not necessarly a mnmal prme deal of A. The example we present s n fact an f -rng, demonstratng that even the addton of a partal order structure does not guarantee that for a mnmal prme deal P of a coordnate rng A,thedeal (P ) s a mnmal prme deal. Example 3. Let A be the fntely -convex f -rng defned n Example 5. Let P ={0} n A and P 2 ={f A 2 : n 0 N such that f (n) = 0 n n 0 } n A 2.ThenP, P 2 are mnmal prme deals of A, A 2 respectvely. Then (P ) ={(a, a 2 ) A: a = 0, a 2 s eventually 0} s a prme deal of A and 2 (P 2) ={(a, a 2 ) A: a x, a 2 s eventually 0} s also a prme deal of A. However, (P ) s a proper subset of 2 (P 2) and so 2 (P 2) s a prme deal, but not a mnmal prme deal of A. In certan homogeneously fnte fbre product rngs, we can characterze all mnmal prme deals n terms of the mnmal prme deals of the coordnate rngs. In the proof of the next theorem we wll mae use of the fact that n a commutatve rng wth dentty element, a prme deal P s a mnmal prme deal f and only f for every p P,thereexstsq / P such that pq = 0. Ths s a re-statement of the fact that a prme deal P n the commutatve rng A wth dentty element s a mnmal prme deal f and only f A P s a maxmal multplcatve system. (A multplcatve system of a commutatve rng A s a set of elements closed under multplcaton.) See Chapter V of [4] for more detal. Theorem 4. Suppose A s a commutatve semprme rng wth dentty element. Suppose A s a homogeneously fnte fbre product constructed from n copes of the rng B and the semprme deal Q. If P s a mnmal prme deal of B, then (P) s a mnmal prme deal of A for =, 2,...,n. Proof. Suppose P s a mnmal prme deal of B. By Lemma, (P) s a prme deal of A. We need only show that (P) s mnmal. We do so by showng for any p (P) there s a q A (P) such that pq = 0. Let p (P). Then π (p) P and snce P s a mnmal prme deal of B, theresab B P such that b π (p) = 0. Consder the case where Q P. Then there s a q Q P.Letq denote the element of A such that π (q) = b q and for all, π (q) = 0. Then q A (P) and pq = 0. Next, consder the case where Q P.Thenforeach, π (p) = π (p) + q for some q Q.Foreach, q P and so there exsts r B P such that r q = 0. Let r = b r.thenr B P and the element q defned by π (q) = r for every satsfes pq = 0, whle q A (P). Suppose A s a commutatve semprme rng wth dentty element that s a fnte fbre product constructed from the rngs A, A 2,...,A n. It should be noted that whle part (7) of Theorem 2 asserts that every prmary and pseudoprme deal of A has the form (I ) for some {, 2,...,n} and prmary and pseudoprme deal I of A, not every prmary deal of A need be pseudoprme and not every prmary deal of A need be of the form (I ) for a prmary deal I of A.Ths s the case even when A has a partal orderng; more specfcally ths s the case even when A s a fntely -convex f -rng constructed from -convex f -rngs as our next example demonstrates. Example 5. Let R[x] denote the rng of polynomals over the reals n one ndetermnate. Totally order R[x] lexcographcally, so that x x 2.NowletB ={ p(x) q(x) : p(x), q(x) R[x], q(0) 0andq > 0} under the usual addton and p multplcaton of quotents of polynomals and under the order nduced by the order on R[x]. That s, q p 2 f and only q 2 f p q 2 p 2 q.thenb s a totally ordered -convex f -rng. Let Q = x n B, andleta ={(a, b) B B: a b Q }. Then A s (homogeneously) fntely -convex. In A, consder the l-deal I ={(a, b) A: a, b nx 2 for some natural number n}. ThenI s an l-deal of A and we wll show that I s prmary. Suppose ( f, g)(h, ) I, and ( f, g) / I. Then ether f nx 2 or g nx 2 for all natural numbers n. We may suppose that f nx 2 for all natural numbers n. Thenh must be n Q and by the defnton of A, Q.So,(h, ) N I for some natural number N. ThusI s a prmary deal. Also, I s not pseudoprme snce (x, 0)(0, x) = (0, 0), whle (x, 0) / I and (0, x) / I. Now we wll show I cannot be wrtten n the form (I ) or 2 (I ) for some prmary deal I of B. Suppose there s a prmary deal I of B such that I = (I ). Then I = π (I) ={a B: a nx 2 for some n}. Then (I ) ={(a, b) A: a nx 2, b nx for some n} I. Smlarly, there s no prmary deal I of B such that I = 2 (I ).

9 896 S. Larson / Topology and ts Applcatons 58 (20) Maxmal and mnmal prme deals n fntely -convex f -rngs Our last two sectons focus on several types of deals n fntely -convex f -rngs. As s the case wth all f -rngs, every l-deal of a fntely -convex f -rng s an deal, but not every deal n a fntely -convex f -rng s an l-deal. However, there are several classes of deals of a fntely -convex f -rng that are necessarly l-deals. These nclude maxmal, prme, semprme, and z-deals. As Theorem 8 ndcates, fntely -convex f -rngs are SV f -rngs and have the bounded nverson property, and t s well nown that maxmal deals n f -rngs wth the bounded nverson property are l-deals (see [4]). By Theorem 5.9 of [2], every prme and every pseudoprme deal of a semprme SV f -rng wth bounded nverson s an l-deal. Snce semprme deals and z-deals are ntersectons of prme deals and ntersectons of l-deals are l-deals, t then follows that semprme and z-dealsofafntely-convex f -rng are also l-deals. In ths secton we focus our attenton on maxmal deals and mnmal prme deals n fntely -convex f -rngs. We wll show that there are many maxmal deals n a fntely -convex f -rng that contan ust one mnmal prme deal. That s, for a large class of fntely -convex f -rngs there s a dense open set of maxmal deals of ran n Max(A) under the hull-ernel topology. Frst, we present a lemma to demonstrate that the deals of the form (M ) need not all be dstnct n a fntely -convex f -rng. Lemma 6. Suppose A s a commutatve semprme f -rng wth dentty element. Suppose A s a fntely -convex f -rng constructed from the -convex f -rngs A, A 2,...,A n such that A s l-somorphc to a sub- f -rng of A A 2 A n and M s a maxmal deal of A. Then for, π (er(π )) M f and only f M = π ( (M )) s a maxmal deal of A and (M ) = (M ). Proof. Suppose and π (er(π )) M. Defne M = π ( (M )). Then M s an deal. We wll show M s a maxmal deal of A. Suppose p A M. Let p A such that π (p) = p. It follows from the defnton of M that π (p) / M.SnceM s a maxmal deal of A,thereexstsr A and m M such that r π (p) + m = π (). Letr,m A such that π (r) = r, π (m) = m.thenπ ( rp m) = 0 M, whch mples π ( rp m) M.Snceπ (m) s also n M, ths tells us that π () s n the deal generated by π (p) and M.HenceM s a maxmal deal of A. Because A s -convex, there s a unque mnmal prme deal call t P ofa contaned n M.Then (P ) s a prme deal of A. Nextwewllshow (P ) (M ).Letp (P ) and suppose π (p) / M. It follows that there s an r,m A such that π (m) M and π (r)π (p) + π (m) = π (). Thenπ ( rp m) = 0 M mples π ( rp m) M. But because p (P ) and π (m) M mples π (m) M, ths shows π () M, a contradcton. So π (p) M and (P ) (M ).Sncetheprmel-deals contanng (P ) form a chan, and l-deals contanng (P ), t must be that (M ) = (M ). Suppose for some, that M = π ( (M )) and (M ) = (M ).Letq π (er(π )). Theresaq er(π ) such that π (q) = q.thenπ (q) = 0 M and hence q (M ) = (M ).Thenq = π (q) M.Thusπ (er(π )) M. (M ), (M ) are both maxmal Our next theorem gves a condton under whch a maxmal deal of a fntely -convex f -rng wll have ran. Theorem 7. Suppose A s a commutatve semprme f -rng wth dentty element. Suppose A s a fntely -convex f -rng constructed from the -convex f -rngs A, A 2,...,A n such that A s l-somorphc to a sub- f -rng of A A 2 A n and for some that P s the mnmal prme deal contaned n the maxmal deal M of A.Ifforevery one of the followng two condtons are met, then ran A ( (M )) = : () π (er(π )) M. (2) π (er(π )) P and π (er(π )) R,whereR s the mnmal prme deal contaned n M = π ( (M )). Proof. We wll show every mnmal prme deal contaned n mnmal prme deal contaned n mnmal prme deal contaned n M, R = P and (M ) s equal to (M ) for some mnmal prme deal R of A.ThenR M and snce P s the unque (R ) = (P ). (P ). Suppose frst that (R ) s a Suppose next that (R ) s a mnmal prme deal contaned n (M ) for some and mnmal prme deal R of A. Note that π (er(π )) M, snce f q π (er(π )), thereexstsq A such that π (q) = q and π (q) = 0 R, whch mples q (R ) (M ) and π (q) = q M. Ths mples that condton (2) n the statement of the theorem must hold. So π (er(π )) P. By Lemma 6, Thus (R ) (M ) = Defne R = π ( (M ) = (M ) where M = π ( (M )) s a maxmal deal of A. (M ) and t follows that R M and R s the mnmal prme deal contaned n M. (R )) and P = π ( So suppose a, b A and a b R. Then there exsts a c (P )). It s easy to see that R s an deal; we now show t s a prme deal. (R ) such that π (c) = a b and there exsts a, b A such

10 S. Larson / Topology and ts Applcatons 58 (20) that π (a) = a, π (b) = b. Then c ab er(π ) and so by condton (2), π (c ab) R. Snceπ (c) R, t follows that π (ab) = π (a)π (b) R. Therefore π (a) R or π (b) R. Ths mples a (R ) or b (R ) and then π (a) = a R or π (b) = b R. A smlar argument shows P s also a prme deal of A. Snce P s the unque mnmal prme deal contaned n M and R M,wehaveP R. Suppose now that P R. Then there s a p R P and there s a p A such that π (p) = p and π (p) R.Ifπ (p) P, then there s an a A such that π (a) = π (p) and π (a) P.Thenπ (a p) = 0 and π (a p) π (er(π )) P.Butsnceπ (a) P,ths would mply π (p) = p P, a contradcton. So, π (p) / P.ThsmplesP R,butP M.SnceR s the unque mnmal prme deal contaned n M,wehaveR P, contrary to π (p) R P. Therefore, P = R. Next we show that (R ) = (R ).Letr and π (s) R.Thenπ (r s) π (er(π )) R.Snceπ (s) R,thenπ (r) R.Sor Now let r (R ) = (P ). (R ). By the defnton of R, π (r) R and r By Theorem 2, every mnmal prme deal contaned n (R ).Thenπ (r) R and so there s an s A such that π (s) = π (r) (R ) and (R ) (R ). (R ).So (R ) = (R ).Wenowhave (R ) = (M ) s of the form (P ) for some and some mnmal prme deal P of A, and hence we have shown that any mnmal prme deal contaned n Thus, ran A ( (M )) =. (M ) s equal to (P ). The prevous theorem allows us to characterze the ran of every maxmal deal n a homogeneously fntely -convex f -rng. Corollary 8. Suppose A s a commutatve semprme f -rng wth dentty element. Suppose A s a homogeneously fntely -convex f -rng constructed from n copes of the -convex f -rng B and the semprme deal Q and that M s a maxmal deal of B. () If Q M, then ran A ( (M)) =. (2) If P s the mnmal prme deal contaned n M and Q Pthenran A ( (M)) =. (3) If Q M, P s the mnmal prme deal contaned n M, and Q Pthenran A ( (M)) = n. Proof. () s a drect consequence of the prevous theorem snce for all, π (er(π )) = Q. (2): Suppose Q P. We wll show that for all, condton (2) of the prevous theorem s satsfed. If, then π (er(π )) = Q P M. Now by Lemma 6, M = π ( (M)) s a maxmal deal of B. Then t s easy to see that M = M (n B). Then P s the mnmal prme deal contaned n M and π (er(π )) = Q P. Thus for all, condton (2) of the prevous theorem s satsfed and hence ran A ( (M)) =. (3): If P s any mnmal prme deal of B dfferent from P,thensnceB s -convex, P M. Lettng p P M, wesee the element p such that π (p) = p for all s contaned n (P ) (M) and hence that (P ) (M) for all. So the only possble deals that could be mnmal prme deals contaned n (M) are of the form (P), where =, 2,...,n. By Theorem 4, each (P) s a mnmal prme deal. Next, note that each of the (P) s contaned n (M). Forfp (P), thenπ (p) π (p) Q M and π (p) P M mples that π (p) M and therefore that p (M). If 2 and q Q P, then the element q of A such that π (q) = q when = and π (q) = 0 when, s contaned n 2 (P) (P). Ths shows that (P) 2 (P) when 2 and the deals of the form (P), for =, 2,...,n are all dstnct. We have found that there are exactly n mnmal prme deals of A contaned n (M). Hence ran A ( (M)) = n. When X s a compact space that s fntely an F-space, there s a dense open set of ponts of X of ran (see 5.6 n [5]). That s to say, there s a dense open set of ponts of X for whch the assocated maxmal deal M x ={f C(X): f (x) = 0} has ran. In a compact space, every maxmal deal of C(X) s of the form M x for some x X and there s a natural homeomorphsm between the maxmal deal space Max(C(X)) and X. Snce a compact space X s fntely an F-space f and only f C(X) s fntely -convex, the followng theorem extends the result that n a compact space that s fntely an F-space, there s a dense open set of ponts of ran. Recall that a commutatve f -rng wth dentty element s semsmple f the ntersecton of all ts maxmal deals s {0} and that every C(X) s semsmple. Theorem 9. Suppose A s a commutatve semprme f -rng wth dentty element. Suppose A s a fntely -convex f -rng constructed from the -convex f -rngs A, A 2,...,A n such that A s l-somorphc to a sub- f -rng of A A 2 A n and that each of the A s semsmple. Then there s a dense open set of maxmal deals of ran n Max(A) (under the hull-ernel topology). Proof. Suppose A s fntely -convex and that each of the A are semsmple. Let V denote the set of maxmal deals of ran. We wll show that nt(v ), the nteror of V,sdensenMax(A) by showng that for each a A wth a 0, h c (a)

11 898 S. Larson / Topology and ts Applcatons 58 (20) meets nt(v ). Soleta A wth a 0. Let (a) d denote the annhlator of the prncpal deal (a). For ease of notaton, for any, {, 2, 3,...,n} we let Q denote the deal Q = π (er(π )). Suppose frst that for all, wth, (Q ) (a) d.sncea 0, there s a {, 2,...,n} and a maxmal deal M of A such that π (a) / M.Hencea / where M s a maxmal deal of A.Thena / (M ) and h c (a). Wewllshowh c (a) V. Suppose (M ) h c (a), (M ). To show that (M ) V, we show condton (2) of Theorem 7 (Q ) (a) ={0} (P ). s satsfed. Let P denote the mnmal prme deal contaned n M. Then for all, Snce (a) deal of A Now f a mply a b (P ), t must be that (Q ) (Q ) (a) ={0} (P ) and hence Q P.IfM = π ( (M )) then M s a maxmal (R ) where R s the mnmal prme deal contaned n M. by Lemma 6 and (R ),thenπ (a) M and so there would exst b A such that π (b) M and π (b) = π (a). Thswould (Q ) (a) d and (a b)a = 0. Then snce a / (M ), a contradcton to the fact that b (M ) and a / (M ) and (M ) s prme, we would have a b (M ).Thusa / (R ).Because (Q ) (a) ={0} (R ) and a / (R ), t follows that Q R. We have shown for every that Q = π (er(π )) P and Q = π (er(π )) R ; that s, we have shown that condton (2) of Theorem 7 s satsfed, and therefore (M ) V. So h c (a) V and because h c (a) s open, h c (a) nt(v ). Next suppose that there s an, wth and (Q ) (a) d.letb ={Q, Q 2 2,...,Q m m } denote a maxmal set of the Q such that (Q ) (Q ) (Q ) (Q m m m ) (a) {0}. Letz (Q ) (Q ) (Q ) (Q m m m ) (a) wth z 0. It follows from the hypothess that each A s semsmple and z 0 that h c (z). We wll show h c (z) V. Suppose (M ) h c (z) for some and maxmal deal M of A. To show that (M ) V, we show one of the two condtons of Theorem 7 s satsfed. Let P denote the mnmal prme deal contaned n M. Then for all such that Q B, wehaveq M snce z (Q ) (M ).Soforall such that Q B, the frst condton of Theorem 7 s satsfed. Suppose now that and Q / B. By our choce of B, we have (Q ) ( (Q ) (Q ) (Q ) (Q m m m ) (a)) ={0} (P ).Snce (P ) s a prme (Q m m ) (a) (M ),tmustbethat (Q ) (P ) and deal and (Q ) (Q ) (Q ) m hence Q P M. Now suppose M = π ( (M ) = (M ).Nowfq Then q r (M ) = means that Q / B, for f Q (M ) = (R ).Butsnce (M )) and R s the mnmal prme deal contaned n M. By Lemma 6, (Q ) then π (q) Q and so there s an r A such that π (r) = π (q) and π (r) = 0. (M ) and snce r (M ),wemusthaveq (M ).Thus (Q ) (M ).Ths were n B, we would have z (Q ) (M ) whch would mply (Q ) (M ), a contradcton. Then (Q ) ( (Q ) 2 (Q 2 2 ) 3 (Q 2 2 ) (Q 3 3 ) m (Q m m ) (a)) ={0} (R ) (M ) = (M ) and (Q ) (Q ) m (R ) and Q R.Soforall such that Q / B, we have shown that Q = π (er(π )) P (Q m m ) (a) (M ), we have (Q ) and Q = π (er(π )) R ; that s we have shown condton (2) of Theorem 7 s satsfed. Now for all, wehave shown that one of the condtons of Theorem 7 s satsfed. Therefore (M ) V and h c (z) V.Snceh c (z) s open, h c (z) nt(v ). By our choce of z, h c (z) h c (a). It follows that h c (z) nt(v ) h c (a); soh c (a) meets nt(v ). We conclude ths secton wth an example to demonstrate that the hypothess that the A be semsmple cannot be left out of the prevous theorem. Example 20. Let A be the fntely -convex f -rng defned n Example 5. The f -rng A satsfes all of the hypotheses of the prevous theorem except the hypothess that B s semsmple. Then x s the unque maxmal deal of B and M = x x s the unque maxmal deal of A. The maxmal deal M contans two mnmal prme deals: {0} x and x {0}. Sothe maxmal deal space of A conssts of a sngle maxmal deal and has no element of ran. 6. Sums of semprme, prme, prmary, and z-deals We now turn to loo at the sums of several types of deals. A commutatve f -rng satsfes the 2nd-convexty property f for any u, v A such that v 0 and 0 u v 2,there exsts a w A such that u = wv. Theorem 2. Let A be a commutatve semprme f -rng wth dentty element. Suppose A s a fntely -convex f -rng constructed such that at each stage of the constructon, the surectve l-homomorphsms map to a semprme f -rng. Then: () A satsfes the 2nd-convexty property. (2) The sum of any two semprme deals of A s a semprme deal. (3) The sum of any two prme deals of A s a prme deal. (4) The sum of any two prmary l-deals of A s a prmary l-deal.