NOETHERIAN SPACES OF INTEGRALLY CLOSED RINGS WITH AN APPLICATION TO INTERSECTIONS OF VALUATION RINGS

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1 NOETHERIAN SPACES OF INTEGRALLY CLOSED RINGS WITH AN APPLICATION TO INTERSECTIONS OF VALUATION RINGS BRUCE OLBERDING Abstract. Let H be an integral domain, and let Σ be a collection of integrally closed T overrings of H. We show that if A is an overring of H such that H = ( R Σ R) A, and if Σ is a Noetherian subspace of the space of all integrally closed overrings of H, then there exists a weakly T Noetherian subspace Γ of integrally closed overrings of H such that H = ( R Γ R) A, and no member of Γ can be omitted from this intersection. Restricting to the case where Σ consists of valuation overrings, we obtain stronger results. 1. Introduction Throughout this article H denotes an integral domain, and F denotes its quotient field. An overring of H is a ring R such that H R F. A classical theorem of Krull states that the domain H is integrally closed if and only if H is an intersection of its valuation overrings (i.e., those overrings of H that are valuation domains). In general, an integrally closed domain need not have an irredundant representation as an intersection of valuation rings. In fact, this property is rather special: see [2], [6], [7], [8], [10], [14] and [15]. An important exception, however, is the class of Krull domains. Recall that a collection Σ of overrings of H has finite character if each nonzero element of H is a unit in all but finitely many members of Σ. A fundamental fact arising from the study of Krull domains is that if an integral domain H is a finite character intersection of Noetherian valuation overrings, then H can be written as an irredundant finite character intersection of Noetherian valuation overrings, meaning that none of these rings can be omitted from the representation. More generally, Heinzer and Ohm in [10] proved that if a domain H is an intersection of an overring A with a finite character intersection of valuation overrings, each having Krull dimension 1, then H can be written as an intersection of A and an irredundant finite character intersection of valuation overrings, each having Krull dimension 1. In another direction, Brewer and Mott proved in [2] that if an integral domain is a finite character intersection of valuation overrings, then the domain can be expressed as an irredundant finite character intersection of valuation overrings Mathematics Subject Classification. Primary 13G05, 13F05, 13F30. June 9,

2 2 BRUCE OLBERDING These results raise the question: Given an overring A of H and a finite character collection Σ of valuation overrings of H such that H = ( V Σ V ) A, does there exist a finite character collection Γ of valuation overrings of H such that H = ( V Γ V ) A, and no member of Γ can be omitted from this intersection? We show that the answer is affirmative, thus generalizing the results of [2] and [10]. We in fact consider irredundant intersections of integrally closed overrings, not just valuation overrings, and we are able to replace the finite character condition with the weaker requirement that the collection of overrings is a Noetherian subspace of the space of all integrally closed overrings of H. (The topology we have in mind here is introduced in Section 2). Restricted to valuation overrings, our main result, Theorem 4.3, states: If H is a domain, A is an overring of H and Σ is a Noetherian space of valuation overrings of H such that H = ( V Σ V ) A, then there exists a Noetherian space Γ of valuation overrings of H such that H = ( V Γ V ) A, and no member of Γ can be replaced by one of its proper overrings. Thus for Noetherian spaces of valuation rings we obtain a strong form of irredundance. For examples and applications of these results to the classification of integrally closed overrings of two-dimensional Noetherian domains, see [14] and [15], where Theorem 4.3 is used heavily. I thank the referee for very helpful comments and corrections that significantly improved the paper. 2. The space of integrally closed overrings In this section we define a topology on the set of integrally closed overrings of the domain H, and establish a correspondence between these overrings and prime semigroup ideals of the Kronecker function ring of H. (2.1) The space of integrally closed overrings. We denote by Z(H) the set of all integrally closed overrings of H. We define the Zariski topology on Z(H) by declaring the basic open sets of Z(H) to be of the form: U H (x 1,..., x n ) := {R Z(H) : x 1,..., x n R}, where x 1,..., x n F. The set Zar(H) of all valuation overrings of H is a subspace of Z(H), and the topology it inherits coincides with the usual one on Zar(H) (see for example [13, 18]). The Zariski topology on Z(H) is for our purposes too coarse, so we define a finer topology. Let I be a fractional ideal of H, and let R Z(H). Denote by cl R (I) the intersection of all the IV, V Zar(R). Then when also I R, it follows that cl R (I) is the integral closure of the ideal IR in R [17, Proposition 6.8.2]. We define the b-topology on Z(H) by declaring the subbasic open sets of Z(H) to be the sets of the form: U H (I, J) := {R Z(H) : I cl R (J)}, where I and J are finitely generated H-submodules of F. (The use of b here in btopology is motivated by the traditional notation used with respect to Kronecker

3 NOETHERIAN SPACES OF INTEGRALLY CLOSED OVERRINGS 3 function rings; see (2.4) below.) In case J = H, we write U H (I) for U H (I, J), and if I = (x 1,..., x n )H, then we write U H (x 1,..., x n ) for U H (I). In case Σ Z(H), we write U Σ (I, J) for Σ U H (I, J). We see later that the b-topology on Z(H) induces on Zar(H) the usual Zariski topology, as defined above. (2.2) Prüfer domains. An integral domain H is a Prüfer domain if every nonzero finitely generated ideal of H is invertible; H is a Bézout domain if every finitely generated ideal is principal. When H is a Prüfer domain (a case that we will need in the proof of Theorem 4.2) the b-topology on Z(H) coincides with the Zariski topology. For suppose that I and J are finitely generated H-submodules of F, with J 0. Then J is an invertible fractional ideal of H and JR = cl R (J) [5, Theorem 24.7]. Hence U H (I, J) = U H (IJ 1 ), where J 1 = {x F : xj H}. (2.3) The space of prime semigroup ideals. We also extend the Zariski topology on the space of prime ideals of a domain H to the space of prime semigroup ideals of H. A subset I of H is a semigroup ideal of H if for all x H, xi I. A semigroup ideal I is prime if I H and for all x, y H \ I, xy H \ I. Observe that a nonempty subset P of H is a prime semigroup ideal of H if and only if H \ P is a saturated multiplicatively closed subset of H. Thus P is a prime semigroup ideal of H if and only if P is a union of prime ideals of H. Following the usual convention for prime ideals, we write H P for H H\P when P is a prime semigroup ideal of H. We define S(H) to be the set of all prime semigroup ideals of H, and we give S(H) the hull-kernel topology by declaring the basic open sets to be those sets of the form: U H (x 1,..., x n ) := {P S(H) : x i P for some i = 1,..., n}, where x 1,..., x n H. For every subset A of H, define V H (A) = {P S(H) : A P }. Then C is a closed subset of S(H) if and only if C = V H ( P C P ); see for example Section 2 of [12]. It follows that S(H) induces on Spec(H) the Zariski topology. (2.4) Kronecker function rings. The passage to the Kronecker function ring of a domain is our main tool in this article, and it relies on the following properties. Let H be an integrally closed domain with quotient field F, and let X be an indeterminate for F. For each V Zar(H) with maximal ideal M V, define a valuation ring V b with quotient field F (X) by V b := V [X] MV [X], and define H b = {V b : V Zar(H)}. The ring H b (the Kronecker function ring of H with respect to the b-operation) is a Bézout domain with quotient field F (X). Also, V b F = V for all V Zar(H). Hence H b F = H. Moreover, if I is a finitely generated H-submodule of F, then cl H (I) = IH b F. These assertions follow from Theorems 32.7 and of [5]. Proposition 2.5. Let H be an integrally closed domain. The mapping h : Z(H) Z(H b ) : R R b is a homeomorphism of Z(H) onto its image in Z(H b ) with respect to the b-topology.

4 4 BRUCE OLBERDING Proof. Since for every R Z(H), R b F = R, it is clear that h is injective. To prove that h is continuous, it suffices to show that the preimages of subbasic open subsets of Z(H b ) are open. By (2.2), since H b is a Bézout domain, we need only consider subbasic open sets of the form U H b(α), where α = f g for some f, g H[X] with g 0. Define I and J to be the ideals of H generated by the coefficients of f and g, respectively. Then for each R Z(H), IR b = fr b and JR b = gr b (see the proof of Theorem 32.7(c) in [5]) and JR b F = cl R (J), so h 1 (U H b(α)) = {R Z(H) : fr b gr b } = {R Z(H) : I JR b R}. Since JR b R = cl R (J), this proves that h 1 (U H b(α)) = U H (I, J). Hence h is a continuous mapping. Since h is injective, to prove that h is open it suffices to show that preimages of subbasic open sets are open. Let A, B be finitely generated H-submodules of F. Then since cl R (B) = BR b F and cl R b(bh b ) = BR b for each R Z(H), we have h(u H (A, B)) = {R b : R Z(H) and A cl R (B)} = {R b : R Z(H) and A BR b } = (Im h) U H b(ah b, BH b ). Therefore h is open, and this proves the proposition. Remark 2.6. In general, the mapping h in Proposition 2.5 need not be surjective. For example, suppose that H is a local Noetherian domain of Krull dimension 2 with quotient field F, and R is the overring of H b that is formed by taking the intersection of the extensions (H P ) b to F (X) of the localizations of H at height 1 prime ideals P. Then R is a PID overring of H b, and it has the property that R F = H. Thus if there exists A Z(H) such that h(a) = A b = R, then A = A b F = R F = H. But h(h) = H b and H b R, since R is a PID and H b has valuation overrings of Krull dimension > 1. Thus R is not in the image of h. In [3] D. Dobbs and M. Fontana showed that Zar(H) is homeomorphic to Zar(H b ) via the mapping V V b. Corollary 2.7. Let H be an integrally closed domain. Then for every pair I and J of finitely generated H-submodules of F, there exist finitely generated H-submodules K 1,..., K n of F such that n Zar(H) U H (I, J) = Zar(H) ( U H (K i )). Proof. The mapping h of Proposition 2.5 has the property that h(zar(h)) = Zar(H b ) [5, Theorem 32.10]. Let U = U H (I, J), where I and J are nonzero finitely generated H-submodules of F. Then as above there exists α F (X) such that h(u) = U H b(ih b, JH b ) = U H b(α). In the proof of Lemma 1 of [3] it is shown that there exist finitely generated H-submodules K 1,..., K n of F such that n U Zar(H) = h 1 (U H b(α) Zar(H b )) = Zar(H) ( U H (K i )). i=1 i=1

5 NOETHERIAN SPACES OF INTEGRALLY CLOSED OVERRINGS 5 Using the corollary, we deduce that the Zariski and b-topologies coincide when restricted to Zar(H): Corollary 2.8. If H is an integrally closed domain, then the subspace topology on Zar(H) induced by the b-topology on Z(H) is the same as the Zariski topology on Zar(H). In order to establish a correspondence between integrally closed overrings of H and prime semigroup ideals of the Kronecker function ring H b of H, we introduce the following notation. For each integrally closed overring R of H, we set P R = {x H : xr R}. Then P R is a prime semigroup ideal of H, as defined in (2.3). Recall that an integral domain H is a QR-domain if every overring R of H is a quotient ring of H, namely, R = H PR. A domain H is a QR-domain if and only if H is a Prüfer domain such that the radical of every finitely generated ideal of H is the radical of a principal ideal [5, Theorem 27.5]. Thus a Bézout domain is a QR-domain, and in particular the Kronecker function ring H b is a QR-domain. Therefore, since ultimately in what follows we are interested in prime semigroup ideals of H b, the restriction to QR-domains in the next two lemmas is not an impediment to our approach here. Lemma 2.9. Let H be a QR-domain, and let R be an overring of H. (1) For every ideal I of H, I P R if and only if IR R. (2) For all 0 x F, x R if and only if H x 1 H P R. Proof. For every prime semigroup ideal P of H, I P if and only if IH P H P. Also, for 0 x F, we have x H P if and only if H x 1 H P. Thus (1) and (2) follow from the fact that since H is a QR-domain, H PR = R. Lemma If H is a QR-domain, then the mapping g : Z(H) S(H) : R P R is a homeomorphism with respect to the b-topology. Proof. Since H is a QR-domain, we have for each R Z(H), R = H PR, so g is a bijection. To show that g is continuous, it suffices to show that the preimages of basic open sets in S(H) are open. Let x 1,..., x n H. Since H is a QR-domain, there exists x H such that U H (x 1,..., x n ) = U H (x), and by Lemma 2.9(2) g 1 (U H (x)) = {R Z(H) : x P R } = {R Z(H) : x 1 R} = U H (x 1 ). Thus g is a continuous mapping. We claim next that g 1 is a continuous mapping. Let U be a subbasic open subset of Z(H). By (2.2) there exists a finitely generated H-submodule I of F such that U = U H (I). Write I = a 1 H + + a m H for some a 1,..., a m I. Then by Lemma 27.2 in [5], the fact that H is a QR-domain implies that for each i = 1,..., m, H[a i ] = H[yi 1 ] for some y i H. Therefore, setting

6 6 BRUCE OLBERDING y = y 1 y m, we have H[I] = H[a 1,..., a m ] = H[y 1 ], and so U H (I) = U H (y 1 ). As above, g(u H (y 1 )) = U H (y). Therefore, g 1 is a continuous mapping. Theorem If H is an integrally closed domain, then the mapping Z(H) S(H b ) : R P R b = {x H b : xr b R b } is a homeomorphism of Z(H) onto its image in S(H b ) with respect to the b-topology on Z(H). Proof. Apply Proposition 2.5 and Lemma Noetherian collections of overrings A topological space X is Noetherian if X satisfies the ascending chain condition for open sets. Let H be a domain and Σ be a collection of integrally closed overrings of H. We say that Σ is a weakly Noetherian collection if Σ is a Noetherian subspace of Z(H) in the Zariski topology. If Σ is a Noetherian subspace of Z(H) in the b-topology, then we say that Σ is a Noetherian collection. Clearly a Noetherian collection is weakly Noetherian, since the b-topology is finer than the Zariski topology on Z(H). The first proposition of this section extends from Spec(H) to S(H) a result of D. Rush and L. Wallace in [16]. It involves some familiar notions that we modify to fit the setting of S(H): For X S(H) and E H, define V X (E) = {P X : E P } and Rad X (E) = P V X (E) P, where if V X(E) is empty, set Rad X (E) = H. Proposition 3.1. (cf. [16, Propositions 1.1 and 1.2]) Let X be a collection of prime semigroup ideals of the domain H. Then the following statements are equivalent. (1) X is a Noetherian subspace of S(H). (2) For every prime semigroup ideal P of H, there exists a finite set E P such that Rad X (P ) = Rad X (E). (3) For every semigroup ideal I of R, there exists a finite subset E I such that Rad X (E) = Rad X (I). Proof. (1) (2) Fix P S(H), and define: F = {Rad X (x 1,..., x n ) : x 1,..., x n P }. As discussed in (2.3), C is a closed subset of S(H) if and only if C = V H ( Q C Q). Thus since X is a Noetherian space, there exists a finite subset E of P such that Rad X (E) is a maximal element in F. Clearly, Rad X (E) Rad X (P ). The reverse inclusion holds also since P Rad X (E). Indeed, if y P, then by the maximality of Rad X (E) in F, we have Rad X (E) = Rad X (E {y}), so that y Rad X (E), whence P Rad X (E). (2) (3) Say that a semigroup ideal I of H has property ( ) if there exists a finite subset E of I such that Rad X (E) = Rad X (I). Let F be the set of semigroup ideals of H that do not have property ( ), and suppose that the set F is nonempty (i.e.,

7 NOETHERIAN SPACES OF INTEGRALLY CLOSED OVERRINGS 7 suppose that (3) fails). Chains in F are closed under unions, so by Zorn s Lemma there exists a maximal element I of F. We claim that I is a prime semigroup ideal of H, thus showing (2) does not hold. Suppose by way of contradiction that I is not a prime semigroup ideal of H, and let y 1, y 2 H \ I such that y 1 y 2 I. Define I 1 = y 1 H I and I 2 = y 2 H I, and note that I 1 and I 2 are semigroup ideals of H properly containing I. Therefore, by ( ) there exists for each i, a finite subset E i of I i such that Rad X (E i ) = Rad X (I i ). Define E = {xy : x E 1, y E 2 }. Then E {xy : x I 1, y I 2 } y 1 y 2 H I I, so Rad X (E) Rad X (I). Since I F, this containment is proper, so there exists P X such that E P but I P. Since P is a prime semigroup ideal of H, E 1 P or E 2 P. Assume that E 1 P. Then I Rad X (I 1 ) = Rad X (E 1 ) P, a contradiction. Therefore, I is a prime semigroup ideal of H, and we conclude that (2) does not hold. (3) (1) To verify (1), it suffices to show that the collection J := {Rad X (I) : I is a semigroup ideal of H} satisfies the ascending chain condition. Let C be a chain of elements in J, and define J = I C I. By (3) there exist x 1,..., x n J such that Rad X (J) = Rad X (x 1,..., x n ). Since x 1,..., x n J, there exists I C such that x 1,..., x n I. Hence Rad X (J) = Rad X (I), and this verifies (1). In the next proposition we make the simple observation that a finite character collection is a Noetherian collection. Proposition 3.2. Let H be a domain. If Σ is a finite character collection of integrally closed overrings of H, then Σ is a Noetherian collection. Proof. Suppose that Σ has finite character, and let W be a nonempty open subspace of Σ. Then there exist finitely generated fractional ideals I and J of H such that U Σ (I, J) is a nonempty open subspace of W. Since Σ has finite character, IR = JR = R for all but finitely many R in Σ. Thus I cl R (JR) (i.e., R U Σ (I, J)) for all but finitely many members R of Σ. Hence Σ \ W is a finite set. This implies that the complement of every nonempty open subspace of Σ is finite, so the collection of open subspaces of Σ satisfies the ascending chain condition. The converse is not generally true even in some narrow circumstances: see for example [15]. However the next theorem shows that if Σ is a Noetherian collection and every member of Σ is a valuation ring of Krull dimension 1, then Σ has finite character. We define for every domain H and collection Σ Z(H), Z(Σ) := R Σ Z(R). If also Σ Zar(H), we write Zar(Σ) for Z(Σ). A consequence of the following lemma is that Zar(Σ) is a spectral space when Σ is Noetherian. Lemma 3.3. Let Σ be a nonempty Noetherian collection of valuation overrings of the domain H, and let B = V Σ V b. Then the mapping f : Zar(Σ) Spec(B) : V M V b B

8 8 BRUCE OLBERDING is a homeomorphism, and the subspace Max(B) of Spec(B) consisting of the maximal ideals of B is a Noetherian space. Proof. The mapping f is the mapping g h : Zar(Σ) Zar(B) Spec(B) induced by the mappings of Lemma 2.10 and Proposition 2.5. Hence f is a continuous open injective mapping. Let X = f(σ). Then X is a Noetherian subspace of Spec(B), so since B is a Bézout domain, we have that for every prime ideal P of B, P Q X Q if and only if P Q for some Q X. For since P is an ideal, there exists by Proposition 3.1, x P such that Rad X (x) = Rad X (P ). Hence if P Q X Q, then x Q for some Q X, so that P Rad X(x) Q. Therefore, since B = V Σ V b and X = f(σ) imply that Q X Q contains every nonunit of B, we conclude that every maximal ideal of B is contained in Q X Q. Hence, by what we have established above, every maximal ideal is contained in some prime ideal in X, and so Max(B) X. We claim that this implies that f is surjective. Let P Spec(B), and let M be a maximal ideal of B containing P. Then since M X = f(σ), we have M = M V b B for some V Zar(Σ). However, since B is a Prüfer domain, B M = V b, which implies then that B P is a localization of V b, and hence of the form W b for some valuation overring W of V [5, Theorem 32.10]. Thus P = M W b B = f(w ), and we conclude that f is surjective. Finally, as a subspace of a Noetherian space, Max(B) is also Noetherian. Theorem 3.4. Let Σ be a Noetherian collection of valuation overrings of the domain H. (1) For each nonempty subset V of Zar(Σ), V is a closed subspace of Zar(Σ) if and only if there exist V 1,..., V n Zar(Σ) such that: V = n {W Zar(Σ) : W V i }. i=1 (2) The set {V Zar(Σ) : V has Krull dimension 1} has finite character. (3) Zar(Σ) is a Noetherian collection if and only if for each V Σ, the prime ideals of V satisfy the ascending chain condition. Proof. (1) Let f and B be as in Lemma 3.3. Since by the lemma Max(B) is Noetherian, every proper ideal of B has only finitely many minimal prime ideals (apply [4, Lemma 5.13] and [16, Corollaries 1.3 and 1.5]). Thus the set V is closed if and only if f(v) is closed; if and only if there exist prime ideals P 1,..., P n of B such that f(v) = n i=1 {Q Spec(B) : P i Q}; if and only if there exist valuation overrings V 1,..., V n Zar(Σ) such that V = n i=1 {W Zar(Σ) : W V i}. (2) Let y be a nonzero element of the quotient field F of H. Then V = {W Zar(Σ) : y W } is a closed subspace of Zar(Σ), so by (1), there exist V 1,..., V n Zar(Σ) such that V = n i=1 {W Zar(Σ) : W V i}. Therefore, the set A y := {W Σ : W has Krull dimension 1 and y W } {V 1,..., V n }, and hence is finite. Thus,

9 NOETHERIAN SPACES OF INTEGRALLY CLOSED OVERRINGS 9 for each 0 x F, A x A x 1 is finite, and so x is a unit in all but finitely many one-dimensional members W of Σ. (3) By Lemma 3.3, the set Zar(Σ) is a Noetherian collection if and only if Spec(B) is a Noetherian space; if and only if Max(B) is a Noetherian space and Spec(B) satisfies ACC on prime ideals (see Theorem 2.7 of [11]). Since Σ is a Noetherian collection, we have by Lemma 3.3 that Max(B) is a Noetherian space. Therefore, Zar(Σ) is a Noetherian collection if and only if Spec(B) satisfies ACC on prime ideals. For V, W Zar(Σ), with f as in Lemma 3.3, we have V W if and only if V b W b ; if and only if M W b M V b; if and only if f(w ) f(v ). (We are using here in this last equivalence that since B is a Prüfer domain, W b = B MW b B, and similarly for V.) Thus since f is a bijection, we have that Spec(B) satisfies ACC on prime ideals if and only if Zar(Σ) satisfies DCC with respect to set inclusion; if and only if the prime ideals of each V Σ satisfy ACC. In [10] Heinzer and Ohm prove that a finite character intersection commutes with localization. The next theorem shows, via a different approach, that for integrally closed rings finite character can be weakened to Noetherian, and localization can be replaced by flat module. Theorem 3.5. Let H be a domain, and let Σ be a nonempty Noetherian collection of integrally closed overrings of H. If Y is a flat H-submodule of F, then Y ( R Σ R) = R Σ Y R. Proof. It suffices to prove that R Σ Y R Y ( R Σ R). Write Σ = {R i : i I} for some index set I, and let x R Σ Y R. Then for each i I, there exists a finitely generated H-submodule J i of Y such that x J i R i. Hence for each i I, R i U Σ (x, J i ). Define F = { i E U Σ(x, J i ) : E is a finite subset of I}. Then F is a nonempty collection of open sets of the space Σ, so since Σ is Noetherian there exists a maximal element U of F. Moreover, since F is closed under finite unions, U is the unique maximal element of F. Now there exist i 1,..., i n I such that: U = U Σ (x, J i1 ) U Σ (x, J in ). Since U is the unique maximal element of F, it follows that for each i I, R i U Σ (x, J i ) U, so that necessarily: Σ = U Σ (x, J i1 ) U Σ (x, J in ). For each k = 1,..., n, define U k = U Σ (x, J ik ) and H k = R U k R. Since the finitely generated fractional ideals J ik H b of H b are principal, and since principal fractional ideals commute with intersections, we have: (J ik R b ) = J ik ( R b ) Y ( R b ). R U k R U k R U k

10 10 BRUCE OLBERDING Since Y is an H-submodule of F, Y F = F. Thus, since flat modules commute with finite intersections [1, I.2.6, Proposition 6], we have: x ( J ik R b ) F Y ( R b ) F R U k R U k = Y (( R b ) F ) = Y ( R) = Y H k. R U k R U k Consequently, x Y H 1 Y H n, and, again since flat modules commute with finite intersections, we conclude that x Y (H 1 H n ) = Y ( R Σ R). This proves the claim. Corollary 3.6. Let H be a QR-domain, and suppose there exists a nonempty Noetherian collection Σ of valuation overrings of H such that H = V Σ V. If the members of Σ are pairwise incomparable, then no member of Σ can be replaced in this intersection with one of its proper overrings. Proof. Suppose the members of Σ are pairwise incomparable. Let V Σ, and define Γ = Σ \ {V }. We may assume Γ is nonempty. By Theorem 3.4(1), {V } is a closed subspace of Σ, so by Lemma 2.10 there exists an ideal I of H such that for := {W Σ : IW W }, we have = {V }. Suppose H = W Γ W. Since H is a Prüfer domain, every H-submodule of F is flat [9], so by Theorem 3.5, I = W Γ IW. Now since = {V }, it must be that IW = W for every W Γ. But then 1 I, so that is empty, a contradiction. Hence H W Γ W, and Σ is an irredundant representation of H. The corollary now follows from the fact that an irredundant intersection of valuation overrings of a Prüfer domain is unique [8, Corollary 1.9]. In addition to commuting with Noetherian collections, flat overrings yield new Noetherian collections: Theorem 3.7. Let Σ be a Noetherian collection of integrally closed overrings of the domain H. For every flat overring B of H, {BR : R Σ} is a Noetherian collection of integrally closed overrings of H. Proof. First we observe that for each integrally closed overring R of H, BR is an integrally closed domain. For since (BR) b = BRH b and R b = RH b, we conclude that (BR) b = B(R b ). Therefore, since flat modules commute with finite intersections [1, I.2.6, Proposition 6], we have (BR) b F = BR b F = B(R b F ) = BR, so BR is an integrally closed domain. By Proposition 2.5 the collection {R b : R Σ} is a Noetherian collection of integrally closed overrings of H b, and if {(BR) b : R Σ} is a Noetherian collection, then so is {BR : R Σ}. Thus, trading H for H b, we assume without loss of generality that H is a Bézout domain. By Lemma 2.10 it suffices to prove that Y := {P BR : R Σ} is a Noetherian subspace of S(H).

11 NOETHERIAN SPACES OF INTEGRALLY CLOSED OVERRINGS 11 We use the following notation. If Q 1 and Q 2 are prime semigroup ideals of H, then Q 1 Q 2 is the largest prime semigroup ideal of H contained in Q 1 Q 2 (in fact, Q 1 Q 2 is the union of all the prime ideals contained in Q 1 Q 2 ). We claim that H Q1 H Q2 = H Q1 Q 2. Since both of these rings are localizations of H, to prove equality it suffices to show that a prime ideal P of H survives in H Q1 H Q2 if and only if it survives in H Q1 Q 2. Let P be a prime ideal of H. Observe that 1 P H Q1 Q 2 if and only if P Q 1 Q 2 ; if and only if P Q 1 Q 2. Now suppose that 1 P H Q1 H Q2. Then there exist p P, b 1 H \ Q 1 and b 2 H \ Q 2 such that b 1 b 2 = p. Hence b 1 or b 2 is in P, so that P Q 1 Q 2, and hence, as we have shown, 1 P H Q1 Q 2. Conversely, if 1 P H Q1 Q 2, then P Q 1 Q 2, so there exists b P \ Q 1 Q 2. Thus 1 = b/b P H Q1 H Q2. This shows that 1 P H Q1 Q 2 if and only if 1 P H Q1 H Q2, and hence H Q1 Q 2 = H Q1 H Q2. Applying the preceding observation, we have that for each R Σ, H PBR = BR = H PB H PR = H PB P R, and hence P BR = P B P R. Therefore, Y = {P B P R : R Σ}. We show via Proposition 3.1 that Y is a Noetherian subspace of S(H). Let P S(H). We claim that Rad Y (P ) = Rad Y (E) for some finite subset E of P. If there exists x P \ P B, then x is not in any member of Y, so Rad Y (P ) = Rad Y (x) = H. Otherwise, suppose P P B. Since by Lemma 2.10, X = {P R : R Σ} is a Noetherian subspace of S(H), there exists by Proposition 3.1 a finite subset E of P such that Rad X (P ) = Rad X (E). Clearly, Rad Y (E) Rad Y (P ). On the other hand, suppose that Q Y and E Q. Then there exists R Σ such that Q = P B P R. Thus E P R, so that since Rad X (P ) = Rad X (E), we have P P B P R = Q. Hence Rad Y (P ) = Rad Y (E), and this proves that Y is a Noetherian subspace of S(H). 4. Irredundant intersections In this section we prove an existence theorem for irredundant intersections. We work in the generality of the set Z(H) rather than Zar(H), and then finally derive stronger results by restricting to Zar(H). For technical purposes, we introduce a sharpening construction for S(H). For each collection X of prime semigroup ideals of H, we let P X P denote the largest prime semigroup ideal of H contained in P X P ; that is, P X P is the union of all the prime ideals of H contained in P X P. For each Q Spec(H) such that V X(Q) is nonempty, we let Q # = {L X : Q L} = V X (Q), so that Q # is a prime semigroup ideal with Q Q # Rad X (Q), and Q # is the largest prime semigroup ideal that contains Q and is contained in Rad X (Q). Then, ranging over the totality of all such prime ideals, we set X # = {Q # : Q Spec(H) such that V X (Q) is nonempty}.

12 12 BRUCE OLBERDING In what follows our strategy is to replace X with X #. The reason for doing so is that the members of X # are radicals of finitely generated ideals, not just finite sets (compare to Proposition 3.1): Lemma 4.1. Let H be a domain. If X is a Noetherian subspace of S(H), then X # is a Noetherian subspace of S(H), and for every prime ideal P of H such that V X (P ) is nonempty, there exists a finitely generated ideal I of H such that P # = Rad X #(I). Proof. To prove that X # is a Noetherian space, it suffices to verify Proposition 3.1(2) for X #. Let P S(H). We find a finite subset E of P such that Rad X #(P ) = Rad X #(E). By Proposition 3.1, since X is a Noetherian space, there exists a finite subset E of P such that Rad X (E) = Rad X (P ). Necessarily, since E P, we have Rad X #(E) Rad X #(P ). On the other hand, suppose that Q Spec(H) so that Q # X # and E Q # = V X (Q) V X (Q). Then for each L V X (Q), we have E L, and hence, since Rad X (E) = Rad X (P ), we have also that P L. Therefore, P V X (Q), and since Q # is the union of all the prime semigroup ideals of H contained in V X (Q) and containing Q, we have that P Q #. This proves that Rad X #(P ) = Rad X #(E), and we conclude that X # is a Noetherian space. Next suppose that P is a prime ideal of H with V X (P ) nonempty. We show there is a finitely generated ideal I P such that P # = Rad X #(I). Since, as we have verified, X # is a Noetherian space, we have by Proposition 3.1 and the fact that P is an ideal that there exists a finitely generated ideal I P such that Rad X #(I) = Rad X #(P ). To complete the proof, we claim that P # = Rad X #(P ). Since P P # X #, it is clear that Rad X #(P ) P #. Conversely, suppose that Q is a prime ideal of H such that V X (Q) is nonempty and P Q #. Since Q # = VX (Q) V X (Q), we have that P L for all L V X (Q). This then implies that V X (Q) V X (P ), and hence P # = V X (P ) V X (Q) = Q #. This proves that any member of X # containing P contains also P #, and hence P # Rad X #(P ). Therefore, P # = Rad X #(P ) = Rad X #(I), and the lemma is proved. We prove now our two main theorems. Theorem 4.2. Let H be a domain, let A be a proper overring of H, and suppose there exists a Noetherian collection Σ of integrally closed overrings of H such that H = ( R Σ R) A. Then there exists a weakly Noetherian collection Γ Z(Σ) such that H = ( R Γ R) A, and no member of Γ can be omitted from this representation. Moreover, if Σ has finite character, then Γ has finite character. Proof. Define C = R Σ Rb, and let F = {B Z(C) : B A = H}. Then F is nonempty since C F, and for every chain C in F, the union of the sets in C is in F. Thus by Zorn s Lemma F contains maximal elements. Let B be a maximal element of F. By Proposition 2.5 the set {R b : R Σ} is a continuous image of Σ, so it is a Noetherian subspace of Z(H b ). Since every overring of a Prüfer domain is flat, we have by Theorem 3.7 that Σ := {BR b : R Σ} is a Noetherian collection

13 NOETHERIAN SPACES OF INTEGRALLY CLOSED OVERRINGS 13 of overrings of H b. Also, by Theorem 3.5 we have B = BC = B( R b ) = R Σ R Σ (BR b ) = S Σ S. By Lemma 2.10 the mapping R P R, R Z(B), is a homeomorphism of Z(B) onto S(B). (Here P R = {x B : xr R}.) Thus X := {P S : S Σ } is a Noetherian subspace of S(B). By Lemma 4.1, X # is a Noetherian subspace of S(B), and it follows that X # contains maximal elements. Denote by Max(X # ) the set of maximal elements of X #. We claim that B = P # Max(X # ) B P #. For let x F (X) \ B. Then B x 1 B Q for some prime ideal Q of B. Observe that V X (Q) is nonempty. Indeed, since B is a Prüfer domain, Q is a flat ideal of B, so by Theorem 3.5, Q = QB = S Σ QS. Thus for some S Σ, S QS, whence Q P S X (Lemma 2.9). Therefore, V X (Q) is nonempty, and we have B x 1 B Q Q # X #. Thus x B Q # with Q # X #, and we conclude that B = P # Max(X # ) B P #. We claim that Γ := {B P # : P # Max(X # )} is an irredundant representation of B. Fix a prime ideal Q of H such that Q # Max(X # ). By Lemma 4.1 there exists a finitely generated ideal I of B such that Q # = Rad Max(X # )(I). In fact, since B is a Bézout domain, I is a principal ideal, say I = xb. Thus since Q # = Rad Max(X # )(x), it follows that x 1 Q # P # Max(X # ) B P # \ B Q #. Hence Γ is an irredundant representation of B. Now we claim that no member of Γ := {B P # F : P # Max(X # )} can be omitted from the representation H = ( R Γ R) A. For suppose that Γ and H = ( R R) A. Then the ring {B P # Γ : B P # F } contains B and is in F. Thus since B is maximal in F, we have B = {B P # Γ : B P # F }, and since Γ is an irredundant representation of B, it must be that Γ = {B P # : B P # F }. Hence Γ = {R F : R Γ } =. This proves that no member of Γ = {B P # F : P # Max(X # )} can be omitted from the representation H = ( R Γ R) A. Note also that our construction of Γ yields Γ Zar(Σ). Indeed, let S be a ring in Γ. Then S = B P # F for some P # X #. The prime semigroup ideal P # is necessarily a subset of some prime semigroup ideal Q X. Thus B Q F B P # F = S. Since Q X, there exists a ring R Σ such that Q = P BR b. Thus, using the fact that B is a QR-domain, we have B Q = B = PBR b BRb. Therefore, R BR b F = B Q F S, which proves that S Zar(Σ). Hence Γ Zar(Σ). Next we claim that Γ is a weakly Noetherian collection. Define f : Z(H b ) Z(H) by f(c) = C F for every C Z(H b ). Then f(γ ) = Γ, so to prove that Γ is a weakly Noetherian collection, it suffices, since Γ is Noetherian, to show that f is continuous with respect to the Zariski topology. We do this by noting that every subbasic open set of Z(H) in the Zariski topology has an open preimage. For let

14 14 BRUCE OLBERDING x 1,..., x n F, and define U = U H (x 1,..., x n ). Then: f 1 (U) = {C Z(H b ) : x 1,..., x n C} = U H b(x 1,..., x n ). Thus f is a continuous mapping. Finally, suppose that Σ is a finite character collection, and let x H. Then {R b : R Σ} is a finite character collection of overrings of C, so x is contained in at most finitely many members of X. Now every element of Max(X # ) is a subset of an intersection of members of X. Hence there are at most finitely many members of Max(X # ) containing x, so x is a unit in all but finitely many members of Γ = {B P # : P # Max(X # )}. Thus x is a unit in all but finitely many members of Γ. This proves that Γ is a finite character collection. To express the valuation case of Theorem 4.2 more succinctly, we use the terminology of [14] and [15]. If H is a domain with overring A and Σ is a collection of valuation overrings with H = ( V Σ V ) A, then Σ is an A-representation of H. If Σ is a Noetherian (resp., finite character) collection, then Σ is Noetherian (resp., finite character) A-representation of H. Also, if no member of Σ can be replaced with one of its proper overrings in the given representation of H, then Σ is a strongly irredundant A-representation of H. Theorem 4.3. If a domain H with overring A has a Noetherian (resp., finite character) A-representation Σ of valuation overrings, then H has a strongly irredundant Noetherian (resp., finite character) A-representation Γ of valuation overrings with Γ Zar(Σ). Proof. We assume the same notation as the proof of Theorem 4.2, and in particular, we have that there exists a Noetherian collection Γ Zar(Σ) such that H = ( V Γ V ) A and no member of Γ can be omitted from this intersection. Note that every member of Γ (where Γ is as in the proof of Theorem 4.2), and hence every member of Γ, is a valuation ring. Since Γ is a weakly Noetherian collection of valuation rings, Γ is a Noetherian collection (Corollary 2.8). Let V Γ, and suppose that W is an overring of V such that for Γ 0 := {W } (Γ \ {V }), it is the case that H = ( U Γ 0 U) A. Then U Γ 0 U b is a member of F, and since B U Γ 0 U b and B is maximal in F, this forces B = U Γ U b = U Γ 0 U b. Since by Corollary 3.6 no member of Γ can be replaced by a proper overring in this representation of B, this forces V b = W b. Hence V = W, and this proves the theorem, save the last assertion, which then follows from Theorem 4.2. Thus when H has a finite character A-representation and each member of the representation has Krull dimension 1, we recover the theorem of Heinzer and Ohm in [10] discussed in the introduction. On the other hand, when A is the quotient field of R and H has a finite character A-representation, we recover a stronger version of the theorem of Brewer and Mott in [2], that asserts only irredunance (rather than strong irredundance) of the representation.

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