BLAISE PASCAL

We consider analogies between the logically independent properties of strong going-between (SGB) and going-down (GD), as well as analogies between the universalizations of these properties. Transfer results are obtained for the (universally) SGB property relative to pullbacks and Nagata ring constructions. It is shown that if A ⊆ B are domains such that A is an LFD universally going-down domain and B is algebraic over A , then the inclusion map A [ X 1 , . . . , X n ] , → B [ X 1 , . . . , X n ] satisﬁes GB for each n ≥ 0 . However, for any nonzero ring A and indeterminate X over A , the inclusion map A , → A [ X ] is not universally (S)GB.


Introduction
All rings considered below are commutative with identity; all ring extensions and ring homomorphisms are unital.Our goal is to further the work that was begun in [9] on the interplay between the going-down (GD) and the strong going-between (SGB) properties of ring homomorphisms.To ease the discussion, we adapt the notation in [15, p. 28], by letting GD and GU denote the going-down and going-up properties, respectively, for ring homomorphisms; we also let Spec(A) denote the set of prime ideals of a ring A.) Following [18], we say that a ring homomorphism f : A → B satisfies SGB in case the following condition is satisfied: if P 1 ⊆ P 2 ⊆ P 3 in Spec(A) and Q 1 ⊆ Q 3 in Spec(B) are such that f −1 (Q 1 ) = P 1 and f −1 (Q 3 ) = P 3 , then there exists Q 2 in Spec(B) such that Q 1 ⊆ Q 2 ⊆ Q 3 and f −1 (Q 2 ) = P 2 .This terminology was chosen in order to avoid confusion with the different "goingbetween" (GB) property introduced by Ratliff [19]: a ring homomorphism f : A → B satisfies GB in case, whenever Q 1 ⊂ Q 3 are prime ideals in B such that there exists a prime ideal P 2 in A such that f −1 (Q 1 ) ⊂ P 2 ⊂ f −1 (Q 3 ), then there exists a prime ideal Q 2 in B such that Q 1 ⊂ Q 2 ⊂ Q 3 .For motivational purposes, it is useful to observe that SGB ⇒ GB.
The most basic fact regarding the interplay between GD and SGB is that these two properties are logically independent.Explicit examples were given in [9,Proposition 2.4] to show this logical independence.In view of the somewhat ad hoc nature of those examples, we think it appropriate to reestablish this fact by appealing to the fundamental structure of the category of commutative rings.For this reason, the logical independence of GD and SGB is proved in Proposition 2.2 by using a realization theorem of Hochster [14,Theorem 6 (a)] for surjective spectral maps of spectral sets.
One of the examples in [9] depended on a result [9, Remark 2.5 (e)] on the transfer behavior of the SGB property relative to the classical D + M construction (in the sense of [12]).This is generalized in Theorem 2.3, which gives a transfer result for SGB in a pullback context that is in the spirit of [10, Section 2] and is, thus, more general than that of the classical D + M construction.Section 2 adds further to the SGB/GD analogy, by giving an additional transfer result (Proposition 2.4) for SGB in the context of Nagata rings.As explained there, this result is motivated by known transfer results for going-down domains [1] and for GB-rings ( [19], [20]).
In view of the preceding comment, it is timely to recall from [4] and [8] that GD has been used to introduce a class of (commutative integral) domains that includes all Prüfer domains and all domains of (Krull) dimension at most 1.Indeed, according to [8, Theorem 1], a domain A, with quotient field K, is a going-down domain if and only if A → B satisfies GD for all domains B that contain A (resp., for all overrings B of A; resp., for all valuation overrings B of A; resp., for all rings of the form B = A[u], u ∈ K).It was shown in [9,Corollary 2.3] that the analogy between SGB and GD can be extended as follows: if one replaces "GD" with "SGB" in each of the preceding four conditions, the resulting conditions are still equivalent and still characterize going-down domains.In other words, what might be called an "SGB-domain" is precisely the same as a going-down domain.(For additional motivation, note that the class of GB-rings was introduced in [19] and studied further in papers such as [20].) It was shown in [7,Theorem 2.6] that if one replaces "GD" with "universally going-down" in each of the above four conditions from [8, Theorem 1], the resulting conditions are still equivalent and characterize what is there called "universally going-down domains".It thus seems natural to ask if the "universally SGB" property can be used to similarly characterize the universally going-down domains.The answer, as given in [9, Corollary 3.2, Corollary 3.9], is essentially "yes and no": "yes" if one restricts attention to overrings B of A; "no" if one considers test extension domains B that are not algebraic over A. As explained below, the work in Section 3 deepens these results from [9] in two ways.
First, [9, Corollary 3.9] is re-obtained in Corollary 3.5, as a consequence of a result (Lemma 3.3) that seems more direct, albeit somewhat intricate, than the method used in [9].As a pleasant bonus, we note, after the proof of Corollary 3.5, that our present approach actually establishes more, in particular, that there does not exist a domain A such that A → B is universally GB for each domain B containing A. This is apparently the first result on "universally GB" in the literature.Second, continuing in the "universally GB" vein, we develop a positive result in Theorem 3.6 that addresses the following question: if A is a universally going-down domain, is it possible to enlarge (beyond the arena of overrings) the class of "test domains" B appearing in the statement of [9, Corollary 3.2]?The proof of Theorem 3.6 depends, in part, on a result concerning the "universally catenarian" property [3,Corollary 6.3].
Our final result, Proposition 3.8, merges several of the above themes by establishing that the universally SGB property transfers to the induced map on Nagata rings.
For the sake of motivation and ease of reference, some results from [9] are restated in this paper.Finally, in addition to the notational conventions indicated above, we mention the following.If A is a ring, then dim(A) denotes the Krull dimension of A; ht A (P ) = ht(P ) denotes the height (in A) of P ∈ Spec(A); and A red denotes A/ √ A, the reduced ring canonically associated to A. If A is a domain, with quotient field K, then A denotes the integral closure of A (in K); and by an overring of A, we mean any ring B such that h red denotes the induced ring homomorphism A red → B red .Any unexplained material is standard, as in [12], [13], [15].

Logical independence and transfer results
We begin by stating some known connections between the SGB and GD properties.The following definitions are needed for Proposition 2.1 (b), (c).
As recalled in the Introduction, a domain A is called a going-down domain in case A ⊆ B satisfies GD for each domain B containing A. Following [5], a ring A is called a going-down ring in case A/P is a going-down domain for each P ∈ Spec(A).Proposition 2.1: Let f : A → B be a ring homomorphism.
(a) [18, Propositions 5.2 and 5.7] The following conditions are equivalent: (1) f satisfies SGB; (2) The induced map A P → B Q satisfies GU for all Q ∈ Spec(B) and and ) The induced map f red : A red → B red satisfies SGB; (5) The induced map A/P → B/Q satisfies SGB for all Q ∈ Spec(B) and Despite the assertions in parts (b) and (c) of Proposition 2.1, we show next that neither SGB nor GD implies the other, even for ring extensions of a domain.

Proposition 2.2: The SGB and GD properties are logically independent. In fact:
(a) There exists an inclusion map A → B that satisfies SGB but does not satisfy GD.It can be arranged that A is a quasilocal one-dimensional domain and that B is a zero-dimensional reduced ring with exactly two prime ideals (that is, B is isomorphic to the direct product of two fields).
(b) There exists an inclusion map A → B that satisfies GD but does not satisfy SGB.It can be arranged that A is a quasilocal two-dimensional treed domain and that B is a quasilocal two-dimensional ring with exactly two minimal prime ideals.
Proof: (a) An ad hoc approach, in the spirit of [9, Proposition 2.4 (a)], is available.Indeed, let X be an indeterminate over a field k, and consider A := k[X] (X) and B := k × k(X).Of course, A is a (quasi)local one-dimensional domain with unique maximal ideal M := XA, and since A = k + M , one can identify A/M = k.Also, the only prime ideals of B are k × 0 and 0 × k(X), which are incomparable.Then the monomorphism i : A → B, given by a → (a + M, a), is easily seen to have the asserted properties, in view of the facts that a i(k × 0) = 0 and a i(0 × k(X)) = M .We next present a more categorical approach to the assertion in (a).Consider a two-element set Y := {Q, N } with the trivial partial order; that is, Q and N are unrelated.Also, consider a two-element linearly ordered set X := {P, M } with P < M .It is easy to see directly that both Y and X are spectral sets (in the sense that each is order-isomorphic to the prime spectrum of some ring).However, we appeal here to a useful more general method, namely, the fact that any finite partially ordered set is a spectral set [16,Theorem 2.10].Identify Y = Spec(D) and X = Spec(C) for some rings D and C; equip the spectral spaces Y and X with the corresponding Zariski topology.
Consider the surjective function g : Y → X given by Q → P and N → M .We claim that g is a spectral map (in the sense of [14, p. 43]), namely, that g is continuous and According to a realization result of Hochster [14, Theorem 6 (a)], Spec is invertible on the subcategory of spectral spaces and surjective spectral maps.Since we have shown that g is a surjective spectral map, there exist a ring homomorphism h : E → F and homeomorphisms α : [14], homeomorphisms of spectral spaces induce order-isomorphisms.Therefore, it is now easy to see that h satisfies SGB but does not satisfy GD (the point being that g was constructed to satisfy the corresponding order-theoretic conclusion).
Next, note via standard homomorphism theorems (and the surjectivity of g) that there is a natural bĳection between Spec(im(h)) and Spec(E).Thus, by abus de langage, we can now replace h : E → F with im(h) → F ; that is, h : E → F is now an inclusion map.Therefore, h red : A → B is also an injection, where A := E red and B := F red .Moreover, h red inherits from h the property of satisfying SGB (resp., of not satisfying GD) by condition (4) in the statement of Proposition 2.1 (a) (resp., by [6, Lemma 3.2 (a)]).Then h red has the asserted properties.Indeed, since Spec(A) is order-isomorphic to X, we see that A is quasilocal and one-dimensional; and since A is a reduced ring with a unique minimal prime ideal, A is a domain.Finally, since Spec(B) is order-isomorphic to Y , we see that B is a zero-dimensional reduced ring with exactly two (incomparable) prime ideals, say q and n.Hence, q + n = B and q ∩ n = 0, and so the Chinese Remainder Theorem yields that B ∼ = B/q × B/n, a direct product of two fields, to complete the proof.
(b) An ad hoc example is available: see [9, Proposition 2.4 (b)].We next sketch how to use the order-theoretic machinery featured in the proof of (a) in order to prove (b).
To that end, consider a four-element set Y := {P 1 , P 2 , Q 1 , M 1 } with the partial order induced by requiring that P 1 < M 1 and P 2 < Q 1 < M 1 .Also, consider a three-element linearly ordered set X := {P, Q, M } with P < Q < M .By [16, Theorem 2.10], Y and X are spectral sets, and so we can identify Y = Spec(D) and X = Spec(C) for some rings D and C; equip the spectral spaces Y and X with the corresponding Zariski topology.
Consider the surjective function g : Y → X given by P 1 → P, P 2 → P, Q 1 → Q, and M 1 → M .We claim that g is a spectral map.As in the proof of (a), it is enough to show that g The rest of the proof now proceeds essentially as in the proof of (a).In other words, one appeals to the realization result of Hochster, then replaces h with im(h) → F , then replaces h with h red : A → B, and then concludes via the order-isomorphisms between Spec(A) (resp., Spec(B)) and X (resp., Y ).
We close the section by considering the transfer of the SGB property in certain pullback constructions and for Nagata rings.To motivate Theorem 2.3, note that [9, Remark 2.5 (e)] concluded with a transfer result on SGB for the special pullbacks of the classical D + M form, while going-down-theoretic results have been obtained for more general pullbacks (cf.[  Proof: Observe the canonical isomorphisms D ∼ = A/M and E ∼ = B/M .Therefore, the "only if" assertion is immediate from condition (5) in Proposition 2.1 (a).For the converse, suppose that D → E satisfies SGB.Our task is to show that if P 1 ⊆ P 2 ⊆ P 3 in Spec(A) and Before proceeding further, we next summarize the order-theoretic impact of the "amalgamated sum" topological description [10,Theorem 1.4] of the prime spectra of the pullbacks A and B. For instance, Spec(A) can be viewed order-theoretically as the quotient space of the disjoint union of Spec(V ) and Spec(D) in which M ∈ Spec(V ) is identified with 0 ∈ Spec(D).In particular, each prime ideal of A is comparable with M , and the set of prime ideals of A that contain (resp., are contained in) M is order-isomorphic to the prime spectrum of D (resp., V ).Of course, Spec(B) admits a similar description.There are now three cases to consider.
If P 1 ∈ im(Spec(D) → Spec(A)), then all the P i (resp., Q j ) are canonical images of prime ideals of D (resp., E) in view of the above comments, and so the existence of a suitable Q 2 follows since D → E satisfies SGB.In the second case, P 3 ∈ im(Spec(V ) → Spec(A)); then all the P i and Q j are canonical images of prime ideals of V in view of the above comments, and so the existence of a suitable Q 2 follows from the triviality that the identity map on V satisfies SGB.In the final case, P 1 ⊆ M ⊆ P 3 .If M ⊆ P 2 (resp., P 2 ⊆ M ), argue as in the first (resp., second) case, attending to the chain M ⊆ P 2 ⊆ P 3 (resp.,P 1 ⊆ P 2 ⊆ M ) in Spec(A).The proof is complete.
Recall that if A is a ring and X is an indeterminate over A, then the Nagata ring A(X) is defined to be the localization of the polynomial ring A[X] at the multiplicatively closed set consisting of all the polynomials in A[X] with unit content.Observe that any ring homomorphism A → B induces a ring homomorphism A[X] → B[X] and, hence, a ring homomorphism A(X) → B(X).Proposition 2.4 records a pair of elementary transfer results for the SGB property that involve Nagata rings.For these assertions, recall that a domain is called a quasi-Prüfer domain in case its integral closure is a Prüfer domain.For additional background on quasi-Prüfer domains, see [11, Section 6.5, especially Corollary 6.5.14].
Proposition 2.4: (a) Let f : A → B be a ring homomorphism, where A and B are each quasi-Prüfer domains.Let g : A(X) → B(X) be the ring homomorphism induced by f .Then f satisfies SGB (resp., GD) if and only if g satisfies SGB (resp., GD).

(b) If
A is both a quasi-Prüfer domain and a going-down domain, then any ring homomorphism A(X) → B satisfies SGB.
Proof: (a) Since A and B are each quasi-Prüfer domains, it follows from [1, Theorem 2.7] that the canonical functions Spec(A(X)) → Spec(A) and Spec(B(X)) → Spec(B) are each order-isomorphisms (in fact, homeomorphisms).The assertions (for SGB, GD, and analogously for any other ordertheoretic property of the prime spectra, such as GU) now follow by easy analyses of diagrams.
(b) By [1, Corollary 2.12], the hypothesis on A ensures (actually, is equivalent to the fact) that A(X) is a going-down domain.Accordingly, an application of Proposition 2.1 (c) completes the proof.For an alternate proof, note first via Proposition 2.1 (c) that the composite function A → A(X) → B satisfies SGB, and then invoke the fact [1, Theorem 2.7] that Spec(A(X)) → Spec(A) is an order-isomorphism.
In view of the above remarks, one should emphasize that some nonoverring extensions of domains do satisfy the universally SGB property.In fact, if a ring homomorphism f : A → B is radiciel (in the sense of [13,Définition 3.7.2,p. 248]) and universally going-down, then f is universally SGB.
For the proof of Lemma 3.3, we need to recall the following material from [15, p. 26] and [3].A domain A is called an S(eidenberg)-domain in case ht A[X] (P A[X]) = 1 for all P ∈ Spec(A) such that ht A (P ) = 1.If A is a one-dimensional S-domain, then [3, Corollary 6.3] ensures that A is a stably strong S-domain (in the sense that for each nonnegative integer n, the polynomial ring A[X 1 , . . ., X n ] has the property that each of its factor domains is an S-domain).Combining this information with [15, Theorem 27], we see that if A is a one-dimensional S-domain, then dim(A[X 1 , . . ., X n ]) = n + 1 for each nonnegative integer n.
The proof of Lemma 3.3 also assumes familiarity with the basic facts about "uppers" in a polynomial ring (cf.[15, p. 25]) and the very convenient < P, α > notation introduced for them in [17,Notation,p. 707].

Lemma 3.3:
Let A be a one-dimensional S-domain and X, Y algebraically independent indeterminates over A. Then A[X] → A[X, Y ] does not satisfy SGB.
Proof: It suffices to produce prime ideals P 1 ⊂ P 2 ⊂ P 3 of A[X] and adjacent prime ideals Let K denote the quotient field of A. Fix a nonzero (necessarily height 1 and maximal) prime ideal M of A. We begin the construction of the desired prime ideals by letting P 1 = 0, P 2 = < 0, X > and P 3 = < M, X >.It follows easily from the definition of the notation for uppers cited above that P 2 = A[X] ∩ XK[X] and P 3 = M + XA[X].Therefore, since M = 0, we have arranged that P 1 ⊂ P 2 = XA[X] ⊂ P 3 .Moreover, ht(P 3 ) = 2 since the above comments about one-dimensional S-domains ensure that dim(A We turn next to the definition of Q 3 .We let In fact, the existence of uppers of P 3 yields that ht(Q 3 ) = 2, since Q 3 is not a maximal ideal of A[X, Y ] and the above comments about one-dimensional S-domains ensure that dim The construction of Q 1 is more intricate.For this purpose, fix any nonzero element m ∈ M , consider the polynomial p := Y − m −1 X ∈ K(X)[Y ], and let Q 1 = < 0, p >.By the definition of the notation for uppers cited above, Using the Factor Theorem, we obtain a convenient description of this prime ideal, namely, Q 1 = {g(X, Y ) ∈ A[X, Y ] | g(X, m −1 X) = 0}.Of course, Q 1 ∩ A[X] = 0 = P 1 and ht(Q 1 ) = 1, since Q 1 is an upper of 0. It remains only to verify that Q 1 ⊆ Q 3 , for then consideration of heights will guarantee that Q 1 and Q 3 are adjacent.We proceed by an indirect argument.Supposing that the assertion fails, choose (a necessarily nonzero element) f ∈ Q 1 \ Q 3 whose degree in Y , say n, is minimal.Note that n ≥ 1 since f = 0. Now, write Multiplying this expression by m n and then solving for f 0 (X)X n , we infer that f 0 (X)X n ∈ mA[X].Consequently, f 0 ∈ mA[X], and so we can write f 0 = mh 0 for some uniquely determined h 0 ∈ A[X].
Consider the polynomial g := (Xh 0 (X)+f 1 (X))Y n−1 +f 2 (X) Observe that g ∈ Q 1 .(This can be accomplished by a straightforward calculation, using the facts that f (X, m −1 X) = 0 and h 0 /m n−1 = f 0 /m n .)As the degree in Y of g is at most n − 1, it follows from the minimality of n that g ∈ Q 3 .In other words, Substituting X → 0 reveals that the constant terms of Xh 0 + f 1 , f 2 , . . ., f n−1 and f n are all in M .However, the constant term of Xh 0 + f 1 is the same as the constant term of f 1 .Moreover, the constant term of f 0 is also in M , since f 0 ∈ mA[X] ⊆ M A[X].Thus, the constant terms of f 0 , f 1 , f 2 , . . ., f n−1 and f n are in M .It follows from the way that we wrote f that f ∈ M A[Y ] + XA[X, Y ] = Q 3 , contradicting the choice of f .The proof is complete.
As X and Y are each finite, the "quasi-compact" conditions can be ignored here.It suffices to consider the only nontrivial open subset of X, namely U := {P }. (According to the usual description of the open sets in the Zariski topology [2, p. 99], this set is the basic open set determined by any element of M \ P .)Evidently, g −1 ({P }) = {Q}, which is Zariski-open in Y , since Y \ {Q} = V (N ) is closed by virtue of the maximality of N [2, Definition 4, p. 99].This proves the claim.
−1 (U ) is open in Y for each open set U in X.It suffices to treat the nontrivial Zariski-open subsets U of X, namely U 1 := {P } and U 2 := {P, Q}.Evidently, g −1

Theorem 2 . 3 :
Let V be a quasilocal ring with nonzero maximal ideal M and residue class field K := V /M ; let ϕ : V → K denote the canonical projection map.Let D ⊆ E be subrings of K, and consider the pullbacks A := ϕ −1 (D) and B := ϕ −1 (E).Then A → B satisfies SGB if and only if D → E satisfies SGB.