Table of Contents Author Guidelines Submit a Manuscript
International Journal of Mathematics and Mathematical Sciences
Volume 2014 (2014), Article ID 315919, 13 pages
http://dx.doi.org/10.1155/2014/315919
Research Article

When an Extension of Nagata Rings Has Only Finitely Many Intermediate Rings, Each of Those Is a Nagata Ring

1Department of Mathematics, University of Tennessee, Knoxville, TN 37996-1320, USA
2Laboratoire de Mathématiques, Université Blaise Pascal, UMR6620, CNRS, Les Cézeaux, 24 avenue des Landais, BP 80026, 63177 Aubière Cedex, France

Received 30 March 2014; Accepted 14 June 2014; Published 2 September 2014

Academic Editor: Kaiming Zhao

Copyright © 2014 David E. Dobbs et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

Let be an extension of commutative rings, with X an indeterminate, such that the extension of Nagata rings has FIP (i.e., has only finitely many -subalgebras). Then, the number of -subalgebras of equals the number of R-subalgebras of S. In fact, the function from the set of R-subalgebras of S to the set of -subalgebras of given by is an order-isomorphism.

1. Introduction and Notation

All rings considered below are commutative and unital; all inclusions of rings and all ring homomorphisms are unital. As usual, if is a ring, then and denote the sets of prime ideals of and of maximal ideals of , respectively; if , then ; if is an (ring) extension, then , the conductor of ; and if is a ring homomorphism, then denotes the canonical map , . As in [1], the support of an -module is the set and . Also as usual, if is an ideal of a ring , then ; and denotes the cardinality of a set .

Let be an (ring) extension. The set of all (unital) -subalgebras of is denoted by . Following [2], the extension is said to have (or to satisfy) FIP (for the “finitely many intermediate algebras property”) if is finite. As usual, by a chain of -subalgebras of , we mean a set of elements of that are pairwise comparable with respect to inclusion. Recall that the extension has (or satisfies) FCP (for the “finite chain property”) if each chain of -subalgebras of is finite. It is clear that FIP implies FCP. We will freely use the characterizations of the FCP extensions and of the FIP extensions that were given in [1].

Minimal (ring) extensions, as introduced by Ferrand and Olivier [3], are our main tool for studying the FIP and FCP properties. Recall that an extension is called minimal if . (Note that since denotes proper inclusion, whenever is a minimal extension.) The key connection between the above ideas is that if has FCP, then each maximal (necessarily finite) chain of -subalgebras of can be written as , with length , where , and results from juxtaposing minimal extensions ,  . For any extension , the length of , denoted by , is the supremum of the lengths of chains of -subalgebras of .

The following notions are also deeply involved in our study.

Definition 1 (see [1, Definition 4.4]). Let be an integral extension. Then is called infra-integral if for each , the residual extension is an isomorphism. Moreover, is called subintegral if is infra-integral and is a bijection.

Consider an extension . Then is called t-closed if the relations , , , and imply . Also, is called seminormal if the relations , , and imply . If is seminormal, is a radical ideal of . The -closure of in , denoted by , is the smallest -subalgebra of such that is t-closed, as well as the greatest -subalgebra of such that is infra-integral. The seminormalization of in , denoted by , is the smallest -subalgebra of such that is seminormal, as well as the greatest -subalgebra of such that is subintegral. The chain is called the canonical decomposition of , where denotes the integral closure of in .

Let be a ring and the polynomial ring in the indeterminate over . (Throughout, we use to denote an element that is indeterminate over all relevant coefficient rings.) Also, let denote the content of any polynomial . Then is a saturated multiplicatively closed subset of , each of whose elements is a non-zero-divisor of . The Nagata ring of is defined to be . Let be an extension. It was shown in [4, Theorem 3.9] that has FCP if and only if has FCP. The analogous assertion does not hold in general for the FIP property. One implication does hold in general, as it was shown in [4, Proposition 3.2] that if has FIP, then must also have FIP. We next recall some partial converses to the preceding assertion.

Let be an FIP extension. By [4, Theorem 3.21], has FIP if and only if has FIP. This last condition holds when for each , by [4, Corollary 3.25]. As for each , it follows that FIP satisfies the following analogue of a result on FCP [4, Corollary 3.10]: has FIP if and only if has FIP for some (resp., each) positive integer .

One can say more along these lines. By using results from [1, 4], we will obtain, in Theorem 2, a new characterization of when FIP holds for . Let us say that a ring extension , with seminormalization , satisfies the property if for each such that , one has that is linearly ordered and , where denotes the nilpotency index of and .

Theorem 2. Let be a ring extension. Then has if and only if has and satisfies .

Proof. Combine [1, Corollary 3.2 and Proposition 3.7(a)] with [4, Corollary 3.25 and Theorem 3.30].

In regard to an extension , our main concern here is, as it was in [4, Section 4], the function , defined by . Our goal, which will be accomplished in Theorem 32, is to show that if has FIP (in which case, must also have FIP), then is an order-isomorphism. Since is known to be an order-preserving and order-reflecting injection [4, Lemma 3.1(d)] in general, it remains only to show that is surjective (assuming that has FIP). Evidence for Theorem 32 was provided in [4, Propositions 4.4, 4.14, 4.17], where it was shown that if has FIP, then is an order-isomorphism in the following three cases: is an integrally closed extension; is a subintegral extension such that has FIP; is a seminormal infra-integral extension. Thus, in view of the steps in the canonical decomposition of an extension , it is clear that [4, Section 4] failed to make much headway for the case of an integral t-closed extension. In fact, as summarized next, our path to Theorem 32 will rely on a deeper study of precisely such extensions.

It is easy to see that any extension of fields is t-closed. We begin Section 2 by showing in Propositions 9 and 11 that if is an FIP field extension (hence, an integral t-closed extension), then is an order-isomorphism. This fact is used in the proof of Theorem 12, which obtains an affirmative answer to our main question in case of an arbitrary integral t-closed extension . The arguments in Section 3 proceed with an eye on the steps in the canonical decomposition of an extension and the four types of minimal extensions (which are reviewed later in this Introduction). In case has FIP, we establish the nature of a minimal subextension of , first for the case of a quasi-local base ring in Proposition 30 and then in general in Proposition 31. Our main result is then obtained in Theorem 32 by an inductive argument.

It is convenient to close the Introduction by stating some results that summarize the fundamental facts about minimal extensions, FCP extensions, and FIP extensions that we will use below.

Theorem 3 (see [5], [6, Theorem 4.1], [3, Théorème 2.2 and Lemme 3.2], [7, Proposition 3.2], [4, Theorem 1.1]). Let be a minimal extension with associated inclusion map . Then,(a)there is some , called the crucial (maximal) ideal of , such that for each . We denote this ideal by ;(b)with and as above, the following three conditions are equivalent:(1)some prime ideal of lies over ;(2);(3) is (module-) finite;(c)the (equivalent) conditions in (b) do not hold if and only if is a flat epimorphism (in the sense of [8]); and, in that case, is a common prime ideal of and that is contained in ;(d)there is a bijection , with when is a flat epimorphism. Moreover, if , then either or .

There are three types of integral minimal extensions, as given in Theorem 4. Thus, by also counting the flat epimorphisms discussed in Theorem 3(c), there are four types of minimal extensions.

Theorem 4 (see [7, Theorem 3.3]). Let be an extension and let . Then, is minimal and finite (i.e., an integral minimal extension) if and only if and (exactly) one of the following three conditions holds:(a)inert case: and is a minimal field extension.(b)decomposed case: there exist such that and the natural maps and are both isomorphisms.(c)ramified case: there exists such that , , and the natural map is an isomorphism.
In each of the above three cases, is the crucial ideal of .

In the context of Theorem 4, consider the field . Recall (as in the proof of [9, Corollary II.2]) that the “decomposed” (resp., “ramified”) case in Theorem 4 corresponds to being isomorphic, as a -algebra, to (resp., to ).

Lemma 5 (see [6, Proposition 4.6]). Let be a ring extension. Then is a minimal extension if (and only if) there is a maximal ideal of such that the induced extension is minimal and for each prime ideal . Moreover, whenever these (equivalent) conditions hold, (resp., ) is the crucial maximal ideal of (resp., ), and the minimal extensions and are of the same type.

The following result will be useful.

Theorem 6 (see [4, Theorem 3.4]). Let be an extension. Then, the natural map is a minimal extension if and only if is a minimal extension. If these (equivalent) conditions hold, then one has the following three conclusions.(a) canonically.(b) and are the same type of minimal extension.(c)If , then .

The next two results recall/develop some facts about integral t-closed FIP extensions that will be used in Section 3.

Proposition 7. Let be an integral t-closed extension. Then,(1)there is a finite chain of minimal extensions, , in which , , and each is inert;(2)the canonical map is a homeomorphism (in the Zariski topology). Moreover, there is a positive integer such that is an intersection of pairwise distinct maximal ideals of and also an intersection of pairwise distinct maximal ideals of .

Proof. (1) is a special case of [1, Lemma 5.6].
(2) Using (1), take to be a finite maximal chain of inert minimal extensions with and . In view of Theorems 4 and 3(d), the canonical continuous map (which is a Zariski-closed map, owing to integrality) is a bijection and hence a homeomorphism for all . By composing these maps, we see that the canonical map is also a homeomorphism. Since any t-closed extension is seminormal, [1, Lemma 4.8] shows that is a radical ideal of (and hence also a radical ideal of ). Hence, by a characterization of integral FCP extensions in [1, Theorem 4.2(a)], is an intersection of finitely many, say, , pairwise distinct maximal ideals of . Since is a bijection and integrality ensures that , it follows that is also an intersection of pairwise distinct maximal ideals of .

Proposition 8 (see [10, Lemme 3.10]). Let be a field and let be an integral ring extension. Then, is t-closed if and only if is a field.

2. -Closed FIP Extensions of Nagata Rings

Consider an FIP field extension and an indeterminate . The first goal of this section is to show that the map defined by is an order-isomorphism. We will need to consider two cases, namely, where is finite and where is infinite. It will be convenient to use the following version of the Primitive Element Theorem: a finite-dimensional field extension has FIP if and only if for some . (Note also that if is any FIP extension of fields, then .)

Proposition 9. Let be an field extension, where is a finite field. Then, the map , given by , is an order-isomorphism, and so has .

Proof. Since is a finite field and , is a finite-dimensional Galois extension of (cf. [11, Proposition 4, Ch. V, Sec. 12, p. 91], taking and ). Hence, is a Galois extension (cf. [11, Théorème 5, Ch. Vl, Sec. 10, p. 68]). Then [11, Corollaire 1, Ch. V, Sec. 10, p. 69] shows that for each , there exists (a unique) such that and . In particular, is surjective and hence an order-isomorphism.

Before getting a result similar to Proposition 9 for the case of an infinite field, we need a lemma. It will use the following definition: if is a field and , let denote the set of all such that exists (for some ).

Lemma 10. Let be an infinite field and such that the set is finite. Then, .

Proof. Write , where and are two relatively prime polynomials in and . Since the set of values is finite and is infinite, we can see (by ignoring the finitely many roots of in ) that there must exist some value that is attained infinitely often, that is, such that is infinite. But , with such that , gives , so that has infinitely many roots in . Thus, , giving that .

Proposition 11. Let be an field extension, where is an infinite field. Then, the map , given by , is an order-isomorphism, and so has .

Proof. By the Primitive Element Theorem, for some . Let denote the (monic) minimal polynomial of over . A standard proof of the Primitive Element Theorem (as given, for instance, in [11, Théorème 1, Ch. V, Sec. 7, p. 39]) shows that the -subalgebras of are of the form , where denotes the -subalgebra of generated by the coefficients of , as runs over the set of monic polynomials in that divide in . (The reader is cautioned that the notation does not refer to a ring of fractions but merely to a -algebra that is constructed from in a certain way.) We will show that each -subalgebra of is of the form for some suitable .
Observe that has FIP since ; and because (by, for instance, [11, Théorème 5, Ch. V, Sec. 10, p. 68] or [4, Lemma 3.1(e)]). Therefore, is also the minimal polynomial of over . We next proceed to describe the -subalgebras of by reapplying the method that was used above to describe the -subalgebras of .
Let be two monic polynomials such that . Write , , and , where and for each , so that we have the following equation: . For a fixed , consider the substitution . Then, gives , supposing for the moment that all the expressions and are meaningful. Under this assumption, it would follow that and each divide in for each . As there are only finitely many such monic polynomials, it must be the case that for each , the sets and are each finite. Since is an infinite field, it follows from Lemma 10 that for each , we have . Consequently, and each divide in . Hence, the -subalgebras of are of the form , where denotes the -subalgebra of generated by the coefficients of , as runs over the set of monic polynomials in that divide in . It follows that , where, as above, denotes the -subalgebra of generated by the coefficients of . In particular, is surjective and hence an order-isomorphism, as asserted.

In the context of the preceding proof, it is interesting to note that since has FIP, we can write for some . Then, . Thus, not only does generate over , but it also generates over .

We can now present this paper’s first contribution to the question under consideration.

Theorem 12. Let be an integral t-closed extension.
Then, , the function is an order-isomorphism, and is an extension.

Proof. Since is an order-preserving and order-reflecting injection, it suffices to prove the first assertion. As has FIP, we have , by [1, Corollary 3.2]. Set and write . By [1, Theorem 3.6], the map defined by is a bijection and . In the same way, we can show that , because and for each and, by [4, Lemma 3.3], . (Note that when we applied [1, Theorem 3.6] to , we did not need to know already that this extension has FIP; it was enough that this extension has FCP, which it indeed inherits from by [4, Theorem 3.9].) Thus, if for each , then . So, without loss of generality, we may assume that is a quasi-local ring which is properly contained in . Note that, in passing from to , the extension has retained the “integral -closed ” hypothesis. Therefore, [4, Lemma 3.17] can be applied, giving that ; necessarily, . Thus, by a standard homomorphism theorem, ; similarly, as , we get (cf. also [1, Proposition 3.7(c)]).
Since and it follows from Propositions 9 and 11 that the FIP field extension satisfies , the equalities that we have collected combine to show that . The proof is complete.

We close this section with some comments about Galois groups and Galois extensions of rings, some of which will be used in the next section. In particular, the isomorphism in Lemma 13 will play a key role in the proof of Lemma 25.

Lemma 13. Let be an algebraic field extension and let (resp., ) be the group of -automorphisms resp., -automorphisms of (resp., of ). Then there is an isomorphism , denoted by , such that for each and each . Moreover, the canonical map is an isomorphism.

Proof. Let and . It is easy to check that we can well define a function by . It is then clear that .
Conversely, for each , let denote the restriction of to . Since is algebraic over and algebraically closed in , it is easy to see that maps (injectively) into itself. (Similarly, so does the restriction of to .) This mapping is, in fact, surjective, for if and we take such that , then . Consequently, .
It is now easy to check that the function defined by is an isomorphism. The final assertion is a special case of [4, Lemma 3.1(e)].

Recall that there is a theory of Galois ring extensions that generalizes the theory of (finite-dimensional) Galois field extensions. A summary of much of that theory appears in Section 1 of a book by Greither [12], with which we will assume familiarity. One may also find many examples in that book. For an extension of rings , let denote the set of all -subextensions of such that is separable (in the usual sense, namely, that is projective over ). Recall also that a ring is said to be connected if its only idempotent elements are and . A ring is connected (if and) only if is connected. Indeed, it was shown in [13, Theorem 2.4] that for any ring , each idempotent element of must belong to .

Proposition 14. Let be a Galois extension of rings with finite Galois group . Then the following assertions hold. (1) is a Galois (ring) extension with Galois group isomorphic to .(2)If, in addition, is connected, then the canonical map , given by , is an order-isomorphism.

Proof. (1) Since is integral (because Galois extensions with finite Galois groups are module-finite), [4, Lemma 3.1(e)] may be applied to show that the natural map is an isomorphism. Next, an application of [12, Lemma 1.11] shows that is a Galois extension with Galois group .
(2) As is connected, the Chase-Harrison-Rosenberg Theorem tells us that has the same (finite) cardinality as the set of all subgroups of (cf. [12, Theorem 2.2]). Also, we noted above that inherits the “connected” property from . Thus, in view of (1), we see similarly that also has the same cardinality as the set of all subgroups of . It therefore suffices to show that if , then (for then, the restriction of to is a necessarily injective function, and an application of the Pigeon-hole Principle would finish the proof). In fact, inherits from the property of being integral over , and so, by another application of [4, Lemma 3.1(e)], canonically. Since is separable over and separability is preserved by arbitrary base changes, it follows that , as desired.

Note that the preceding result gives another proof of the special case of Lemma 13 where is a finite-dimensional Galois field extension.

3. The General Case

The aim of this section is to prove that for any ring extension such that has FIP, the map , given by , is an order-isomorphism. Since is known to be an order-preserving and order-reflecting injection [4, Lemma 3.1(d)], it remains only to show that is surjective. As an FIP extension has FCP, we will prove this result by induction on the length of a maximal chain of minimal extensions. That induction will begin in Proposition 31 by showing that if has FIP and is such that is a minimal extension, there exists such that . We will need to first treat the case of a quasi-local base ring in Proposition 30, for which the following lemmas will be useful.

Lemma 15. Let be a field, with and minimal (ring) extensions such that the composite exists. If is ramified and is decomposed, then is the only such that is a decomposed minimal extension satisfying .

Proof. Since is ramified, the comments following Theorem 4 provide an element such that and so that is a local ring with maximal ideal . Since is decomposed, those same comments provide an element such that and so that has exactly two maximal ideals, say, and . As is a field, the extensions and necessarily have the same crucial maximal ideal (i.e., ), and so [6, Proposition 7.6] can be applied. There are three cases.
(1) Assume that . Then, by [6, Proposition 7.6(a)], is a decomposed minimal extension and is a ramified minimal extension. Since , we have , and so . Thus, is a -basis of . We show next that, given and , if is such that is a decomposed minimal extension with , then . As above, we can write , for some such that . Write , for some . As , we get so that () and . By , either or . Thus, by , in any event. Hence, is either or , and so . Then , whence by the minimality of .
(2) Assume that . Then one can reason as in case (1), with replacing .
(3) Finally, assume that . By the assumptions of (3), the elements and that were introduced above satisfy and . Notice that and are linearly independent over . (Otherwise, for some ; multiplication by leads to , whence and , a contradiction.) In fact, we claim that is not in the -span of . If the claim fails, for some . Multiplication by leads to . Then, by the above comment about linear independence, , whence , contradicting that same comment. This proves the claim. Hence, is a -basis of . We proceed to prove that, given and , if is such that is a decomposed minimal extension with , then .
As for some element , we have . Write , with . As , we get so that , ,   and . By , either or , and so by , in any event.
Suppose first that . Then gives and gives . From and , we deduce that either (in which case, and , a contradiction) or and (in which case, , and so the minimality of forces , as desired).
Lastly, suppose that . Then gives and gives . Combining and , we get either (in which case, and , a contradiction) or and (in which case, , and so the minimality of forces ).

Proposition 16. Let and be minimal ring extensions of a quasi-local ring , such that the composite exists and such that is ramified and is decomposed. Then is the only such that is a decomposed minimal extension satisfying .

Proof. Since and are integral minimal extensions, we get that (the crucial maximal ideal) is a common ideal of , and and hence also an ideal of . Put , , , and . We are reduced to the situation of Lemma 15, since is a minimal extension of the same type as , for . Hence, by Lemma 15, is the only ring such that is a decomposed minimal extension satisfying . Now, suppose that is such that is a decomposed minimal extension satisfying . Then is such that is a decomposed minimal extension and , and so . With denoting the canonical surjection, it follows that .

The next lemma uses the notion of the ideal-length of an ideal of a ring , in the sense of [14, Definition, p. 233]. For the sake of completeness, we recall that definition: , where denotes the length of an -module (with such length taken to be if the module does not have a composition series) and denotes the complement of the union of the associated primes of .

To motivate Lemma 17, note that the following is a consequence of the classification of the minimal extensions of a field [3, Lemme 1.2]. If is a field and is a minimal (hence integral FIP) extension, then is not a reduced ring if and only if . In view of the comment following Theorem 4, the preceding assertion is the special case of Lemma 17 in which is a field and is a minimal extension.

Lemma 17. Let be an integral extension, , and . Then is not a radical ideal of if and only if there exists such that is a ramified minimal extension such that .

Proof. Since is an integral extension with FCP, [1, Theorem 4.2(a)] shows that both and are Artinian rings and hence of (Krull) dimension . In particular, . Set , , , and for each that lies over in ; necessarily, .
Suppose first that there exists such that is a minimal ramified extension with . By Theorem 4(c), there exists with . Consequently, is an -primary ideal of . We proceed to derive a contradiction from the assumption that is a radical ideal of . Note that is an Artinian ring and, hence, has only finitely many prime (necessarily maximal) ideals. Thus, since is being assumed radical, we have , where is the (finite) list of (pairwise distinct) maximal ideals of that contain . Therefore, by the Chinese Remainder Theorem, , a direct product of finitely many fields. Thus, is also isomorphic to a direct product of finitely many fields and hence is a reduced ring. So is a radical ideal of . But the radical of in is . Since , we have the desired contradiction.
For the converse, assume that is not a radical ideal of ; equivalently, is not a radical ideal of . As is zero-dimensional and Noetherian (i.e., Artinian), a classic result [14, Theorem 9, page 213] gives a unique primary decomposition of in , namely, , where is a -primary ideal of for each . As is not a radical ideal of , there exists an index such that . Fix .
As is zero-dimensional, no ideal of has embedded components. Hence, by [14, Theorem 24, p. 234 and Theorem 26, p. 235], is a -primary ideal of finite ideal-length, with ; moreover, there exists a -primary ideal of such that , with and adjacent ideals. Thus, by [14, Corollary 2, p. 237]. It follows that is a quasi-local Artinian ring with maximal ideal such that .
Consider the canonical surjection . Put and . Note that and is a -primary ideal of . By a standard homomorphism theorem, the above “adjacency" assertion implies that and are adjacent ideals of ; and is a quasi-local Artinian ring with maximal ideal () such that .
We have so that and the field can now be viewed as a subring of . We next apply a piece of the structure theory of complete local rings. By the proof of [15, Corollaire 19.8.10, p. 113], , where and is the vector-space dimension of over . It follows that has a field of representatives which contains .
Let be the canonical surjection and set so that is an ideal of satisfying . It follows that . Also, recall that . Thus, if we wish to establish that is a minimal ramified extension with by appealing to Theorem 4(c), it suffices to prove that . To that end, note first that is a one-dimensional vector space over . But is also a one-dimensional vector space over () because and are adjacent ideals of . This completes the proof.

We can now give the first and second of the results in this section wherein a suitable ring in is shown to take the desired form.

Proposition 18. Let be a quasi-local ring, a ramified minimal extension, and a decomposed minimal extension. Let be such that is a ramified minimal extension. Then there exists such that .

Proof. We know that is a quasi-local ring and that and each have exactly two maximal ideals. We claim that the integral extension has FIP. To see this, note first that is subintegral (since it is ramified) and is seminormal (since it is decomposed: cf. [1, Lemma 5.3(a)]). Therefore, . As and necessarily each have FIP (being minimal extensions) and is infra-integral (as a consequence of parts (b) and (c) of Theorem 4), it therefore follows from [1, Proposition 5.5] that has FIP, thus proving the above claim. Hence, by Lemma 17, is not a radical ideal of .
We next claim that is not a radical ideal of . Since is ramified and is decomposed, it follows from Theorem 4 (and integrality) that there are exactly two prime ideals of lying over . Denote these prime ideals by and , and note that . Thus, has exactly two maximal ideals, namely, and . Suppose that the claim fails; that is, is a radical ideal of . Then is an intersection of some prime ideals of  , and each of these primes must be maximal (because it lies over the maximal ideal ). Thus, either or is of the form . Hence, is either or of the form . Thus, is a radical ideal of , the desired contradiction, thus proving the above claim. Hence, by another application of Lemma 17, there exists such that is a ramified minimal extension.
Consider and . We claim that and are incomparable ideals of . To see this, first observe that is the intersection of the two maximal ideals of , while Theorem 4(c) ensures that is a primary nonmaximal ideal of whose radical is a maximal ideal of . It is now clear that . On the other hand, if , then (which, by the Chinese Remainder Theorem, is isomorphic to a direct product of two fields) would map homomorphically onto , which is a nonzero quasi-local ring but not a field. This contradiction establishes the above claim. Thus, [6, Proposition 6.6(a)] can be applied to the base ring . It follows that is a minimal extension. Thus, the minimality of gives , and then [6, Proposition 6.6(a)] shows that inherits the “decomposed minimal extension” property from . Then, by Theorem 6, is also a decomposed minimal extension.
It remains only to prove that . This, in turn, is a consequence of the uniqueness assertion in Proposition 16. To check the applicability of that result here, it suffices to note that . Notice that can be obtained by “composing” the ramified minimal extension and the decomposed minimal extension . As is therefore infra-integral, it follows from [1, Lemma 5.4] that each maximal chain of rings going from to has length . Therefore, the integral extension must be a minimal extension, necessarily decomposed since two distinct prime ideals of lie over in . Next, since , the minimality of gives that , and so . It therefore suffices to prove that . This, in turn, follows from the minimality of , since . The proof is complete.

Proposition 19. Let be a quasi-local ring and let be an extension such that is an extension with . Let be such that is a decomposed minimal extension. Then there exists such that .

Proof. Set and . Then , , and by [4, Lemma 3.15]. As is infra-integral, it follows that . Since is an FIP extension, each maximal chain of rings going from to must be finite. Set and . We will inductively construct two increasing chains and , with , for some integer , such that and also such that the following induction hypothesis is satisfied for each : is a decomposed minimal extension and either and are both ramified minimal extensions or we have both and .
We begin with the induction basis, that is, the case . As , we can choose such that is a ramified minimal extension. Consider the two cases identified in [6, Proposition 7.6] corresponding to the choices , , and (along with ). In the first case, the induction hypothesis holds for if we take and (). In the second case, [6, Proposition 7.6] provides certain rings and . If we could find such that in this second case, then the induction hypothesis would hold for if we take and (). To that end, it suffices, by [4, Theorem 3.4], to show that is contained in , the seminormalization of in . As is a ramified minimal extension by [6, Proposition 7.6], it follows that is subintegral, whence , thus completing the proof of the induction basis.
Next, for the induction step, suppose that the induction hypothesis holds for some . If , the inductive construction of the chains of rings is terminated. Assume, instead, that . We will sketch how to adapt the argument that was given for the induction basis. First, choose such that is a ramified minimal extension. Next, consider the two cases identified in [6, Proposition 7.6], corresponding to the choices , , and , along with . (Note that the meanings of the symbols , and have changed in this paragraph.) The analysis in the preceding paragraph carries over, mutatis mutandis, to provide rings , , and with the desired behavior. This completes the proof of the induction step. Since has FCP, we thus find (and fix) a positive integer such that .
We will show, by a decreasing induction proof, that for each with , there exists such that . Once this has been established, taking will complete the proof, for then , as desired.
We turn to the basis for the decreasing induction, that is, the case . Since is decomposed, we get that . As the FIP extension is seminormal and infra-integral, we can now apply [4, Proposition 4.17], thus finding some such that . This completes the proof for the case .
Next, for the induction step of the decreasing induction, assume that we have a positive integer , along with some such that . Suppose first that and . Then satisfies . We proceed to accomplish the same in the remaining case.
In that remaining case, and . Then is a ramified minimal extension, while is decomposed minimal. Moreover, and are incomparable, because is a primary nonmaximal ideal of whose radical is a maximal ideal while is a nonmaximal ideal of that is the intersection of two maximal ideals of . Therefore, by [6, Proposition 6.6(a)], is such that is a ramified minimal extension. This implies that , since [4, Proposition 4.13] ensures that is linearly ordered by inclusion. Applying Proposition 18 to the chain produces such that .
The preceding two paragraphs show that, in all (i.e., both) cases, there exists such that . We wish to find such that . If , we may reason as two paragraphs ago, namely, by now taking . On the other hand, if , we may reason as in the preceding paragraph by applying Proposition 18 to produce such that . This completes the proof of the induction step.

Our next results with a conclusion of the form “ for some ” will begin with Lemma 25. Several preparatory results are needed first. As usual, if is a ring, then will be used to denote the ring .

Lemma 20. Let be an integral extension with conductor . Let and assume that there exist distinct which each lie over . Then and are isomorphic as -algebras if and only if there exists such that is a decomposed minimal extension with conductor .

Proof. Since is an integral extension with FCP, it follows as above from [1, Theorem 4.2(a)] that and are Artinian rings so that and . Assume first that and are isomorphic ()-algebras. Identify with , and let denote this field. With , we see via the Chinese Remainder Theorem that . Identifying with in this way and viewing as a subring of via the diagonal map (given by ) then allows us to view as a subring of . Consider the canonical surjection , and set . Note that is a subring of which contains and and that . It follows that and . Hence, by the criterion in Theorem 4(b), is a decomposed minimal extension with conductor .
Conversely, suppose that there exists such that is a decomposed minimal extension with conductor . Then and are the only prime (in fact, maximal) ideals of that lie over in . By the characterization of “decomposed extensions” in Theorem 4(b), the canonical map is an isomorphism for . In particular, , as desired.

Lemma 21. Let be an integral extension. Let be a t-closed subextension of and a decomposed minimal subextension of . Set . Then, is a decomposed minimal extension and is an integral t-closed extension.

Proof. Since is an integral FIP extension, it follows from Proposition 7(a) that can be obtained via a finite (maximal) chain of inert minimal extensions. Also, note that since (cf. [1, Lemma 5.6]); similarly, .
Assume first that is quasi-local. Pick a maximal (finite) increasing chain consisting of () inert extensions going from to . Set and . We will prove by induction on that is a decomposed minimal extension with and that can be obtained via a chain of inert extensions.
We begin with the induction basis. Then, , and so is a t-closed minimal extension and hence inert (cf. [1, Lemma 5.6]). Thus, it follows from [6, Proposition 7.1(a)] that is a decomposed minimal extension with ; and it follows from [6, Proposition 7.1(b)] that can be obtained (in two ways) via a chain of inert extensions.
For the induction step, the induction hypothesis states that, for some , with , is such that is a minimal decomposed extension with and can be obtained via a chain of inert extensions. Note that is a quasi-local ring, with being inert. Consider (). By another application of [6, Proposition 7.1(a)], is a decomposed minimal extension with ; and, by [6, Proposition 7.1(b)], can be obtained (in two ways) via a chain of inert extensions. Thus, can be obtained via a chain of inert extensions (while is obtained via a chain of inert extensions). This completes the proof of the induction step and establishes the result in case is quasi-local.
Finally, suppose that is not quasi-local. Set . Pick with . Then, since . It follows that . Also, by [6, Proposition 4.6], is a decomposed minimal extension. We wish to use the above case of a quasi-local base ring to conclude that is a decomposed minimal extension and is an integral t-closed extension. To do so, one must also address the possibility that ; but, in this degenerate case, , and the assertions follow. Next, note that inherits the t-closed property from . Hence, by [10, Théorème 3.15], the integral extension is t-closed. Lastly, the assembled information combines with [6, Proposition 4.6] to show that is decomposed.

As usual, it will be convenient to call a field extension simple if for some element .

Lemma 22. Let and be ring extensions of a field , such that the composite exists and has . Assume, in addition, that is a decomposed minimal extension and is an integral -closed extension. Then,(1) is a simple field extension of . One can identify as -algebras and as -algebras;(2)let be such that is t-closed with . Then there exist and such that and , where . Thus, the diagonal map , given by , allows to be identified with as a -algebra.

Proof. (1) By Proposition 8, is a field. Hence, since inherits FIP from , the Primitive Element Theorem applies to show that the field extension is simple. Next, by applying Lemma 21, we get that is a decomposed minimal extension (and that is an integral t-closed extension). By the classification of the minimal extensions of a field [3, Lemme 1.2], as -algebras and, similarly, as -algebras. There is no harm in identifying and ; nor is there any harm in viewing (resp., ) as a subring of (resp., of ) via the diagonal map.
(2) The case follows from the first assertion in (1) (with ). Thus, without loss of generality, . By reasoning as above, Proposition 8, Lemma 21, and the Primitive Element Theorem can be used to show that is a field, is a decomposed minimal extension, and for some element