Acta Math Vietnam https://doi.org/10.1007/s40306-018-0271-2

Absolutely lq-Finite Extensions El Hassane Fliouet1

Received: 14 April 2017 / Revised: 2 March 2018 / Accepted: 13 March 2018 © Institute of Mathematics, Vietnam Academy of Science and Technology (VAST) and Springer Nature Singapore Pte Ltd. 2018

Abstract We describe the lower quasi-finite extensions K/k of characteristic p > 0, which −n are defined as follows: for every n ∈ N, k p ∩ K/k is finite. We are especially interested in examining the absolute case. In this regard, we give necessary and sufficient condition for an absolutely lq-finite extension to be of finite size. Moreover, we show that any extension that is at the same time modular and lq-finite is of finite size. Furthermore, we construct an example of extension K/k of infinite size such that for any intermediate field L of K/k, L is of finite size over k. Keywords Purely inseparable · Irrationality degree · Modular extension · q-finite extension · lq-finite extension · Absolutely lq-finite extension Mathematics Subject Classification (2010) 12F15

1 Introduction Let K/k be a purely inseparable extension of characteristic p > 0. We first introduce the size measurement of a purely inseparable extension by di(K/k) = supn∈N (|Bn |) where Bn −n is a minimal generator of k p ∩ K/k, and for any family k ⊆ L ⊆ L ⊆ K of purely inseparable extensions, we then establish the inequality di(L /L) ≤ di(K/k). −n An extension K/k is said to be lq-finite if, for any n ∈ N, k p ∩ K/k is finite, and if moreover di(K/k) is finite, K/k will be called a q-finite extension. It is clear that K/k is q-finite when [k : k p ] is finite. For reasons of noncontradiction, we give examples that differentiate these notions. However, our main objective is to describe the absolutely lqfinite extensions; in this regard, we show that the following properties are equivalent:

 El Hassane Fliouet

[email protected] 1

Regional Center for the Professions of Education and Training, Agadir, Morocco

E. H. Fliouet

(1) For any subextension L/k of K/k, K/L is lq-finite. (2) Any subextension L/k of K/k satisfies L/k(Lp ) is finite. (3) For any subextension L/k of K/k, we have L/rp(L/k) is finite where rp(L/k) is the relatively perfect closure of L/k. (4) Any decreasing sequence of subextensions of K/k is stationary. An extension that satisfies one of the above conditions is called an absolutely lq-finite extension. It is about an absolute distribution of lq-finitude. Below is a diagram relating the some notions in the paper: For a purely inseparable extension K/k, finite ⇒ q-finite ⇒ absolutely lq-finite ⇒ lq-finite. In particular, any extension that is at the same time modular and lq-finite is q-finite. More generally, we show that K/k is q-finite if and only if K/k is absolutely lq-finite and for any intermediate field L of K/k, the smallest subextension m of L/k such that L/m is modular is nontrivial (m = L). Finally, we approve this notion by constructing an extension K/k of infinite size such that for every intermediate field L of K/k, L/k is q-finite. Furthermore, this example makes it possible to distinguish between simple and absolute lq-finitude. It should be noted that in Sections 2 and 3, we establish fundamental inequalities concerning size and exponents, and which will then be used as key tools for studying the properties of lq-finitude.

2 Irrationality Degree Throughout this note, k always designates a commutative field of characteristic p > 0,  an algebraic closure of k, and all fields under consideration will be purely inseparable extensions of a common ground field k. They are to be viewed as contained in . It is also convenient to denote sometimes [k, K] the set of intermediate fields of an extension K/k, −n and for any n ∈ N, we will often put kn = k p ∩ K. Definition 2.1 Let K/k be a purely inseparable extension. A subset G of K is said to be an r-generator of K/k if K = k(G), and if moreover for each x ∈ G, x  ∈ k(G\x), then G will be called a minimal r-generator of K/k. Definition 2.2 Given an extension K/k of characteristic p > 0, and a subset B of K. We say that B is an r-basis (relative p-basis) of K/k if B is a minimal r-generator of K/k(K p ). B is said to be r-free (or r-independent) over k if B is an r-basis of k(B)/k, in the opposite case B is said to be r-dependent over k. Any r-basis of k/k p is called a p-basis of k, and any subset of elements of k which is r-free over k p will be called p-independent (or p-free) over k p . As in linear algebra, the r-dependence in K/k(K p ) is a dependence relation (cf. [12, Lemma 6.1]), and consequently, according to (cf. [12, Theorem 1.3]) we have: 1. Every extension K/k has an r-basis, and any two r-bases of K/k have the same cardinality. 2. Any relatively r-independent set in K/k(K p ) can be extended to an r-basis of K/k. 3. Any r-generator G of K/k(K p ) contains an r-basis of K/k. Recall that K is said to have an exponent (or, to be of bounded exponent) over k if there e exists e ∈ N such that K p ⊆ k, and the smallest integer that satisfies this relation will

Absolutely lq-Finite Extensions

be called the exponent (or height) of K/k. Taking into account [14, Corollary], if K/k has an exponent, a subset B of K is an r-basis of K/k if and only if B is a minimal rgenerator of K/k. However, a minimal r-generator may not exist in the general case (cf. [14, Lemma 1.16 and Proposition 1.23]). We now have the required tool for defining the irrationality degree of an extension K/k, let us consider a purely inseparable extension K/k −n of characteristic p > 0, clearly for any n ∈ N, k p ∩ K/k has an exponent, and in addition, −n p ∩ K/k depends only on n. the cardinality of any minimal r-generator of k Definition 2.3 The invariant di(K/k) = supn∈N (|Bn |) where Bn is a minimal r-generator −n of k p ∩ K/k will be called the irrationality degree of K/k. Here, the sup is used in the sense of [1, III, p. 25, Proposition 2]. Moreover, for reasons of thematic specificity and coherence, di(k/k p ) will designate the imperfection degree of k, and it is denoted by di(k). Theorem 2.1 Let k ⊆ L ⊆ K be purely inseparable extensions. Then, we have di(L/k) ≤ di(K/k). Furthermore, di(K/k) = sup(di(L/k))L∈[k,K] . Proof By virtue of [1, III, p. 25, Proposition 2], it is enough to establish the theorem for extensions of bounded exponent. We note that if the special case [K : L] < +∞ holds, by [14, Proposition 1.8], the cardinality of a minimal r-generator of L/k is not greater than that of a minimal r-generator of K/k, i.e., di(L/k) ≤ di(K/k). Now, we will give the proof in the general case. If K/k has an exponent 1, i.e., K p ⊆ k; therefore, Lp ⊆ k. According to the dependence relation, any relatively r-basis of L/k can be extended to an r-basis of K/k, and consequently, di(L/k) ≤ di(K/k). Now, we consider a natural integer e distinct from 0 and 1. We proceed by induction, assuming that the theorem holds for any extension of exponent < e, and let K/k be a purely inseparable extension of exponent e. Let also BL and BK be two minimal r-generators respectively of L/k and K/k. It is clear that k(K p ) ⊆ L(K p ) ⊆ K, and therefore, there exists B1 ⊆ BL and B2 ⊆ BK such that B1 and B2 are two r-bases, respectively of L(K p )/k(K p ) and K/L(K p ). From the transitivity of the r-independence (cf. [14, Proposition 1.7]), B1 ∪ B2 is an r-basis of K/k(K p ). We then put k1 = k(B1 ) and BL = BL \ B1 ; we immediately check that L ⊆ k1 (K p ) = k1 (B2 p ), and k1 (K p )/k1 has an exponent < e. By applying the recurrence relation, we get the cardinality of a minimal r-generator of L/k1 is not greater than that of a minimal r-generator of k1 (K p )/k1 which is in turn not greater than that of an r-generator of k1 (K p )/k1 , and hence, |BL | ≤ |B2 p | = |B2 |. Since B1 ∩ BL = ∅ and B1 ∩ B2 = ∅, then |B1 ∪ BL | ≤ |B1 ∪ B2 |. But, as B1 ∪ B2 and Bk are two r-bases of K/k, we get |B1 ∪ B2 | = |Bk |, and consequently |BL | ≤ |BK |, or again di(L/k) ≤ di(K/k). One of the applications of Theorem 2.1 is the following result: Theorem 2.2 For any purely inseparable extension K/k, we have di(K/k) ≤ di(k). Proof It is sufficient to note that a subset B of k  is a p-basis of k if and only if for every −n −n −∞ −n n ≥ 1, B p is an r-basis of k p /k. As k p = n≥1 k p , then for each n ≥ 1, we have −n −∞ −∞ k p ∩ K ⊆ k p , and consequently, di(K/k) ≤ di(k p /k) = di(k). Proposition 2.3 Let (Kn /k)n∈N be an increasing  family of purely inseparable subextensions of an extension /k. Then, we have di( n∈N (Kn )/k) = supn∈N (di(Kn /k)).

E. H. Fliouet

 Proof We denote n∈N Kn by K, and let j be a positive integer. It is immediate that kj =  −j −j k p ∩ K = n∈N (k p ∩ Kn ). In the following, two cases can be distinguished: −j

Case 1 If di(kj /k) is finite, or again kj /k is finite. As for every n ∈ N, we have k p ∩ −j −j −j Kn ⊆ k p ∩ Kn+1 ⊆ k p ∩ K, then the sequence of integers ([k p ∩ Kn : k])n∈N is increasing and bounded, so it is stationary from a rank n0 ; and therefore for all n ≥ −j −j −j −j n0 , k p ∩ Kn = k p ∩ Kn+1 . Furthermore, di(k p ∩ K/k) = di(k p ∩ Kn0 /k) = −j supn∈N (di(k p ∩ Kn /k)). −j

−j

Case 2 If di(k p ∩ K/k) is infinite, i.e., supn∈N (di(k p ∩ Kn /k)) is not finite. Since   −j −j −j j j k p ∩ K = n∈N (k p ∩ Kn ), thus if Bn is an r-basis of k p ∩ Kn /k, then n∈N Bn is an  −j −j j r-generator of k p ∩ K/k. Accordingly, di(k p ∩ K/k) ≤ | n∈N Bn |, and from [1, III,  −j j j p. 49, Corollary 3], | n∈N Bn | ≤ supn∈N (|Bn |) = supn∈N (di(k p ∩ Kn /k)). In view of these two cases, we are able to deduce that di(K/k) ≤ supn∈N (di(Kn /k)). But as Kn ⊆ K for every n ≥ 1, by Theorem 2.1, we obtain supn∈N (di(Kn /k)) ≤ di(K/k), and consequently, di(K/k) = supn∈N (di(Kn /k)). Let K1 and K2 be two intermediate fields between k and K that are k-linearly disjoint. For every L1 ∈ [k, K1 ] and L2 ∈ [k, K2 ], it is well known that L2 (K1 ) and L1 (K1 ) are k(L1 , L2 )-linearly disjoint. In particular, L2 (K1 ) ∩ L1 (K2 ) = k(L1 , L2 ). Define {Fi | i ∈ J } of field extensions of k to be linearly disjoint over k if every finite subfamily  is linearly disjoint over k (cf. [11, p. 36]). It is trivially obvious that k((Fi )i∈J ) = i∈J Fi ⊗k (⊗k Fi )i∈J (the tensor product is used according to [2, III, p. 42, Definition 5] if and only if the family (Fi /k)i∈J is k-linearly disjoint. Moreover, the properties of linear disjointness of the finite case naturally extend to any linearly disjoint k-linearly disjoint, by family. In particular, for all i ∈ J , let Li ∈ [k, Fi ], if (Fi /k)i∈J is  transitivity of linear disjointness, we have (L /k) (respectively, (( i i∈J n∈J Ln )Fi /k)i∈J ) is  k (respectively, n∈J Ln ) linearly disjoint. Assume now that K1 /k and K2 /k have an exponent. We verify immediately that if B1 and B2 are two r-bases, respectively of K1 /k and K2 /k, then B1 and B1 ∪ B2 are two r-generators, respectively of K1 (K2 )/K2 and K1 (K2 )/k. Furthermore, di(K1 (K2 )/K2 ) ≤ di(K1 /k) and di(K1 (K2 )/k) ≤ di(K1 /k) + di(K2 /k). More precisely, we get the following result. Proposition 2.4 Under the above conditions, if K1 /k and K2 /k are k-linearly disjoint, then we have: (i) B1 ∪ B2 is an r-basis of K1 (K2 )/k. (ii) B1 is an r-basis of K1 (K2 )/K2 . Proof Here, we only present the proof of the first item; the other is similar. It is clear that K1 (K2 ) = k(B1 ∪ B2 ), and it is enough to show that B1 ∪ B2 is minimal. For this, we assume for example the existence of an element x ∈ B1 such that x ∈ k((B1 \ {x}) ∪ B2 ) = K. As K1 /k and K2 /k are k-linearly disjoint, by transitivity, we have k(B1 ) = K1 and K2 (B1 \{x}) = K are also k(B1 \{x})-linearly disjoint, and thus K1 = K ∩K1 = k(B1 \{x}). However, this contradicts the fact that B1 is an r-basis of K1 /k.

Absolutely lq-Finite Extensions

Corollary 2.5 Let K1 and K2 be two intermediate fields of a purely inseparable extension /k. Then: (i) di(K1 (K2 )/k) ≤ di(K1 /k) + di(K2 /k), and the equality holds if K1 and K2 are klinearly disjoint. (ii) di(K1 (K2 )/K2 ) ≤ di(K1 /k), and the equality holds if K1 and K2 are k-linearly disjoint. Proof It is sufficient to note that K1 (K2 ) = −j (k p



j ∈N (k

p−j

∩ K1 )(k p

−j

∩ K2 ) =



j ∈N K2

∩ K1 ), and if K1 and K2 are k-linearly disjoint, by transitivity of linear disjointness, −j −j for each j ≥ 1, k p ∩ K1 and k p ∩ K2 are also k-linearly disjoint. We thus reduce to the case where K1 /k and K2 /k have an exponent in which case the result immediately follows from the previous proposition. As a special case, we obtain the following result. Corollary 2.6 For every subextension L/k of a purely inseparable extension K/k, we have di(L(K p )/k(K p )) ≤ di(L/k(Lp )), and the equality holds if k(K p ) and L are k(Lp )linearly disjoint. Proof Due to Corollary 2.5. The following result naturally extends Theorem 2.1 Theorem 2.7 For any family k ⊆ L ⊆ L ⊆ K of purely inseparable extensions, we have di(L/L ) ≤ di(K/k).  Proof It’s clear that K = j ∈N L(kj ), and by Proposition 2.3, and Theorem 2.1, we have di(L /L) ≤ di(K/L) = supj ∈N (di(L(kj )/k)) ≤ supj ∈N (di(kj /k)) = di(K/k). Corollary 2.8 For any purely inseparable extension K/k, we have di(K) ≤ di(k). Proof It is sufficient to note that K ⊆ k p di(k).

−∞

, and di(K) = di(K/K p ) ≤ di(k p

−∞

/k p ) =

3 Quasi-finite Extensions Definition 3.1 Any extension whose irrationality degree is finite is called a q-finite extension. In other words, the q-finitude is synonymous of the horizontal finitude. However, the finitude is translated by the horizontal and vertical finitude; it is a finitude in terms of height and length, i.e., K/k is finite if and only if K/k is q-finite of bounded exponent. Furthermore, it is verified that the irrationality degree of an extension K/k is 1 if only the set of intermediate fields of K/k is totally ordered. Then, every extension that satisfies the previous statement will be called a q-simple extension. Let from now on rp(K/k) denote the relatively perfect closure of K/k (cf. [6]).

E. H. Fliouet

Remark 3.1 We recall that when di(k) is finite, in [4] we first showed that K/k(K p ) is finite and di(K) ≤ di(k), and we then defined the irrationality degree of a purely inseparable extension K/k by the formula di(K/k) = di(k) − di(K) + di(K/k(K p )). Moreover, any extension is q-finite if di(k) is finite. With some slight modifications, we can always extend this definition to the case where di(k) is unbounded. We start by choosing a q-finite relatively perfect extension K/k. Given a p-basis B of k, then k = k p (B), and consequently k(K p ) = K p (B). As K/k is relatively perfect, then K = k(K p ) = K p (B). By [12, Corollary 6.3], there exists B1 ⊆ B such that B1 is a p-basis of K. Thus, we will have −∞ −∞ −∞ −∞ −∞ −∞ k p = k(B p ) = k(B1 p ) ⊗k k((B \ B1 )p ) K p K ⊗k k(B1 p ). Notably, −∞ −∞ −∞ −∞ p p p by Corollary 2.5, di(K/k) = di(K ⊗k k(B1 )/k(B1 )) = di(k /k(B1 p )) = −∞ p di(k((B \ B1 ) )/k) = |B \ B1 |. If we interpret (by abuse of language) |B \ B1 | as a difference of the imperfection degree of k and K, by writing |B \ B1 | = di(k) − di(K), we will get di(K/k) = di(k) − di(K). In the general case, suppose that K/k is an arbitrary q-finite extension, so K/rp(K/k) is finite, whence di(K) = di(rp(K/k)), and consequently di(K/k) = di(rp(K/k)/k) + di(K/k(K p )) = di(k) − di(K) + di(K/k(K p )) (cf. Proposition 3.4 below). It should be noted in this regard that all the results of the articles [5, 7, 8] can naturally be generalized by translation to any q-finite extension. Let L/k be a subextension of a q-finite extension K/k, for each n ∈ N, we recall that kn −n designates k p ∩ K. We immediately check that n (i) The q-finitude is transitive, in particular, for each n ∈ N, K/k(K p ) and kn /k are finite. (ii) There exists n0 ∈ N, for each n ≥ n0 , di(kn /k) = di(K/k). In addition, here are some applications of [6, Lemma 2.1] that will be needed in the rest of this paper. n

Proposition 3.1 Let K/k be a q-finite extension. The sequence (k(K p ))n∈N stops over rp(K/k) from a n0 . In particular, K/rp(K/k) is finite. As a consequence, we have: Corollary 3.2 The relatively perfect closure of a q-finite extension K/k is not trivial. More precisely, rp(K/k)/k has unbounded exponent if the same is also true for K/k. Proof Immediate. Proposition 3.3 For every q-finite extension K/k, there exists n ∈ N such that K/kn is relatively perfect. Moreover, kn (rp(K/k)) = K. Proposition 3.4 The irrationality degree of a q-finite extension K/k satisfies the following equality: di(K/k) = di(K/k(K p )) + di(rp(K/k)/k) = di(K/rp(K/k)) + di(rp(K/k)/k). Proof Let G be an r-basis of K/k and Kr = rp(K/k), so k(G)/k has an exponent denoted by m and K = Kr (G). In particular, for each n ≥ m, k(G) ⊆ kn . Having regard of rindependence of G over k(K p ) and since that k(kn p ) is a subset of k(K p ), we deduce that G is r-free over k(kn p ) for each n ≥ m. We complete G into an r-basis of kn /k by

Absolutely lq-Finite Extensions

a subset Gn of kn . Under these conditions, for a sufficiently large integer n, we should have |G| + |Gn | = supj ≥m (|G| + |Gj |) = di(K/k) = di(Kr (G)/k) ≤ di(Kr /k) +  m di(k(G)/k) = di(Kr /k) + |G|, and so |Gn | ≤ di(Kr /k). However, as n≥m k(kn p ) =   m m m m m p p p p ) = K (K p ), by Theorem 2.1, r n≥m k(Gn , G ) = n≥m k(Gn ) = k(K m for a sufficiently large integer n, we will also have di(Kr /k) ≤ di(Kr (K p )/k) = m m di(k(kn p )/k) ≤ |Gn p | = |Gn |. Hence, |Gn | = di(Kr /k) for a sufficiently large integer n, and consequently di(K/k) = di(Kr /k) + di(K/k(K p )). The following result is a special feature of finite extensions. Corollary 3.5 For a q-finite extension K/k to be finite, it is necessary and sufficient that di(K/k) = di(K/k(K p )). Theorem 3.6 For all q-finite extensions k ⊆ L ⊆ K, we have di(K/k) ≤ di(K/L) + di(L/k). The equality holds if and only if L/k(Lp ) and k(K p )/k(Lp ) are k(Lp )-linearly disjoint.  p−n ∩ K and K/k are q-finite, by Theorem 2.1, for a suffiProof As K = n∈N L −n ciently large integer n, we have di(K/k) = di(Lp ∩ K/k); therefore we are led to the case where K/L is finite, or again rp(K/k) = rp(L/k). In the following, we will put Lr = Kr = rp(K/k). By Proposition 3.4 above and [14, Proposition 1.7], we will have di(K/k) = di(Kr /k) + di(K/k(K p )) = di(Lr /k) + di(K/L(K p )) + di(L(K p )/k(K p )) = di(Lr /k) + di(K/L) +di(L(K p )/k(K p )). Taking into account Corollary 2.6, we will have di(L(K p )/k(K p )) ≤ di(L/k(Lp )), and so di(K/k) ≤ di(Lr /k) + di(K/L) + di(L/k(Lp )), = di(L/k) + di(K/L). However, the equality holds if and only if di(L/k(Lp )) = di(L(K p )/k(K p )), or again [L : k(Lp )] = [L(K p ) : k(K p )], i.e., L/k(Lp ) and k(K p )/k(Lp ) are k(Lp )-linearly disjoint. Remark 3.2 The condition of linear disjointness in the above proposition is expressed in terms of r-independence by saying that any r-basis of L/k can be completed to an r-basis of K/k. As an immediate application, we have: Corollary 3.7 Any relatively perfect subextension L/k of a q-finite extension K/k satisfies di(K/k) = di(K/L) + di(L/k). This can be formulated in a fairly general way by the following result. Proposition 3.8 For any sequence of relatively perfect subextensions k = K 0 ⊆ K1 ⊆ n−1 · · · ⊆ Kn of a q-finite extension K/k, we have di(K/k) = i=0 di(Kn+1 /Kn ) + di(K/Kn ). Proof Immediately follows from the preceding Corollary.

E. H. Fliouet

In the following, we will study more closely the properties of exponents of a q-finite extension.

3.1 Exponents of a q-Finite Extension We distinguish two cases:

Case: K/k Is a Finite Purely Inseparable Extension In this paragraph, we will us some basic definitions and notations as it is mentioned in [3]. For x ∈ K, put o(x/k) = inf{ m m m ∈ N | x p ∈ k} and o1 (K/k) = inf{m ∈ N | K p ⊂ k}. An r-basis B = {a1 , a2 , . . . , an } of K/k is said to be canonically ordered if for j = 1, 2, . . . , n, we have o(aj /k(a1 , a2 , . . . , aj −1 )) = o1 (K/k(a1 , a2 , . . . , aj −1 )). The integer o(aj /k(a1 , . . . , aj −1 )) thus defined satm isfies o(aj /k(a1 , . . . , aj −1 )) = inf{m ∈ N | di(k(K p )/k) ≤ j − 1} (cf. [4, Lemma 1.3]). We immediately deduce the result [15, Satz 14] which confirm the independence of the integers o(ai /k(a1 , . . . , ai−1 )), (1 ≤ i ≤ n), with respect to the choice of canonically ordered r-bases {a1 , . . . , an } of K/k. Next, we put oi (K/k) = o(ai /k(a1 , . . . , ai−1 )) if 1 ≤ i ≤ n, and oi (K/k) = 0 if i > n where {a1 , . . . , an } is a canonically ordered r-basis of K/k. The invariant oi (K/k) defined above is called the i-th exponent of K/k.

Case: K/k Is q-Finite of Unbounded Exponent Let now K/k be a q-finite extension. −n Recall that for each n ∈ N∗ , kn always designates k p ∩ K. By virtue of [3, Proposition 6], for each j ∈ N∗ , the sequence of natural integers (oj (kn /k))n≥1 is increasing, and thus (oj (kn /k))n≥1 converges to +∞, or (oj (kn /k))n≥1 becomes constant after a certain rank. It is trivially obvious if (oj (kn /k))n≥1 is bounded, then by construction, for each t ≥ j , (ot (kn /k))n≥1 is also bounded (and therefore stationary). Definition 3.2 Let K/k be a q-finite extension and j a positive integer. We call j -th exponent of K/k the invariant oj (K/k) = limn→+∞ (oj (kn /k)). Lemma 3.1 Let K/k be a q-finite extension. Then, os (K/k) is finite if and only if there n exists a natural integer n such that di(k(K p )/k) < s, and we have os (K/k) = inf{m ∈ m N | di(k(K p )/k) < s}. In particular, os (K/k) is infinite if and only if for each m ∈ N, m di(k(K p )/k) ≥ s. Proof To simplify the writing, we put et = ot (K/k) if ot (K/k) is finite. By taking into account of [4, Lemma 1.3], we easily verify that os (K/k) is infinite if and only if for m each m ∈ N, di(k(K p )/k) ≥ s, then one reduces to the case where os (K/k) is finite. Hence, there exists an integer n0 , for each n ≥ n0 , es = os (kn /k). By [4, Lemma 1.3], es es −1 es di(k(kn p )/k) < s and di(k(kn p )/k) ≥ s. By virtue of Theorem 2.1, di(k(K p )/k) < e −1 m s s and di(k(K p )/k) ≥ s. In other words, os (K/k) = inf{m ∈ N | di(k(K p )/k) < s}. The result below makes it possible to reduce the study of the properties of the exponents of a q-finite extension to a finite extension through the relatively perfect closure. Theorem 3.9 Let Kr /k be the relatively perfect closure of irrationality degree s of a qfinite extension K/k (di(Kr /k) = s). Then, we have: (i) For each, t ≤ s, ot (K/k) = +∞.

Absolutely lq-Finite Extensions

(ii) For each, t > s, ot (K/k) = ot−s (K/Kr ). In addition, ot (K/k) is finite if and only if t > s. Proof For each t ∈ N∗ , we denote r ) by et . As for each natural integer e, we  ot (K/K e p e ), so s = di(K /k) ≤ di(k(K pe )/k) = have k(K p ) = Kr (K pe ) = k(k n r n∈N e di(k(kn p )/k) for a sufficiently large integer n. First, by Lemma 3.1, for each t ≤ s, en−s ot (K/k) = +∞. Likewise, for each n > s, we have di(Kr (K p )/k) = di(Kr /k) + e e −1 n−s n−s )/Kr ) < s + n − s = n and di(Kr (K p )/k) = di(Kr /k) + di(Kr (K p en−s −1 )/Kr ) ≥ n. In particular, for each n > s, on (K/k) = on−s (K/Kr ). di(Kr (K p Furthermore, on (K/k) is finite if and only if n > s. Here is a list of immediate consequences of Theorems 2.1 and 3.9. Proposition 3.10 Let K and L be two intermediate fields of a q-finite extension M/k. For every j ∈ N∗ , we have oj (L(K)/L) ≤ oj (K/k). Proof Due to Lemma 3.1, and to the following inequality resulting from Corollary 2.5: n n n n di(L(Lp , K p )/L) = di(L(K p )/L) ≤ di(k(K p )/k) for each n ∈ N. Proposition 3.11 Given the q-finite extensions k ⊆ L ⊆ L ⊆ K, then for each j ∈ N, we have oj (L/L ) ≤ oj (K/k). Proof Immediate application of Lemma 3.1, and the following inequality resulting from n n n Theorem 2.7: di(L (Lp )/L ) ≤ di(L (K p )/L ) ≤ di(k(K p )/k) for each n ∈ N. Moreover, the size of a relatively perfect extension remains invariant up to finite extension as indicated by the following result. Proposition 3.12 Given a relatively perfect subextension K/k of a q-finite extension M/k, then for every finite extension L/k of M/k, we have di(L(K)/L) = di(K/k). Proof By virtue of Corollary 2.5, it is enough to show that di(L(K)/L) ≥ di(K/k). Let us first put e = o1 (L/k) and t = di(K/k). According to Theorem 3.9, for each s ∈ {1, . . . , t}, os (K/k) = +∞, so for a sufficiently large integer n, we will have ot (kn /k) > e + 1; moreover, L ⊆ kn and di(kn /k) = di(K/k). Let {α1 , . . . , αt } be a canonically ordered r-basis of kn /k, if there exists s ∈ {1, . . . , t} such that αs ∈ L(kn p )(α1 , . . . , αs−1 ), by Proposition 3.11, we will have e < ot (kn /k) ≤ os (kn /k) = o(αs , k(α1 , . . . , αs−1 )) ≤ o1 (L(kn p )(α1 , . . . , αs−1 )/k(α1 , . . . , αs−1 )) ≤ sup(o1 (L/k), os (kn /k) − 1) = os (kn /k) − 1, and so os (kn /k) ≤ os (kn /k) − 1, a contradiction. From where {α1 , . . . , αt } is an r-basis of L(kn )/L, and so t = di(K/k) = di(L(kn )/L) ≤ di(L(K)/L).

4 Lower Quasi-finite Extensions In the following, we propose a renovated framework that naturally generalizes the q-finite case. Definition 4.1 Let K/k be a purely inseparable extension of characteristic p > 0. We say −n that K/k is lq-finite if, for any natural integer n, k p ∩ K/k is finite.

E. H. Fliouet

It is obvious that q-finite ⇒ lq-finite. However, here is a nontrivial example of an lqfinite extension of infinite irrationality degree, and thus, the lq-finitude differs completely from the q-finitude. It is also evident that the lq-finitude is stable under the lower inclusion; more precisely, any subextension of an lq-finite extension is lq-finite.

4.1 Nontrivial Example of a lq-Finite Extension Example 4.1 Let Q be a perfect field of characteristic p > 0, (X, (Yi , Zi )i∈N∗ ) an algebraically independent family over Q, and let k be the field of rational fractions Q(X, (Yi , Zi )i∈N∗ ) in indeterminates (X, (Yi , Zi )i∈N∗ ). For each integer n ≥ 2, for j = 1, . . . n − 1, −n+j −n+j −n −n n ). Since K ⊆ K X p +Yj p and Kn = k(X p , α1n , . . . , αn−1 we put αjn = Zj p n n+1  and Kn /k is purely inseparable, then K = n≥2 Kn is a purely inseparable field extension −∞ over k. By convention, we set K0 = k and K1 = k(X p ). p

We verify immediately that for every integer n ≥ 2, for all j = 1, . . . n − 1, (αjn+1 ) p = (αjn ). In particular, for all n ≥ 2, k(Kn+1 ) = Kn , and consequently K/k is relatively perfect. Theorem 4.1 The extension K/k above is lq-finite of infinite irrationality degree. For the proof, we will need the following results. Throughout this section, for every integer t ≤ s, ζst designates the set {1, . . . , s} \ {t}. It is clear that for every integer s ≥ 2, for −s −s −s −s −s all i ∈ {1, . . . , s −1}, αis ∈ k(X p , Zi p , Yi p ); and consequently Ks ⊆ k(X p , (Yj p , −s Zj p )j ∈ζ i , αis ). s−1

Lemma 4.1 For every j ≥ 1, Zj is p-independent over K p , or again Zj p Proof Suppose that Zj p Ks . As

−1

∈K = Yj p

then Yj

p−1

∈ −1

−1

−1



n≥2 Kn .

= (αjs )p

s−j −1

−1

 ∈ K.

Then, there exists s > j such that Zj p −1

− Zj p X p

Ks and Ks is included in k(X −1

−1

−1

p−s

−j −1

−1

∈

, −s

−s

, (Yi p , Zi p )i∈ζ j , αjs ), so −s

−s

s−1 −s

k(X p , Y1 p , . . . , Ys−1 p , Z1 p , . . . , Zs−1 p ) ⊆ k(X p , (Yi p , Zi p )i∈ζ j , αjs ). s−1

However, the family (X, (Yi , Zi )i≥1 ) is algebraically independent over Q; thus, 2(s − 1) + −1 −1 −1 −1 −s −1 −s 1 = di(k(Xp , Y1 p , . . . , Ys−1 p , Z1 p , . . . , Zs−1 p )/k) ≤ di(k(Xp , (Yi p , −s Zi p )i∈ζ j , αjs )/k) ≤ 2(s −2) +2. It follows that 1 ≤ 0, a contradiction, and consequently Zj p

−1

s−1

∈ K.

Lemma 4.2 For every n ≥ 2, the family {X p Kn+1 /Kn .

−n−1

, α1n+1 , . . . , αnn+1 } is an r-basis of

−n−1

−n−1

Proof It is immediate that Kn (X p , α1n+1 , . . . , αnn+1 ) = Kn+1 , and Xp  ∈ Kn −n−1 p since n + 1 = o(X , k) > n = o1 (Kn /k). If there exists i ∈ {1, . . . , n} such −n−1 n+1 n+1 , α1n+1 , . . . , αi−1 , αi+1 , . . . , αnn+1 ), as k(Kn+1 p ) = Kn and that αin+1 ∈ Kn (X p Kn+1 /k has an exponent, we deduce that k(X p

−n−1

n+1 n+1 , α1n+1 , . . . , αi−1 , αi+1 , . . . , αnn+1 )

Absolutely lq-Finite Extensions

= Kn+1 . Specially, αin+1 ∈ k(X p (Zj

p −n−1 ,

−1

Yj

p −n−1 )

j ∈ζni ),

−n−1

n+1 n+1 , α1n+1 , . . . , αi−1 , αi+1 , . . . , αnn+1 ) ⊆ k(X p

and therefore, Zi

p −1

belongs to

−n−1 k(X p ,

(Zj

p−n−1 ,

Yj

−n−1

,

p −n−1 )

p ). Since the family (X, Y , Z ) i i i∈N∗ is algebraically independent over Q, then j ∈ζni , Yi −1

−1

−1

−1

−1

−n−1

−n−1

2n + 1 = di(k(X p , Y1 p , Z1 p , . . . , Yn p , Zn p )/k) ≤ di(k(Xp , (Zj p , −n−1 −1 −n−1 p p p Yj )j ∈ζni , Yi )/k) = 2(n − 1) + 2, so 1 ≤ 0, a contradiction. Hence, {X , α1n+1 , . . . , αnn+1 } is an r-basis of Kn+1 /Kn .

According to above proposition, for every j > 2, {Xp j +1 p

−j −1

j +1 p

j +1

, α1

j +1

, . . . , αj

−j

} and {X p ,

(α1 ) , . . . , (αj −1 ) } are two r-bases of Kj +1 /Kj and Kj /Kj −1 respectively. As Kj = k(Kj +1 p ) and Kj −1 = k(Kj p ), by Lemma 3.1, for every j > 2, for all i ∈ {1, . . . , j − 1}, −j −1 j +1 j +1 j +1 j +1 j +1 we get o(αi , Kj −1 (X p , α1 , . . . , αi−1 , αi , . . . , αj −1 )) = o1 (Kj +1 /Kj −1 ) = 2. In particular, Kj −1 ( X p

−j −1

j +1 Kj −1 (αi−1 ).

j +1

, α1

j +1

, . . . , αj −1 ) Kj −1 (X p

−j −1

j +1

) ⊗Kj −1 Kj −1 (α1

)

⊗Kj −1 · · · ⊗Kj −1 We recall that an extension K/k is said to be modular if and only if for each n ∈ N, n n K p and k are K p ∩ k-linearly disjoint. This notion has been introduced for the first time by Swedleer in [16], where she characterized the purely inseparable extensions which are tensor product of simple extensions over k. Moreover, any r-basis B of K/k such that K ⊗k (⊗k k(a))a∈B will be called modular r-basis (or subbase) of K/k. In particular, according to Swedleer’s theorem, if K/k has an exponent, it is equivalent to say that (i) (ii)

K/k has a modular r-base. K/k is modular.

A purely inseparable extension K/k of exponent e is equiexponential if K/k satisfies one of the following equivalent conditions: (1) There exists an r-basis G of K/k verifying K ⊗k (k(a))a∈G , and for each a ∈ G, o(a, k) = e. (2) Any r-basis G of K/k satisfies K ⊗k (k(a))a∈G , and o1 (K/k) = e. (3) There exists an r-basis G of K/k such that for each a ∈ G, o(a, k(G\{a})) = o(a, k) = e. (4) For every r-basis G of K/k, for each a ∈ G, o(a, k(G \ {a})) = o(a, k) = e. In particular, K/k is equiexponential if and only if all r-basis G of K/k is a modular r-basis of K/k. Lemma 4.3 For each m ∈ N, for all n ∈ N, we have k p −n k p ∩ Km = Km if m ≤ n.

−n

∩ Km = Kn if m ≥ n, and

Proof The lemma is verified for n = 0. Let n ≥ 1, and suppose that the result holds for −n every natural integer i ≤ n−1. Since o1 (Km /k) = m, it is immediate that k p ∩Km = Km if m ≤ n. We are led to the case where n < m. Taking account of recurrence hypothesis, we −1 −n+1 −n have k p ∩Km = Kn−1 ; therefore, k p ∩Km = Kn−1 p ∩Km . Since o1 (Kn /Kn−1 ) = 1 −1 −1 p and Kn ⊆ Km (n < m), we have Kn ⊆ Kn−1 ∩ Km . If Kn−1 p ∩ Km  = Kn , or again if −1 there exists θ ∈ Kn−1 p ∩ Km such that θ ∈ Kn , let j be the largest integer satisfying θ  ∈ Kj ; thus, θ ∈ Kj +1 . In particular, n ≤ j < m, and o(θ, Kj −1 ) = 1 (namely θ p ∈ Kn−1 ⊆ −j −1 Kj −1 ). To simplify the writing, we put temporarily α1 = X p , and for all i = 2, . . . , j +

E. H. Fliouet j +1

1, αi = αi−1 . We will next show that {θ, α1 , . . . , αj } is an r-basis of Kj +1 /Kj . By assumption, θ ∈ Kj and for every i ∈ {1, . . . , j }, αi ∈ Kj (θ, α1 , . . . , αi−1 ); otherwise, we have 1 < o(αi , Kj −1 (α1 , . . . , αi−1 )) ≤ o1 (Kj (α1 , . . . , αi−1 , θ )/Kj −1 (α1 , . . . , αi−1 )) ≤ o(Kj (θ )/ Kj −1 ) = 1 (o(θ, Kj −1 ) = 1 and o1 (Kj /Kj −1 ) = 1), a contradiction. Whence {θ, α1 , . . . , αj } is r-free over Kj . As di(Kj +1 /Kj ) = j + 1 and k(Kj +1 p ) = Kj , then {θ, α1 , . . . , αj } is an r-basis of Kj +1 /Kj . On the other hand, we have o(θ, Kj −1 ) = 1 and Kj −1 (α1 , . . . , αj )/Kj −1 is equiexponential of exponent 2; this implies that Kj +1 Kj −1 (α1 , . . . , αj ) ⊗Kj −1 Kj −1 (θ ),

Kj −1 (α1 ) ⊗Kj −1 · · · ⊗Kj −1 Kj −1 (αj ) ⊗Kj −1 Kj −1 (θ ), hence, Kj +1 /Kj −1 is modular. But, the definition equation of αj +1 over Kj −1 (α1 , α2 , −j . . . , αj ) is written: αj +1 p = Zj X p + Yj = Zj α1 p + Yj , by the criterion of modularity (cf. [4, Proposition 1.4]), we obtain Zj , Yj ∈ Kj +1 p ∩ Kj −1 ⊆ Kj +1 p . We deduce that −1 −1 −n Zj p , Yj p ∈ Kj +1 , contradicting Lemma 4.1. Hence, k p ∩ Km = Kn .  Proof of Theorem 4.1 For each n ≥ 1, we have Kn ⊆ Kn+1 and K = m≥1 Km , by virtue  −n −n of Lemma 4.3, k p ∩ K = m≥1 k p ∩ Km = Kn . As for every n ≥ 1, we have Kn /k is finite, we deduce that K/k is not trivially lq-finite (i.e., is not q-finite). Proposition 4.2 Let K/k be a purely inseparable extension of unbounded exponent, and H the set of subextensions of K/k of unbounded exponent. If K/k is lq-finite, then H is inductive for the order relation defined by K1 ≤ K2 if and only if K2 ⊆ K1 . Proof H = ∅, since K ∈ H . Let (Kn /k)n∈I be a totally ordered family of subextensions −j of unbounded exponent of K/k, so for each j ∈ N∗ , the family (k p ∩ Kn )n∈I is also −j totally ordered. As K/k is lq-finite, therefore, the family ([k p ∩ Kn : k])n∈I of natural integers is totally ordered, and hence, it admits a smallest element which is denoted by −j [k p ∩ Knj : k]. Let I1 = {s ∈ I | Ks ⊆ Knj } and I2 = {s ∈ I | Knj ⊆ Ks }, so for each m ∈ I1 , [k p

−j

∩ Km : k] ≤ [k p

−j

∩ Knj : k]. Taking account of characteristic property of

−j kp

for all m ∈ I1 , we have  p−j ∩ K = k p−j m m∈I1 k

−j

∩ Km : k] = [k p

−j kp

Since K/k is of unbounded exponent and k p −j (k p

−j (k p

−j

∩ Knj : k], and consequently  −j ∩ Km = ∩ Knj . From where k p ∩ ( m∈I1 Km ) =   ∩ Knj . If we put L = i∈I Ki , it is clear that L = m∈I1 Km .

the smallest element, we deduce that [k p

−j

∩ Knj ⊆ Km for every m ∈ I1 , therefore

∩ L/k) = o1 ∩ Knj /k) = j . Whence L/k is of unbounded exponent, and for o1 each n ≥ 1, Kn ≤ L. It follows that H is inductive. Corollary 4.3 Any lq-finite extension of unbounded exponent admits a minimal subextension m/k of unbounded exponent. In particular, m/k is relatively perfect. Proof Immediate. Remark 4.1 The condition of the lq-finitude is necessary for H to be inductive, as shown in the following counterexample. Example 4.2 Let Q be a perfect field of characteristic p > 0, and let k = Q((Xi )i∈N∗ ) be the field of rational fractions in indeterminates (Xi )i∈N∗ . For each n ≥ 1, we put Kn = −i k(((Xi )p )i≥n ).

Absolutely lq-Finite Extensions

We immediately check that the family (Kn )n∈N∗ of subextensions of unbounded expo−i nent of K1 /kis totally ordered. Since ((Xi )p )i∈N∗ is a modular r-basis (subbase) of K1 over k, then n≥1 Kn reduces to k, and therefore, H is not inductive. As an another consequence of Proposition 4.2, we obtain the following result. Proposition 4.4 The relatively perfect closure Kr of an lq-finite extension K/k of unbounded exponent is not trivial. More precisely, if K/k is of unbounded, then the same is true for Kr /k. Proof By Proposition 4.2, the set of subextensions of unbounded exponent of K/k has a minimal subextension m/k. Necessarily, m/k is relatively perfect; otherwise, k(mp )/k would be better than m, a contradiction, and therefore Kr /k is of unbounded exponent (m ⊆ Kr ). When K/k is q-finite, we can locate rp(Kk) with respect to K; in particular, K/rp(K/k) is finite. But, in the lq-finitude case, the location of the relatively perfect closure varies from one extension to the other. Theorem 4.5 Any lq-finite modular extension is q-finite. For this proof, we need the following results. Firstly, given a modular r-basis B of a modular extension K/k, and for each a ∈ B, we put na = o(a, k). Consider now the subsets B1 and B2 of B defined by B1 = {a ∈ B | na > j }, B2 = B \ B1 = {a ∈ B | na ≤ j } (j being a natural integer not exceeding o(K/k)). Proposition 4.6 Under the conditions specified above, for any positive integer j less than na −j )a∈B1 , B2 ). o1 (K/k), we have kj = k((a p Proof Since K/k is an inductive union of modular extensions generated by finite subsets of B, and taking into account distributivity of the intersection with respect to the union, we can assume without loss of generality that K/k is finite and has an exponent e. Let {α1 , · · · , αn } be a canonically ordered modular r-basis of K/k, and mj the j -th exponent pm1 −j

of K/k. Let us denote by s the greatest integer such that ms > j , and L = k(α1

,...,

pms −j , αs

αs+1 , . . . , αn ). It is immediately verified that (i) L ⊆ kj , (ii) K k(α1 ) ⊗k · · · ⊗k k(αn ) L(α1 ) ⊗L · · · ⊗L L(αs ). Thus, for all x ∈ K, there are unique constants Cε ∈ L such that x =  ε m1 −j , . . . , 0 ≤ C ε∈ ε (α1 , . . . , αs ) where  = {(i1 , . . . , is ) such that 0 ≤ i1 < p  ε j j j j is < p ms −j }, and therefore, x p = ε∈ Cε p (α1 p , . . . , αs p ) . In view of [4, Proposij j j tion 5.3], we deduce that x p ∈ k (i.e., x ∈ kj ) if and only if x p = C(0,...,0) p , or again x = C(0,...,0) . Consequently, x ∈ kj if and only if x ∈ L, i.e., kj = L. In the case of modularity, the following result expresses a property of stability of the size of certain intermediate fields. More precisely, Corollary 4.7 For every modular extension K/k, and for each n ∈ N, we have di(kn /k) = di(k1 /k). In particular, di(K/k) = di(k1 /k).

E. H. Fliouet

Proof For each n ∈ N, kn /k is modular of bounded exponent (cf. [13, Proposition 3]), so kn /k has a modular r-basis, and by virtue of the previous theorem, we have di(kn ) = di(k1 /k). Proof of Theorem 4.5 It is enough to note that for each n ≥ 1, we have di(k p −1 di(k p ∩ K/k) < +∞ from the previous corollary.

−n

∩ K/k) =

5 Absolutely Lower Quasi-finite Extensions Let K/k be a purely inseparable extension. It is clear that if K/k is q-finite, then for every subextension L/k of K/k, we have that K/L is also q-finite. However, this result is generally false in the lq-finitude case. Let us go back to Example 4.1 again. If we put −∞ −1 L = k(X p ), then we check immediately that k((αs−1 s )s>1 ) ⊆ Lp ∩ K, and the family s ) p−1 ∩ K/L is not finite, and consequently, K/L is not (αs−1 s>1 is r-free over L. Hence, L lq-finite. This leads us to study closely the stability in the absolute sense within the lq-finite extensions. The study of this property is the subject of this section. Proposition 5.1 Let K/k be a purely inseparable extension. The following assertions are equivalent: (1) For every subextension L/k of K/k, K/L is lq-finite. (2) Any subextension L/k of K/k satisfies L/k(Lp ) is finite. (3) For every subextension L/k of K/k, we have L/rp(L/k) is finite. −1

Proof 1 ⇒ 2 ⇔ 3 is immediate: it is sufficient to note that L ⊆ (k(Lp ))p ∩ K and K/k(Lp ) is lq-finite for every subextension L/k of K/k, and apply [6, Lemma 2.1]. Like−n wise, if (2) holds, then for each n ∈ N∗ , Ln /k(Ln p ) is finite where Ln = Lp ∩ K. In particular, di(Ln /L(Ln p )) is finite. Since Ln /L has an exponent, therefore, di(Ln /L) = di(Ln /L(Ln p )) < +∞, or again Ln /L is finite for each n ∈ N∗ . Definition 5.1 An extension that satisfies one of the equivalent conditions above is called an absolutely lq-finite extension. It is immediately verified that “to be q-finite” ⇒ “to be absolutely lq-finite” ⇒ “to be lq-finite”, and K/k is absolutely lq-finite if and only if any subextension L2 /L1 of K/k is lq-finite. In particular, every subextension of an absolutely lq-finite extension is also absolutely lq-finite. Theorem 5.2 A purely inseparable extension K/k is absolutely lq-finite if and only if any decreasing sequence of subextensions of K/k is stationary. Proof Suppose that the sufficient condition is satisfied, and consider a subextension L/k of K/k. Let G be an r-basis of L/k. If |G| is not finite, then there exists a sequence (αn )n∈N∗ of distinct elements in G. For each n ≥ 1, we put Kn = k(G \ {α1 , . . . , αn })). It is clear that the sequence (Kn /k)n≥1 of subextensions of K/k is strictly decreasing (for inclusion), contradicting the assumption of the sufficient condition. Hence, |G| is finite, or again L/k(Lp ) is finite, and therefore K/k is absolutely lq-finite. Conversely, suppose that K/k is absolutely lq-finite, and let (Kn /k)n≥1 be a decreasing sequence of subextensions of K/k. For every n ≥ 1, let Ln = rp(Kn /k). Since K/k is absolutely lq-finite, then for each n ≥ 1,

Absolutely lq-Finite Extensions

we have Kn /k(Kn p ) is finite. Hence, according to [6, Lemma 2.1], there exists en ∈ N,  en i such that Ln = k(Kn p ) = i≥1 k(Kn p ). Since the sequence (Kn /k)n≥1 is decreasing, the same is true for (Ln /k)n≥1 . We distinguish two cases: Case 1 If there exists n0 ∈ N such that Ln = Ln0 for each n ≥ n0 , then Kn /Ln0 will be finite for each n ≥ n0 . By virtue of the monotonicity of (Kn )n≥1 , we deduce that the sequence of natural integers ([Kn : Ln0 ]n≥n0 is decreasing, and therefore stationary, i.e., there exists e ≥ n0 such that [Kn : Ln0 ] = [Ke : Ln0 ] for each n ≥ e. But for every n ≥ 1, Kn+1 ⊆ Kn , therefore ,Kn+1 = Kn for each n ≥ e. Case 2 If (Ln )n≥1 is not stationary, we are led to consider the situation that the sequence (Ln )n≥1 is strictly decreasing, and consequently, we will construct by recurrence a sequence (αi )i∈N∗ of elements in K such that αi ∈ Li and αi  ∈ k(α1 , . . . , αi−1 )(Li+1 ) for each i ≥ 1. The case where i = 1 is trivial, just choose α1 in L1 \ L2 . Consider now a natural integer n distinct from 0 and 1, and assume that this property holds for all i = 1, . . . , n. Since Li /k is relatively perfect for each i ≥ 1 and (Li )i≥1 is strictly decreasing, we deduce that (k(α1 , . . . , αn )(Li ))i≥1 is also strictly decreasing, and thus, there exists αn+1 ∈ Ln+1 such that αn+1 ∈ k(α1 , . . . , αn )Ln+2 . In the following, we put G = (α1 , . . .  , αn , . . . ) and Gn = (α1 , . . . , αn ), and we will show that G is r-free over k. Since G = n≥1 Gn and Gi ⊆ Gi+1 , it is enough to show that Gn is r-free over k for all n ≥ 1. If there exists i ∈ {1, . . . , n} such that αi ∈ k(Gn \ {αi }), in particular, αi ∈ k(α1 , . . . , αi−1 )Li+1 (namely for each j ≥ i + 1, αj ∈ Lj ⊆ Li+1 ), a contradiction by construction. As a result, di(k(G)/k) = di(k(G)/k((k(G))p )) = +∞, and so K/k is not absolutely lq-finite, a contradiction. Let K/k be a purely inseparable extension. It is clear that if K/k is q-finite, then any subextension L/k of K/k is finite over k(Lp ), so it is very likely that the converse is also true. In other words, the absolute lq-finitude is synonymous with the q-finitude. However, using the characteristic property of lq-finitude, we will construct an example of a purely inseparable extension K/k of infinite size such that any subextension of K/k is q-finite, and consequently, K/k is absolutely lq-finite. In particular, K/k is not q-finite. Moreover, the two notions coincide in the modularity case.

5.1 Example of an Absolutely lq-Finitude Extension of Infinite Irrationality Degree Example 5.1 Given a perfect field Q of characteristic p > 0, and a family j

j

(X, (Zi , Yi ))(i,j )∈N∗ ×(N∗ \{1}) j

j

algebraically independent over Q. Let k be the field Q((X, (Zi , Yi ))(i,j )∈N∗ ×(N∗ \{1}) ) of j j rational fractions in indeterminates ((Zi , Yi )(i,j )∈N∗ ×(N∗ \{1}) , X). Let (an )n∈N∗ be a family −∞ −n of elements of Q(Xp ) such that Q(an ) = Q(Xp ) for each n ∈ N∗ . In particular, −n −n k(an ) = k(X p ), and consequently for each n ∈ N ∗ , o(an , k) = o(X p , k) = n. The following notations will be used throughout this section: K1 = i≥1 k(θi1 ) with  p−1 p−1 θi1 = ai for each i ∈ N ∗ , K2 = i≥1 K1 (θi2 ) with θ12 = (Z12 ) θ21 + (Y12 ) , and p−1

p−1

p−1

1 + (θ 2 ) for each i ≥ 2, θi2 = (Zi2 ) θ2i + (Yi2 ) . Proceeding by induction, we i−1  −1 −1 −1 n−1 put Kn = i≥1 Kn−1 (θin ) where θ1n = (Z1n )p θ2n−1 +(Y1n )p , and θin = (Zin )p θ2i

E. H. Fliouet −1

−1

n )p +(θi−1 +(Yin )p for each i ≥ 2. We also set K = convention, K0 = k, and for each i ∈ N, θ0i = 0.



i≥1 Kn ,

and finally we put, by

It is immediately verified that for each (i, n) ∈ N∗ × (N \ {0, 1}), −1

n−1 n + (θi−1 )p θin = (Zin )p θ2i −1

−1

−1

+ (Yin )p , −i

n−1 + · · · + (Z1n )p (θ2n−1 )p = (Zin )p θ2i

−i+1

+ (Yin )p

−1

−i

+ · · · + (Y1n )p ,

n ), and thus in particular, for each n ∈ N∗ , for each i ∈ N, we have Kn−1 (θin ) ⊆ Kn−1 (θi+1 Kn /Kn−1 is q-simple.

Theorem 5.3 (Ki /k)i≥1 are the only subextensions of unbounded exponent of K/k up to finite extension, i.e., for any proper subextension L/k of unbounded exponent of K/k, there exists i ∈ N∗ such that Ki ⊆ L and L/Ki is finite. As a typical application of this theorem we have: Proposition 5.4 Any decreasing sequence of subextensions of K/k is stationary. Proof Let (Fi /k)i≥1 be a decreasing sequence of subextensions of K/k. From the preceding theorem, for every j ∈ N∗ , there exists ij ∈ N such that Kij ⊆ Fj and Fj /Kij is finite.   n n As Fj +1 ⊆ Fj , then n∈N k(Fj +1 p ) = Kij +1 ⊆ n∈N k(Fj p ) = Kij . Hence, there exists s ∈ N∗ , for each j ≥ s, Kij = Kis . In particular, for each j ≥ s, Fj /Kis is finite, and thus, for each j ≥ s, [Fj +1 : Kis ] ≤ [Fj : Kis ] ≤ [Fs : Kis ], or again the sequence of natural integers ([Fj : Kis ])j ≥s is decreasing, it is therefore stationary starting at n0 , and consequently for all n ≥ n0 , Fn = Fn0 . The following result is a natural consequence of this. Theorem 5.5 Under the same assumptions as above, K/k is an absolutely lq-finite extension. Proof Results from Theorem 5.2. Before we present a proof of Theorem 5.3, we need the following results. From now on, for each (i, s) ∈ (N\{0, 1})×N∗ , we set is = {(j, l) ∈ N2 | 2 ≤ j ≤ i and 1 ≤ l ≤ 2i−j s}. i−j s

−2 j p

Lemma 5.1 For each i ≥ 2, for each s ≥ 1, θsi is an element of Q(θ21i−1 s , ((Zl )

,

i−j s

−2 j p (Yl )

)(j,l)∈is ).

p−1 1 θ2s p−s 1 p−s+1 p−s p−s+1 p−1 2 2 1 1 2 2 + · · · + (Z1 ) (θ2 ) + (Ys ) + · · · + (Y1 ) , hence θs ∈ Q(θ2s , . . . , (θ2 ) , p−s p−s p−s p−1 p−1 p−s p−s 2 2 1 2 2 2 2 2 (Zs ) , . . . , (Z1 ) , (Ys ) , . . . , (Y1 ) ) ⊆ Q(θ2s , (Zs ) , . . . , (Z1 ) , (Ys ) , p−s . . . , (Y12 ) ), and so the lemma holds for the first-order. Suppose that the result is true p−1 i + ··· + up to order i > 1. Also, for all s ∈ N∗ , we have θsi+1 = (Zsi+1 ) θ2s −s −s p−s+1 p−s+1 p−1 i+1 p i+1 p i i i+1 i+1 + (Ys ) + · · · + (Y1 ) , and so θs ∈ Q(θ2s , . . . , (θ2i ) , (Z1 ) (θ2 )

Proof We will prove by induction. By construction, for all s ∈ N∗ , θs2 = (Zs2 )

Absolutely lq-Finite Extensions

(Zsi+1 )

p−1

p−s

, . . . , (Z1i+1 )

, (Ysi+1 )

p−1

, . . . , (Y1i+1 ) j

i ∈ Q(θ 1 , ((Zl ) each r ∈ {1, . . . , s}, θ2r 2i−1 .2r

fect, for each n ∈ N, we have Qp Q((θ21i r )

p−(s−r)

−n

i+1−j r−(s−r)

,

j

• (Zl )

) = Q(X p

i+1−j r−(s−r) p−2

From where θsi+1 ∈

−2i r−(s−r)

i−j 2r

−2 j p

2r

p−(s−r)

is an element of

)(j,l)∈i ) for each r ∈ {1, . . . , s}. 2r

and = + (s − r)) = 2i+1−j r + then, we have the above relations: 2i+1−j s

) ⊆ Q(X p

∈ Q((Zl )

)(j,l)∈i ). As Q is per-

, (Yl )

−2 j p , (Yl )

i+1−j s p−2

j

). But by induction hypothesis, for

i+1−j r−(s−r)

−2 j p ((Zl )

Nonetheless, Q(θ21i r ) = Q(X ) 2i+1−j (s − r) ≥ 2i+1−j r + (s − r); p−(s−r)

i−j p−2 2r

i ) = Q, and consequently, (θ2r

i p−2 r

• Q((θ21i r )

p−s

−2i s

2i+1−j (r

) = Q((θ21i s )), i+1−j r−(s−r)

−2 j p

) and (Yl )

i+1−j s

−2 j p Q(θ21i s , ((Zl )

i+1−j s

−2 j p , (Yl )

i+1−j s

−2 j p

∈ Q(( Yl )

).

)(j,l)∈i+1 ). s

−∞ j p

−∞

−∞

j

In the following, for each n ≥ 2, we put Sn = k(X p , ((Zi ) , (Yi )p )(i,j )∈n ) −∞ where n = N∗ × {2, . . . , n}, and by convention S1 = k(X p ). From the above lemma, for every n ≥ 2, we verify immediately that Kn ⊆ Sn . Moreover, for all (i, n) ∈ N∗ × (N \ −1 −1 {0, 1}), (Z1n )p ∈ Sn−1 ((Y1n )p , θin ) by construction. j

Proposition 5.6 The family (Zi )(i,j )∈N∗ ×(N\{0,1}) is p-independent over K p . Throughout this demonstration, we will use the following notations: for all (i, j ) ∈ N∗ × j

(N \ {0, 1}), we set Ai = ((Zss12 )

p−1

j p ((Zl )

j Ci

−i

)(s1 ,s2 )<(i,j ) where ≤ is the hexadecimal order relation, −i j p

= , (Yl ) )l
j p

j

j

show that Zi ∈ K p ((Ai ) ), or again (Zi )

∈ K(Ai ). For this, we suppose the existence −1  j p j of a couple (i, j ) ∈ N∗ × (N \ {0, 1}) such that (Zi ) ∈ K(Ai ). As K = n≥1 Kn , there −1 j p

exists n ∈ N∗ such that (Zi ) ∈ Kn (Ai ). Let n be the smallest integer that satisfies this relation. By construction, it is clear that n > 1. In addition, we distinguish two cases: j

−1 j p

Case 1 If n ≤ j , therefore (Zi ) −1 j p j −1 θ2i

j

θi = (Zi )

−i j p

+ · · · + (Z1 )

j

that

∈ j

j Sj −1 (Kj )(Ci ). j

di(Sj −1 (Ci , (Zi )

p−1

j

, (Yi )

−i+1 j −1 p

(θ2

(r, l) ∈ {1, . . . , i − 1} × N∗ , we have −1 j p (Yi )

j

j

∈ Kn (Ai ) ⊆ Kj (Ai ) ⊆ Sj −1 (Kj )(Ci ). But, we have )

j p (Zr )

−i

−1 j p

+ (Yi )

j p , (Yr )

−i

∈

−i j p

+ · · · + (Y1 )

j Ci

and

j −1 θl

, and for each

∈ Sj −1 , we deduce

This leads by virtue of Theorem 2.1 to

p−1

j

j

j

)/Sj −1 (Ci )) = 2 ≤ di(Sj −1 (Kj , Ci )/Sj −1 (Ci )) = 1;

and so 2 ≤ 1, a contradiction. −1 j p

Case 2 If n ≥ j + 1, we have (Zi ) j

j

j

−1 j p

j

∈ Kn−1 (Ai ) and (Zi )

∈ Kn (Ai ), as

j

Kn (Ai )/ Kn−1 (Ai ) is q-simple and θ1n ∈ Kn−1 (Ai ) (otherwise we will have (Z1n )p Sn−1 ((Y1n )

p−1

), a contradiction), then

j Kn−1 (Ai )(θ1n )

=

p−1

j j Kn−1 (Ai )((Zi )

−1

∈

) ⊆ Sn−1 . In

E. H. Fliouet

particular, θ1n ∈ Sn−1 , and therefore (Z1n )p −1

−1

−1

−1

∈ Sn−1 ((Y1n )p ), (θ1n = (Z1n )p θ2n−1 +

(Y1n )p ), a contradiction. Lemma 5.2 For each (i, s) ∈ N × N∗ , and for each j > i, we have o(θs , Ki ) = 2j −i−1 s. j

Proof It is immediate that o(θs1 , K0 ) = s for each s ∈ N∗ , so we reduce to the case in which j ≥ 2. In the following, we will use an induction on j . By construction, we have

p−s p−s p−s+1 p−1 i p−1 θ2s + · · · + (Z1i+1 ) (θ2i ) + (Ysi+1 ) + · · · + (Y1s+1 ) , and p−1 p−1 ps ps−1 therefore (θsi+1 ) ∈ Ki . If (θsi+1 ) ∈ Ki , we will have (Z1i+1 ) ∈ Ki ((Y1i+1 ) ), ps−1 ∈ Ki , and so o(θsi+1 , Ki ) = s. Now let j be an a contradiction. From where (θsi+1 ) −1 j j p j −1 integer such that j ≥ i + 2. Also, for every s ≥ 1, we have θs = (Zs ) θ2s + · · · + −s+1 −s −1 −s j p j −1 p j p j p (Z1 ) (θ2 ) + (Ys ) + · · · + (Y1 ) . But, by induction hypothesis, for each j −1 s ∈ N ∗ , for each n ∈ {0, . . . , s − 1}, we have o(θ2(s−n) , Ki ) = 2j −1−i−1 .2(s − n) = p−n j −1 2j −i−1 (s − n), hence for all n ∈ {1, . . . , s − 1}, o((θ2(s−n) ) , Ki ) = 2j −i−1 (s − n) + n < j −1 j j −1 2j −i−1 s = o(θ2(s) , Ki ). It follows that o(θs , Ki ) = o(θ2s , Ki ) = 2j −i−1 s.

θsi+1 = (Zsi+1 )

Lemma 5.3 For each i ∈ N, Ki p

−1

∩ K = Ki (θ1i+1 ). p−1

p−1

Proof It is clear that Ki (θ1i+1 ) ⊆ Ki ∩ K. If there exists α ∈ Ki ∩ K such that α ∈ Ki (θ1i+1 ), as Ki+1 /Ki is q-simple, we have α ∈ Ki+1 ; otherwise, Ki (θ1i+1 ) = Ki (α). Let j be the largest integer such that α ∈ Kj , therefore α ∈ Kj +1 with i + 1 ≤ j . In j +1 particular, we have o(α, Kj −1 ) = 1, and as a result Kj (α) = Kj (θ1 ) Kj ⊗Kj −1 j +1

Kj −1 (α); thus, Kj (θ1

j p (1, (θ2 ) )

j

)/Kj −1 is modular. But, o(θ2 , Kj −1 ) = 2, therefore the system

is linearly independent over Kj −1 , and in particular, it is linearly independent j +1 p

over Kj −1 ∩ [Kj (θ1

j +1 p

)] . We extend this system to a linear basis B of [Kj (θ1

)]

j +1 p j +1 j +1 p over Kj −1 ∩ [Kj (θ1 )] . As Kj (θ1 )/Kj −1 is modular, thus, [Kj (θ1 )] and Kj −1 j +1 p are Kj −1 ∩ [Kj (θ1 )] -linearly disjoint, and consequently, B is also a linear basis of j +1 p j +1 over Kj −1 is writKj −1 ([Kj (θ1 )] ) over Kj −1 . But the defining equation of θ1 p p j +1 j +1 j j +1 j +1 j +1 ∈ ten as (θ1 ) = Z1 (θ2 ) + Y1 . By identification, we will have Z1 , Y1 −1 j +1 p p p ∩K = Kj −1 ∩[Kj (θ1 )] ⊆ K , but this contradicts Proposition 5.6, and therefore, Ki Ki (θ1i+1 ). −n

Corollary 5.7 For each (i, n) ∈ N × N∗ , we have Ki p −n particular, Ki p ∩ K ⊆ Kn+i for every natural number n. −n

Proof It is sufficient to note that Ki p ∩K = (Ki p As a consequence, we have: Corollary 5.8 For each j ∈ N, K/Kj is lq-finite.

−1

∩ K ⊆ Ki+1 p

p−n+1

∩ K)

−n+1

∩K ⊆ Ki+1 p

−n+1

∩ K. In

∩K.

Absolutely lq-Finite Extensions

Proof Immediate application of the previous result and the fact that for all j, n ∈ N, di(Kn+j /Kj ) = n. Let n be a positive integer. Consider the sequence (ns )s∈N∗ of integers defined by the following recurrence relation: n1 = n, and for each s ∈ N∗ \ {1}, ns+1 = [ n2s ] where [.] is the integer part. The sequence (ni )i≥1) of natural numbers thus obtained is decreasing, and so stationary. The first integer e such that ne = 1 is called the lower parity length of n and is denoted by lp(n). It is immediately verified that – For every s > lp(n), ns = 0. – If n ≤ m, then lp(n) ≤ lp(m). – For each i ∈ N∗ , 2ni+1 + εi = ni , with εi = 0 or 1 according to the parity of ni . In particular, 2ni+1 ≤ ni and ni + 1 ≤ 2(ni+1 + 1), and consequently for all j ≥ i, we deduce that 2j −i nj ≤ ni and ni + 1 ≤ 2j −i (nj + 1). Notably, if we put p1 = lp(n), we will have 2p1 −1 ≤ n < 2p1 . In other words, p1 is the smallest integer such that n < 2p1 . The following theorem play an important role in proving that K/k is absolutely lq-finite. −n In particular, it offers a very useful tool for constructing a canonical r-basis of Ki p ∩ K/Ki . Theorem 5.9 For each (i, n) ∈ N × N∗ , Ki p

−n

∩ K ⊆ Ki+1 p

−[ n ] 2

.

To prove this theorem, we will need the following results: Proposition 5.10 Let n be a positive integer. Under the above notations, for all s ∈ N∗ , for each n ∈ N, we obtain the following equality: o(θns+r , Kr (θn1+r , . . . , θns−1+r )) = ns = sup (o(θni+r , Kr (θn1+r , . . . , θns−1+r )). s i 1 s−1 1 s−1 i∈N∗

p +r

, . . . , θnp11 } is a canonically ordered r-basis of Kr ((θni i )i∈N∗ )/Kr In particular, {θn1+r 1 where p1 = lp(n). Proof Since the Ki /k (1 ≤ i) are constructed in the same way, and for clarity of representation, we only present the proof in the case where r = 0. By convention, for each s > p1 , o(θnss , k) = o(θ0s , k) = 0, therefore, we reduce to the case in which s ≤ p1 . From Lemma 5.2, we have for each i ∈ {1, . . . , p1 }, o(θni i , k) = o(θni i , K0 ) = 2i−1 ni ≤ n1 = o(θn11 , k), and so o(θn11 , k) = supi∈N∗ (o(θni i , k)). Let s ∈ {2, . . . , p1 − 1}. Assume, j

j −1

as an induction hypothesis, that for every j ∈ {1, . . . , s}, o(θnj , k(θn11 , . . . , θnj −1 )) = j −1

nj = supi∈N∗ (o(θni i , k(θn11 , . . . , θnj −1 )). We next put L1 = k(θn11 , . . . , θnss ), L2 = p p k(θn11 , . . . , θnp11 ) and L3 = L2 (Ks ) = Ks (θns+1 , . . . , θnp11 ). Firstly, L3 /k is q-finite, so by s+1 virtue of Lemma 5.2, for each i ≥ s + 1, o(θni i , Ks ) = 2i−s−1 ni ≤ ns+1 = o(θns+1 , Ks ). s+1 s+1 , L ) ≤ Secondly according to Proposition 3.10 and 3.11, ns+1 = o(θns+1 , K ) ≤ o(θ s 1 ns+1 s+1 s+1 , L ) = o1 (L2 /L1 ) = os+1 (L2 /k) ≤ os+1 (L3 /k) = o(θns+1 , K ) = n , and so o(θ s s+1 1 ns+1 s+1 os+1 (L2 /k) = o1 (L2 /L1 ) = ns+1 . Thence, for each s ∈ N ∗ , o(θnss , k(θn11 , . . . , θns−1 )) s−1 = ns = supi∈N∗ (o(θni i , k(θn11 , . . . , θns−1 )). In particular, by applying the r-basis completion s−1

E. H. Fliouet p

algorithm (cf. [4, Proposition 1.3]), we will have {θn11 , . . . , θnp11 } is a canonically ordered r-basis of k((θni i )i∈N∗ )/k. Let n be a positive integer. Consider the sequence (ns )s∈N∗ of integers defined by the following recurrence relation: (i) n1 = n if n is even, and n1 = n + 1 otherwise. n

n +1

(ii) For each s ∈ N∗ , ns+1 = 2s if ns is even, and ns+1 = s2 if ns is odd. The sequence of natural numbers thus obtained is decreasing and is therefore stationary. The smallest integer r such that nr = 1 is called the upper parity length of n and is denoted up(n). It is immediately verified that • For each s ≥ up(n), ns = 1. • For each i ∈ N∗ , 2ni+1 = ni + εi , with εi = 0 or 1 according to the parity of ni . In particular, ni ≤ 2ni+1 and 2(ni+1 − 1) ≤ ni − 1. Consequently, for all j ≥ i, ni ≤ 2j −i nj and 2j −i (nj − 1) ≤ ni − 1. Furthermore, if we put p2 = up(n), we will have 2p2 −2 < n ≤ 2p2 −1 . In other words, p2 − 1 is the smallest integer such that n ≤ 2p2 −1 . Similarly, it is easy to obtain the expression ni ≤ ni for all i ∈ N∗ . However, we show by induction that ni ≤ ni + 1. Indeed, for i = 1, the result holds, and if ni ≤ ni + 1, or again n +1

ni + 1 ≤ 2(ni+1 + 1) + 1, we will have ni+1 ≤ i 2 ≤ (ni+1 + 1) + 12 ; and consequently ni+1 ≤ [ni+1 +1+ 12 ] = ni+1 +1. In addition, ni = ni +αi , with αi = 0 or 1; and as a result up(n) = lp(n) + α, with α = 0 or 1. Also we have 2p2 −2 ≤ 2p1 −1 ≤ n ≤ 2p2 −1 ≤ 2p1 where p1 = lp(n) and p2 = up(n), and as consequences, p2 = p1 if and only if n is power of 2. In the following, we set = {(f, g) ∈ N2 | f ≥ 2 and 1 ≤ g ≤ nf }, and Qn0 = Q(θn1 , f p ((Zg )

−n2

,

f p (Yg )

−n2

1

)(f,g)∈ ). Since for each i ∈

N∗

\ {1}, and for every j ∈ {2, . . . , i},

i−j n

−2 j p {(Zl )

i

i−j n

−2 j p

i

, (Yl ) } ⊆ Qn0 for every 2i−j ni ≤ nj ≤ nj ≤ n2 , we deduce that l ∈ {1, . . . , 2i−j ni }. By Lemma 5.1, for each i ∈ N∗ , we have θni i ∈ Qn0 ; and so Q((θni i )i∈N∗ ) ⊆ Qn0 . For each i ∈ N, put also Fin = Qn0 (Ki ) and F n = Qn0 (K). It is immediately verified that n , and F n =  n (1) For each i ∈ N, Fin ⊆ Fi+1 i∈N Fi . n n (2) For each i ∈ N, Fi+1 /Fi is q-simple. (3) For each 1 ≤ i, (θni  −1 )

p−1

i

∈ Qn0 . j p

Indeed, θni  −1 ∈ Q(θ21i−1 (n −1) , ((Zl ) i

−2i−j (ni −1)

j p

−2i−j (ni −1)

, (Yl )

i

)(j,l)∈i

ni −1

) by virtue of

Lemma 5.1. As θn1 −1 ∈ (Qn0 )p (Q being perfect), and for each j ∈ {2, . . . , i}, 2i−j (ni − i

1) ≤ nj − 1 ≤ n2 − 1 (and in particular, 2i−1 (ni − 1) ≤ n1 − 1), then θni  −1 ∈ j p Q(θn1 −1 , ((Zl ) 1

−(n2 −1)

, p−1

Moreover, (θni  −1 ) p−1

+(θni  −1 ) i

i

+ (Yni  ) i

j p (Yl )

i

−(n2 −1)

)(j,l)∈i

ni −1

) ⊆

(Qn0 )p ;

or again

p (θni  −1 ) i

−1

n (namely, θ i = (Z i ) ∈ F0n , and for all i ≥ 1, θni  ∈ Fi−1 n n

p−1

with (Zni  )

Lemma 5.4 For each (s, j ) ∈

i

N∗

p−1

, (Yni  )

× N∗ ,

i

p−1

i

i

i

∈ Qn0 . p−1 i−1 θ2n i

i−1 n ∈ F0n , and θ2n  ∈ Ki−1 ⊆ Fi−1 ).

n ) o(θns +j , Fs−1 s

i

= j.

Absolutely lq-Finite Extensions

We first consider L2 = k((Qn0 )p (Yn2 +1 )

p−∞

−∞

p−∞

, (Yn2 +1 ) 2

) = k(X p

−∞

, ((Zls )p

−∞

, (Yls )p

−∞

)(s,l) ,

) where (s, l) runs through the set , and

2

Lt = St−1 ((Qn0 )p = k(X p

−∞

−∞

, (Yn2 +1 )

p−∞

t

p−∞

, ((Zih )

, (Yih )

),

p−∞

−∞ j p

)(i,h)∈t−1 , ((Zl )

−∞ j p

, (Yl )

)(j,l)∈ , (Ynt  +1 )

p−∞

t

),

where t > 2 and t−1 = N∗ × {2, . . . , t − 1}. Clearly, Fsn ⊆ Ls . Furthermore, as Q is −1

−i

s−1 perfect, and θis = (Zis )p θ2i + · · · + (Z1s )p (θ2s−1 ) ∗ s > 1 and i ∈ N , then for each j ∈ N,

(θis )p

−j

∈ k(X p

In particular, (θns )p

−j

s

−∞

p−∞

, ((Zrh )

, (Yrh )

p−∞

p−i+1

+ (Yis )p

)(r,h)∈s−1 , ((Zls )p

−∞

−1

+ · · · + (Y1s )p

, (Yls )p

−∞

−i

for

)1≤l≤i ).

∈ Ls .

Proof For s = 1, the result is trivial, so we are led to the case s ≥ 2. By construc−1 s−1 −1 s−1 p−1 s p−1 + (Y s tion, θns +1 = (Zns  +1 )p θ2(n = (Zns  +1 )p θ2(n  +1) + (θn )  +1) + · · · + n +1 ) s

s

s



s

s

s

s

−ns −1 −1 −ns −1 p−ns p s−1 (θ2s−1 ) +(Yns +1 )p + · · · + (Y1s )p , so (θns +1 )p = Zns  +1 (θ2(n (Z1s )p  +1) ) s s s s n . However, if θ s n +θns + Yns +1 , and consequently (θns +1 )p ∈ Fs−1 ns +1 ∈ Fs−1 , we will s s s −1 have (Zns  +1 )p ∈ Ls . Note that for all h ≤ s − 1 and for all i ∈ N∗ , Zih  = Zns  +1 , also s s j j j for all 1 ≤ l ≤ nj , Zns  +1 = Zs , and since the family (Zi , Yi )(i,j )∈N∗ 2 is p-independent s

−1

n ) = 1. Let over k p , then (Zns  +1 )p ∈ Ls , a contradiction. From where o(θns +1 , Fs−1 s s j be a natural number different from 0 and 1. By inductive hypothesis, assume that p−1 s−1 n ) = i for all i ∈ {1, . . . , j }. Since θ s s θ2(n +j +1) + o(θns +i , Fs−1 n +j +1 = (Zn +j +1 ) s

(θns +j )p

−1

s

−1

+ (Yns +j +1 )p , we will have (θns +j )p s

−1

−1

s −1

s

s

s

−1

p n (θ s s ∈ Fs−1 , n +j +1 , (Zn +j +1 ) s

s

−1

p n ) ≤ o (F n (θ s s (Yns +j +1 )p ). Thus, j + 1 = o((θns +j )p , Fs−1 , 1 s−1 n +j +1 , (Zn +j +1 ) s

s

−1

s

s

n ) ≤ o(θ s n s n (Yns +j +1 )p ) / Fs−1 ns +j +1 , Fs−1 ) ≤ j + 1, and consequently o(θns +j +1 , Fs−1 ) s = j + 1. s Corollary 5.11 For each s ∈ N∗ , (θ2(n 

s+1 +1)

p

n (θ s )p is not in Fs−1 ). In particular, 2n s+1

n ( θs ) ∈ Fs−1 ). (θns+1  2n +1 s+1

s+1

s Proof If (θ2(n 

s+1 +1)

n (θ s s )p ∈ Fs−1 ), then o((θ2(n  2n

s+1 +1)

s+1

p n ) = o((θ s )p , Fs−1 n +εs +2 ) , s

n ) = ε + 2 − 1 ≤ o(θ s n ) = o(θ s n Fs−1 , Fs−1 s n +εs , Fs−1 ) = εs , which implies that 2n s+1

s

s 1 < 0, a contradiction. From this, it implies that (θ2(n 

s+1 +1) p−1

p ) = Zns+1 (θ s  )p + (Zns+1 ) hand, as (θns+1    +1 s+1 s+1 +1 2(ns+1 +1) s+1 −1 −1 −1 −1 p p p p ) , and (θns+1 ) , (Zns+1 ) , (Yns+1 ) (Yns+1     s+1 s+1 −1 s+1 s+1 p n ( θs ) ∈ Fs−1 ). (θns+1  2ns+1 s+1 +1

n (θ s )p  ∈ Fs−1 ). On the other 2n

s θ2n  s+1

s+1

p−1 + (θns+1 )  −1 s+1

+ Yns+1 +  +1 s+1

are elements of F0n , we deduce that

E. H. Fliouet p−1

Lemma 5.5 For each s ≥ 1, (Zns+1 )  +1

∈ F n .

−∞ f p

−∞ f p

s+1

, (Yg ) )(f,g)∈ ), m = k(m0 ), Mi = m(Ki ), Proof By setting m0 = Q(((Zg )  j j j j j j M = m(K) = i≥1 Mi , and for each i ∈ N∗ , Z  i = Zn +i , Y  i = Yn +i and αi = θn +i j

j

j

we find ourselves under similar conditions of Proposition 5.6 where X, Z  i , Y  i , and αi play j j j the same roles as X, Zi , Yi , and θi , and consequently, the result immediately follows. j

j

j

n ((α i ) However, it is immediately verified that for every i ∈ N∗ , Fin = Fi−1 j j ≥1 ), and for i n each j ≥ 1, o(αj , Fi−1 ) = j .

Lemma 5.6 (F0n )p

−1

∩ F n = F0n (α11 ).

Proof It is clear that F0n (α11 ) ⊆ (F0n )p

F0n (θn1 +1 ), 1

thus there exists θ ∈

−1

∩ F n . Suppose that (F0n )p

−1 (F0n )p

Fsn ,

∩

Fn

−1

∩ F n  = F0n (α11 ) = n θ  ∈ F0 (θn1 +1 ). Let s be the 1 n /F n is q-simple and As Fs+1 s

such that

n Fs+1

so θ ∈ and 1 ≤ s. largest integer such that θ ∈ n ) = 1, and taking into account the transitivity of linear disjointness, o(θ, Fsn ) = o(θ, Fs−1 n s n F n (θ ) F n ⊗F n (θ s we will have Fsn (θ ) = Fsn (α1s+1 ) Fsn ⊗Fs−1 ) Fs−1 (θ2n )(θ ). s s−1  s−1 2ns+1

s+1

n (θ s n (θ s Whence, Fsn (θ )/Fs−1 ) is modular, so (Fsn (θ ))p and Fs−1 ) are (Fsn (θ ))p ∩ 2n 2n s+1

s+1

n (θ s s Fs−1 )-linearly disjoint. As (θ2(n  2n

n (θ s )p ∈ Fs−1 ) by Corollary 5.11, then 2ns+1 s+1 s+1 +1) p s n s s (1, (θ2(n +1) ) ) is linearly free over Fs−1 (θ2n ), and in particular, (1, (θ2(n )p )  s+1 s+1 s+1 +1) n (θ s ). Extending this system to a linear basis B is linearly free over (Fsn (θ ))p ∩Fs−1 2ns+1 p p n s n (θ s n n )) of (Fs (θ )) over (Fs (θ )) ∩Fs−1 (θ2n ), B is also a linear basis of (Fs−1 2ns+1 s+1 p n s n ((Fs (θ )) ) over Fs−1 ( θ2n ) from linear disjointness. But, by construction, we have s+1 p

(α1s+1 )

p

= (θns+1 )  +1 s+1

= Zns+1 (θ s  +1 2(n s+1

s+1 +1)

)p + (Zns+1 )  s+1

p−1 s θ2n s+1

+ (θns+1 )  −1

p−1

s+1

+ Yns+1 + (Yns+1 )   +1 s+1

p−1

s+1

p−1 s p−1 p−1 n (θ s with (Zns+1 ) θ2n + (θns+1 ) +Yns+1 + (Yns+1 ) belonging to Fs−1 )      2ns+1 s+1 s+1 s+1 −1 s+1 +1 s+1 −1 −1 −1 p p p ) , (Yns+1 ) , and (θns+1 ) are elements of F0n . By identification, we since (Zns+1    s+1 s+1 s+1 −1 p−1 n (θ s ) is in (Fsn (θ ))p ∩ Fs−1 ), and thus, (Zns+1 ) ∈ Fsn (θ ) ⊆ F n , a get (Zns+1   2n +1 +1 s+1

s+1

contradiction by Lemma 5.5 above, and consequently (F0n )p Lemma 5.7 For all n ∈ N∗ \ {1}, o2 ((k(θn11 , . . . , θnp11 )) lp(n) and n1 = n, and for each s ∈ N∗ , ns+1 = [ n2s ]. p

p

Proof We first put L = k(θn11 , . . . , θnp11 ), L1 = Lp distinguish two cases

−1

p−1

s+1 −1

∩ F n = F0n (α11 ).

∩ K/k) = [ n12+1 ] where p1 =

∩ K, and L2 = k(θn11 +1 , θn22 +1 ). We

Absolutely lq-Finite Extensions

Case 1 If n is even, then n1 = n and n2 = n2 = [ n2 ]. By virtue of Lemma 5.6, L1 ⊆ (F0n )p

−1

∩ K ⊆ (F0n )p −n j p 2

−1

∩ F n = F0n (α11 ).

−n j p 2

But, F0n (α11 ) is included in K1 (((Yi ) , (Zi ) )(i,j )∈N∗ ×(N\{0,1}) ), thus for each β ∈ −n n −1 F0n (α11 ), we have β p 2 ∈ K1 ; and consequently Lp ∩ K ⊆ F0n (α11 ) ∩ K ⊆ K1 p 2 ∩ K. Taking into account of Propositions 3.10 and 3.11, we have n2 = o2 (L/k) ≤ o2 (L1 /k) ≤ −n −n o2 (K1 p 2 ∩ K/k) = o1 (K1 p 2 ∩ K/K1 ) = n2 , and therefore o2 (L1 /k) = n2 = [ n+1 2 ]. Case 2 If n is odd, as k(L1 p ) ⊆ L ⊆ L1 , according to Proposition 3.11, we get oi (L1 /k) p = oi (L/k) + εi , with εi = 0 or 1. On the one hand, we have by construction (θn22 +1 ) = p

1 Zn22 +1 (θ2(n ) + θn22 + Yn22 +1 . On the other hand, 2n2 + 1 = n = n1 , and for each i ∈ N∗ , 2 +1) −i

p

p

1 1 Q(θi1 ) = Q(Xp ), so in particular Q(θ2n ) = Q((θ2(n ) ); therefore, (θn22 +1 ) ∈ L. 2 +1 2 +1) We deduce that L2 ⊆ L1 , and by Propositions 5.10 and 3.10, we have o2 (L2 /k) = n2 +1 ≤ o2 (L1 /k) ≤ n2 + 1. Whence, o2 (L1 /k) = n2 + 1 = [ n12+1 ]. −n

−n

Lemma 5.8 Let n and r be two natural integers with n  = 0. If Kr p 1 ∩ K ⊆ Kr+1 p 2 ∩ −np K ⊆ · · · ⊆ Kr+p1 −1 p 1 ∩ K ⊆ Kr+p1 where p1 = lp(n), n1 = n, and for each s ∈ N∗ , −n p1 ). ns+1 = [ n2s ], then we have Kr p ∩ K = Kr (θn1r+1 , . . . , θnr+p 1 Proof Since the (Ki )1≤i are constructed in the same manner, therefore, we reduce to −n the case where r = 0. For each n ∈ N∗ , recall that kn always designates k p ∩ K. p1 1 Firstly, according to Proposition 5.10, we have o1 (k(θn1 , . . . , θnp1 )/k) = n1 , and so p

k(θn11 , . . . , θnp11 ) ⊆ kn1 . On the other hand, as kn1 ⊆ Ks−1 p

−ns

∩ K, by virtue of Proposition

p

5.10, we will have ns = os (k(θn11 , . . . , θnp11 )/k) ≤ os (kn1 /k) ≤ os (Ks−1 p p −ns

p−ns

−ns

∩ K/k) =

∩ K/Ks−1 ) = ns (namely Ks−1 ∩ K/k is q-finite and Ks−1 /k is relo1 (Ks−1 atively perfect of irrationality degree s − 1), and so os (kn1 /k) = ns . From this kn1 = p k(θn11 , . . . , θnp11 ). Proof of Theorem 5.9 In what follows, we continue to use the notations introduced in above −1 lemma, and let n be a positive integer. According to Corollary 5.7, for each i ∈ N, Ki p ∩ −2 −1 K ⊆ Ki+1 and Ki p ∩K ⊆ Ki+1 p ∩K, so the theorem holds if, n ∈ {1, 2}. Consider now a natural number n > 2 such that the considered statement is true for any natural number less then n, i.e., for every j ≤ n, for each i ∈ N, Ki p p −n−1

−[ n+1 ] 2 p

−j

∩ K ⊆ Ki+1 p

j −[ ] 2

, and show that

∩ K ⊆ Ki+1 for each i ∈ N. Henceforth, we will denote e = [ n12+1 ]. −n −n Special case: By recurrence hypothesis, we have k p 1 ∩ K ⊆ K1 p 2 ∩ K ⊆ · · · ⊆ −np p Kp1 −1 p 1 ∩K ⊆ Kp1 where p1 = lp(n). From Lemma 5.8, kn1 = k(θn11 , . . . , θnp11 ),

Ki

−1

and by Lemma 5.7, o2 (kn1 p ∩ K/k) = o2 (kn1 +1 /k) = e. As o(θn11 +1 , k) = n1 + 1 = o1 (kn1 +1 /k), therefore θn11 +1 ∈ kn1 +1 , and by applying the r-basis completion algorithm, we extend θn11 +1 to an r-basis of kn1 +1 /k. Thus for each β ∈ kn1 +1 , we have e

−e

β p ∈ k(θn11 +1 ) ⊆ K1 . It follows that kn1 +1 ⊆ K1 p ∩ K. General case: Let s ≥ 1, by repeatedly applying the recurrence relation, we obtain −np −n −n Ks p 1 ∩ K ⊆ Ks+1 p 2 ∩ K ⊆ · · · ⊆ Ks+p1 −1 p 1 ∩ K ⊆ Ks+p1 . We want to prove that

E. H. Fliouet

Ks p

−n1 −1

−e

−n

s+p

∩K ⊆ Ks+1 p . From Lemma 5.8, Ks p ∩K = Ks (θns+1 , θns+2 , . . . , θnp1 1 ). For 1 2 −∞

j

p−∞

j

p−∞

)(i,j )∈s+1 \{(1,s+1)} , (Yi ) )(i,j )∈s+1 ) where simplicity, we set h0 = Q(X p , ((Zi ) s+1 = N∗ × {2, . . . , s + 1}. It is clear that h0 is perfect, and let us put h = k(h0 ). From j Lemma 5.1, for each (i, j ) ∈ s , we have θi ∈ h0 , and consequently Ks ⊆ h. Furthermore, p−1 s θ2i

as for every i ≥ 1, θis+1 = (Zis+1 ) p−i

+ · · · + (Z1s+1 )

p−i

p−i

(θ2s )p

−i+1

+ (Yis+1 ) + · · · +

(Y1s+1 ) , we deduce that h0 (θis+1 ) = h0 ((Z1s+1 ) ). By slightly modifying the notations above, we will reduce to the conditions of the example above. To do this, for every  i ≥ 1, we put Hi = h(Ks+i ), and by convention H0 = h. Also we set H = h(K) = i≥1 Hi , j +s

j +s

X  = Z1s+1 , and for each j ≥ 2, Z  i = Zi and Y  i = Yi . It is immediately verified −1 −1 −1  j j j +s j +s p j +s−1 j +s p j +s p that Hj = i≥1 Hj −1 (βi ) where βi = θi = (Zi ) θ2i +(θi−1 ) +(Yi ) j

j

−1 j −1 j p

p−1

p−1

for every j ≥ 1. In particular, for all j ≥ 2, βi = (Z  i ) β2i +(βi−1 ) + (Y  i j ) . We find ourselves again in the conditions of the previous example, and the rest of the proof follows from the following lemma. j

j

j +1

s+j +1

Lemma 5.9 For each (i, j ) ∈ N∗ × N, we have o(βi , Hj ) = o(θi s+j +1 , Kj +s ). In particular, K and h are Ks -linearly disjoint. o(θi j

, h(Ks+j )) = i =

j

Proof Since the family (X, (Zi , Yi )(i,j )∈N∗ ×(N\{0,1}) is algebraically free over Q and h0 = Q(X p

−∞

−∞ j p

, ((Zi )

−∞ j p

)(i,j )∈s+1 \{(1,s+1)} , (Yi )

)(i,j )∈s+1 ),

then the family (X  , (Z  i , Y  i )(i,j )∈N∗ ×(N\{0,1}) ) remains algebraically independent over h0 . Hence, in the same way as Lemma 5.2, for each (i, j ) ∈ N∗ × N, we have j

j +1

o(βi

j

s+j +1

, Hj ) = o(θi

s+j +1

, h(Ks+j )) = i = (θi

, Kj +s ),

j +1

or again for each (i, j ) ∈ N∗ × N, Hj and Ks+j (βi ) are Ks+j -linearly disjoint. As j +1 Kj +s+1 is an union of increasing extensions (Kj +s (βi ))i≥1 , we deduce that Kj +s+1 and Hj are Ks+j -linearly disjoint for every j ∈ N. By repeated application of the transitivity of linear disjointness, we see that Kj +s+1 and H0 are Ks -linearly disjoint. Since K is a union of increasing subextensions (Kn )n≥s , so K and h are Ks -linearly disjoint. p

Continuation of proof of Theorem 5.9 For simplicity, put N = h(βn11 , . . . , βnp11 ). We proceed −1

just in the same way as we did before (cf. Lemma 5.7) and we have o2 (N p ∩ H / h) = e. −n −n −1 −1 p But, Ks p 1 ∩K = Ks (βn11 , . . . , βnp11 ) ⊆ N , therefore Ks p 1 ∩K ⊆ N p ∩H . However −n −1

−n −1

K and h are Ks -linearly disjoint; then, h(Ks p 1 ∩ K) h ⊗Ks Ks p 1 ∩ K. Conse−n −1 quently, according to [4, Proposition 7] and Proposition 3.11, o2 (h(Ks p 1 ∩K)/ h) = −n −1 −n −1 −1 o2 (h ⊗Ks Ks p 1 ∩ K/ h) = o2 (Ks p 1 ∩ K/Ks ) ≤ o2 (N p ∩ H / h) = e. As s+1 o(θn1 +1 , Ks ) = n1 + 1, by applying the r-basis completion algorithm, we extend θns+1 1 +1 −n −1

−n −1

to a canonically ordered r-basis of Ks p 1 ∩ K/Ks , and so o2 (Ks p 1 ∩ K/Ks ) = −n −1 −n −1 e o1 (Ks p 1 ∩ K/Ks (θns+1 )) ≤ e. Hence, for each x ∈ Ks p 1 ∩ K, x p ∈ Ks (θns+1 )⊆ 1 +1 1 +1 Ks+1 ; it follows that x ∈ Ks+1 p

−e

∩ K. Whence, Ks p

−n1 −1

∩ K ⊆ Ks+1 p

−e

∩ K.

Absolutely lq-Finite Extensions −n

Corollary 5.12 For every (n, s) ∈ N ∗ × N, Ks p ∩ K = Ks (θns+1 , θns+2 , . . . , θnp1 1 ) 1 2 ns ∗ where n1 = n, lp(n) = p1 , and for each s ∈ N , ns+1 = [ 2 ]. s+p

Proof Similar to that of Lemma 5.8. By repeatedly applying Theorem 5.9, for each n1 ∈ N∗ , we obtain the following decom−np −n −n position Ks p 1 ∩ K ⊆ Ks+1 p 2 ∩ K ⊆ · · · ⊆ Ks+p1 −1 p 1 ∩ K ⊆ Ks+p1 where −n −n n lp(n1 ) = p1 , and nj +1 = [ 2j ], and consequently Ks p 1 ∩ L ⊆ Ks+1 p 2 ∩ L ⊆ · · · ⊆ −np

Ks+p1 −1 p 1 ∩ L ⊆ Ks+p1 ∩ L for any subextension L/k of K/k. In particular, for every −2n −n −n n n n ∈ N, Ks p ∩ L ⊆ Ks+1 p ∩ L, so Ks (Ks p ∩ Lp ) ⊆ Ks (Ks+1 ∩ Lp ) ⊆ Ks+1 , and −n n as Ks+1 /Ks is q-simple, then Ks (Ks p ∩ Lp )/Ks is simple. Proof Let L/k be a subextension of K/k. If for any natural number s, Ks ⊆ L, then L = K and the proof is finished at this point. So let s be the largest integer such that Ks ⊆ L. If −n L/Ks has an exponent n, or again L ⊆ Ks p ∩ K, then, as K/Ks is absolutely lq-finite, −n we deduce that L/Ks is finite since the same holds also for Ks p ∩ K/Ks . Otherwise, −2n −n n for all n ∈ N∗ , we get o1 (Ks p ∩ L/Ks ) = 2n, and so o1 (Ks (Ks p ∩ Lp )/Ks ) = n. −n n Since Ks+1 /Ks is q-simple and Ks (Ks p ∩ Lp ) ⊆ Ks+1 for every n ∈ N∗ , it follows that  −n n Ks+1 = n∈N Ks (Ks p ∩ Lp ) ⊆ L, a contradiction. According to Waterhouse (cf. [17, Proposition 1.2]), there exists the smallest subextension m/k of K/k such that K/m is modular. As of now, lm(K/k) denotes the smallest subextension such that K/ lm(K/k) is modular. We mention here an application of Theorem 5.3. Theorem 5.13 The smallest subextension m/k of K/k such that K/m is modular, is trivial. In other words, lm(K/k) = K Proof If m = K, by virtue of Theorem 5.3, there exists i ∈ N∗ such that Ki ⊆ m and m/Ki is finite. As di(K/Ki ) is infinite, the same is true for di(K/m). By Corollary 4.7, −1 −1 di(mp ∩ K/m) = di(K/m), and therefore, di(mp ∩ K/m) is infinite. Furthermore, as −1 K/m is also absolutely lq-finite, we have di(mp ∩ K/m) < +∞, a contradiction. Hence m = K. Corollary 5.14 For each j ≥ 1, lm(K/Kj ) = K. Proof Immediate.

5.2 Characterization of q-Finite Extensions Proposition 5.15 Any absolutely lq-finite extension K/k is q-finite over lm(K/k). In particular, any absolutely lq-finite modular extension is q-finite. Proof We first set m = lm(K/k). The case m = K is trivial, so assume that m  = K. Note −1 that di(mp ∩ K/m) = di(K/m) (cf. Corollary 4.7) and K/k is absolutely lq-finite, then the same is true for K/m, and therefore di(K/m) is finite.

E. H. Fliouet

Theorem 5.16 An absolutely lq-finite extension K/k is q-finite if and only if for every proper intermediate field L between k and K, the smallest subextension m of L/k such that L/m is modular is nontrivial, i.e., lm(L/k) = L. Proof The necessary condition results from [10, Theorem 3.3]. Conversely, consider the decreasing sequence of subextensions (mi /k)i of K/k defined by: m0 = K, and for each i ≥ 1, mi = lm(mi−1 /k). By hypothesis, if mi−1 = k, then mi  = mi−1 . However, since any decreasing sequence of subextensions of K/k is stationary, there exists a natural integer j such that mj = k. Hence, we have K = m0 ←− m1 ←− . . . ←− mj −1 ←− mj = k. In particular, for each i ∈ {1, . . . , j }, mi−1 /mi is absolutely lq-finite, and consequently by Proposition 5.15, mi−1 /mi is q-finite. From where K/k is q-finite. As an immediate consequence of Theorems 5.16 and 5.2, we have: Corollary 5.17 For a purely inseparable extension K/k to be q-finite, it is necessary and sufficient that K/k satisfy the following two conditions: (i) Any decreasing sequence of subextensions of K/k is stationary. (ii) For every intermediate field L of K/k, we have lm(L/k)  = L. Furthermore, here is a characteristic property that identifies the modular extensions that are q-finite. Theorem 5.18 Let K/k be a purely inseparable modular extension. The following assertions are equivalent: (1) Any decreasing sequence of subextensions of K/k is stationary. (2) Any set of subextensions of K/k has a minimal element. (3) K/k is q-finite. Proof Immediately follows from Theorem 5.2 and Proposition 5.15. An interesting application of Theorem 5.18 is given by the following result. Corollary 5.19 Let k be a commutative field of characteristic p > 0, and let  be an algebraic closure of k. The following assertions are equivalent: (1) di(k) is finite. (2) Every set of purely inseparable subextensions of /k has a minimal element. (3) Any decreasing sequence of purely inseparable subextensions of /k is stationary. −∞

of /k is modular Proof It is enough to remark that the purely inseparable closure k p −∞ over k, and di(k) = di(k p /k). Consequently, the result follows from Theorem 5.18. Let K/k be a modular extension of unbounded exponent. For K/k to be q-finite, it suffices to modify the sufficient condition of Theorem 5.2 in the following way: Theorem 5.20 For a modular extension K/k of unbounded exponent to be q-finite, it is necessary and sufficient that any decreasing sequence of subextensions of unbounded exponent of K/k be stationary. In the proof, we will need the following results.

Absolutely lq-Finite Extensions

Proposition 5.21 For any relatively perfect modular extension K/k, for every natural number n, kn /k is equiexponential of exponent n. Proof From Proposition 4.6, it suffices to show that k(kn p ) = kn−1 . Having regard of n n modularity of the K/k, K p , and k are k ∩ K p -linearly disjoint for each n ≥ 1, and n−1 n n−1 n by virtue of the transitivity of linear disjointness, k p (K p ) and k are k p (k ∩ K p )n−1 n n−1 linearly disjoint. But K/k is relatively perfect, so k p (K p ) = K p , and consequently n−1 n−1 n = k p (k ∩ K p ), or again kn−1 = k(kn p ). k ∩ Kp Lemma 5.10 Let K/k be a purely inseparable extension of unbounded exponent and infinite irrationality degree. If K/k is modular and relatively perfect, then K/k contains a proper modular subextension L/k of unbounded exponent. Proof We will build by induction a strictly increasing sequence (Kn /k)n≥1 of modular subextensions of exponent n of K/k. As K/k is relatively perfect, according to Proposition −n −1 5.21 and Corollary 4.7, for each n ≥ 1, di(k p ∩ K/k) = di(k p ∩ K/k) = di(K/k) −n −1 and k p ∩ K/k is equiexponential of exponent n. Let G1 be an r-basis of k p ∩ K/k, −1 it follows that k p ∩ K ⊗k (k(a))a∈G1 . Let us choose an element x of G1 , since G1 is infinite, there exists a finite subset G1 of G1 such that x  ∈ k(G1 ). We put K1 = k(G1 ), it is clear that K1 /k is modular. We suppose that we have constructed a sequence of finite subextensions k ⊆ K1 ⊆ K2 ⊆ · · · ⊆ Kn of K/k such that (1) For each i ∈ {1, . . . , n}, Ki /k is modular. (2) For every i ∈ {1, . . . , n}, o1 (Ki /k) = i. (3) x ∈ Kn . −n−1 −n−1 ∩ K/k, then k p ∩ K ⊗k (k(a))a∈Gn+1 . Since Let Gn+1 be an r-basis of k p −n−1 ∩ K. But Kn /k is finite and Gn+1 is infinite, o1 (Kn /k) = n, we deduce that Kn ⊆ k p therefore there exists a finite subset Gn+1 of Gn+1 such that Kn ⊆ k(Gn+1 ). Two cases can occur: Case 1 If x ∈ k(Gn+1 ), then Kn+1 = k(Gn+1 ) is suitable. Case 2 We first put B = Gn+1 \ Gn+1 . If x ∈ k(Gn+1 ), then as kp

−n−1

∩ K ⊗k (⊗k (k(a))a∈G

n+1

) ⊗k (⊗k (k(a))a∈B ),

we would have x ∈ k(B). Otherwise, since k(Gn+1 ) and k(B) are k-linearly disjoint, then x ∈ k(Gn+1 ) ∩ k(B) = k, a contradiction. Let y be an element of B (y exists because Gn+1 is infinite and Gn+1 is finite). We put Kn+1 = Kn (y); it is immediately verified that • Kn+1 /k is finite, and o1 (Kn+1 /k) = o(y, k) = n + 1. • Kn+1 Kn ⊗k k(y) (Application of the transitivity of linear disjointness of k(Gn+1 ) and k(B)), and as Kn /k is modular, by [6, Lemma 3.4], Kn+1 /k is modular. −n−1 ∩ K k(Gn+1 ) ⊗k k(B) Kn (Gn+1 ) ⊗Kn Kn (B), • x ∈ Kn+1 . Otherwise, since k p   then x ∈ k(Gn+1 ) ∩ Kn (y) ⊆ Kn (Gn+1 ) ∩Kn (B) = Kn , a contradiction. Whence, Kn+1 /k is suitable, and consequently L = i≥1 Ki satisfies the conditions of the above result. Proof of Theorem 5.20 It is enough to show that 2 ⇒ 1. To do this, we will use a contrapositive proof to establish this result. Suppose that K/k is not q-finite, and let G be an r-basis of K/k(K p ). If |G| is infinite, there exists a family (ai )i∈N∗ of distinct elements in G. For

E. H. Fliouet

each n ∈ N ∗ , we put Kn = k(K p )(G \ {a1 , . . . , an }), and K0 = K. It is clear that the sequence (Kn )n∈N of subextensions of unbounded exponent of K/k is decreasing, but as G is an r-basis of K/k(K p ), so for each n ∈ N, Kn = Kn+1 ; hence, the sequence (Kn /k)n∈N is nonstationary, so we are led to the case where |G| is finite. We put e = o1 (k(G)/k) and e L1 = k(K p ). Firstly, according to [13, Proposition 7], L1 /k is modular, and secondly, as e e+1 e e+1 K = k(K p )(G), then k(K p ) = k(K p )(Gp ) = k(K p ); or again L1 /k is relatively perfect. It is verified that: • If L1 /k is q-finite, the same is true for K/k since di(K/k) ≤ di(K/L1 ) +di(L1 /k). • If L1 /k is not q-finite, by Lemma 5.10, L1 /k contains a proper modular subextension L2 /k of unbounded exponent. We distinguish two cases: Case 1 If L2 /k is q-finite, as L1 /k is modular and not q-finite, by Corollary 4.7, −1 −1 di(k p ∩ L1 /k) is infinite. We set M = L2 (k p ∩ L1 ). It is immediate that di(M/k) ≥ −1 di(k p ∩L1 /k), and as L2 /k is q-finite, then di(M/L2 ) is infinite. Hence, if G is an r-basis of M/L2 , then |G| is infinite. As in the previous case, we construct a strictly decreasing sequence of subextensions of K/k. Case 2 If L2 /k is not q-finite, we find ourselves exactly in the conditions of K/k, and by repeating the same procedures, we construct a nonstationary decreasing sequence (Ln /k)n≥1 of subextensions of K/k.

5.3 The Absolute lq-Finitude and the w0 -Generated Recall that an extension K/k is called w0 -generated if it does not admit any proper subextension of unbounded exponent. For more details on this topic, refer to the articles [9] and [4]. Theorem 5.22 For any absolutely lq-finite extension K/k, there exists a family of subextensions k = K0 ⊆ K1 · · · ⊆ Kn · · · ⊆ K of K/k such that Ki+1 /Ki is w0 -generated. Suppose furthermore that K/Ki is of unbounded exponent. Then, Ki+1 /Ki is also of unbounded exponent. In particular, if K/k is q-finite, the sequence (Ki )i is stationary. Proof If K/k is finite or w0 -generated, then the result holds immediately, and we put K1 = K. Otherwise, let H be the set of subextensions of unbounded exponent of K/k. By Proposition 4.2, H contains a minimal subextension of unbounded exponent of K/k that we will denote K1 . K1 /k is necessarily w0 -generated. Suppose that we have constructed a sequence k = K0 ⊆ K1 ⊆ · · · ⊆ Kn of subextensions of K/k such that Ki /Ki−1 is w0 generated for each i ∈ {1, . . . , n}, and if K/Ki is of unbounded exponent, then the same is also true for Ki+1 /Ki . If K = Kn , we stop. Otherwise, as K/Kn is absolutely lq-finite, we return to the starting situation, and so there exists a subextension Kn+1 /Kn of K/Kn such that Kn+1 /Kn is w0 -generated and if K/Kn is of unbounded exponent, then the same is also true for Kn+1 /Kn . Hence, by repeating this procedure, it is easy to show that there exists a sequence k = K0 ⊆ K1 ⊆ · · · ⊆ Kn ⊆ · · · ⊆ K such that Ki /Ki−1 is w0 -generated for each i ≥ 1. By construction, it is clear that if K/Ki is of unbounded exponent, then the same is also true for Ki+1 /Ki . As an application of Proposition 5.15, we give a necessary and sufficient condition for an w0 -generated extension to be q-finite.

Absolutely lq-Finite Extensions

Proposition 5.23 Let K/k be an absolutely lq-finite extension. If moreover K/k is w0 generated, then lm(K/k) = K if and only if K/k is q-finite. Proof Immediate.

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