Some generalizations of Abhyankar lemma. K.N.Ponomaryov. Abstract. ramied extensions of discretely valued elds for an arbitrary (not

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1 Some generalizations of Abhyankar lemma. K.N.Ponomaryov Abstract We prove some generalizations of Abhyankar lemma about tamely ramied extensions of discretely valued elds for an arbitrary (not nite, not tamely ramied) extension of valued elds. The paper is based on the article [1]. We use notations from this paper and prove some generalizations of theorem 2 for an arbitrary extension of discretely valued elds of a nonzero characteristic p (may be not nite and not separable extension). We have to extend the notion of strongly solvable extension from the paper. Let us include in the class of strongly solvable extension the class of unseparable nite extensions. We call a nite extension F=K as strongly solvable in wide sense extension if it is a strongly solvable separable extension over some nite purely unseparable extension. Let us note that classes of strongly solvable extensions, of purely unseparable extensions and of nite extensions are marked classes of eld extensions. So the class of strongly solvable in wide sense extensions is a marked class too (see [6]). Any strongly solvable in wide sense extension K 2 =K 1 can be presented as a tower of quasicyclic extensions K 2 = L n =:::=L 0 = K 1 of the height n. Any extension L i+1 =L i is a decomposition eld of a polynomial f(x)? a; a 2 L i. We denote by f or a separable monomial X m ; (m; p) = 1 or an unseparable monomial X p or Artin - Schreier polynomial }(X) = X p? X. We assume that any eld under consideration contains all roots from unit. We use the notion of strongly solvable extension in the wide sense only. Let us remember that we denote a kernel of perfection \l pn as l e. We denote an algebraic closure of a eld l as l a. The eld l e and any of its algebraic extension are perfect eld. So any algebraic closure of a kernel of perfection is a separable closure l ea = l es. 1

2 A key statement of the paper is the following extension of theorem 1 from [1]. Theorem 1 Let L=K be some extension of elds. Let L 0 be an intermediate discretely valued eld with the property that extension L=L 0 is a nite extension and L 0 =K is a tamely unramied extension. Let l 0 =k be a corresponding extension of residue elds for L 0 =K. Let we have an inclusion l ea 0 k. Then there is a solvable totally ramied extension K 0 =K in algebraic closure of K with the following property: the extension LK 0 =K 0 is a tamely unramied extension for any continuation of the valuation from L 0 to a composite of extensions LK 0. It is easy to see that in the proof of theorem 1 from [1] statement of theorem above was proved in a particular case of nite and wild ramied Galois extension L=L 0. We reduce the proof of theorem to this particular case. We have in mind to apply the theorem to nite coverings of algebraic varieties. So we deduce some generalization of theorem 2 from [1]. Corollary 1 Let L be some nite extension of a discretely valued eld K and a residue eld of K be a functional eld. Then there is a totally ramied solvable extension K 0 =K in algebraic closure of K with the property that extension LK 0 =K 0 is a tamely unramied extension for any continuation of valuation of K to a composite of extensions LK 0. Abhyankar lemma can be reformulated in the general form for an arbitrary (not nite) extension of discretely valued elds. It reduces any extension of valued eld to a wild ramied extension (see proposition 2). From theorem 1 we deduce a generalization of Abhyankar lemma. Theorem 2 Let L=K be some extension of discretely valued elds. Let l=k be an extension of residue elds corresponding to L=K and we have an inclusion l ea k. Then there is a solvable totally ramied extension K 0 =K in algebraic closure of K that the extension LK 0 =K 0 is a tamely unramied extension for any continuation of valuation of L to LK 0. 2

3 This theorem takes a nal form when the extension of the residue elds l=k is of nite type, i.e. when l is a nitely generated extension of k. This is a suitable form for an application of the statement to an investigation of arbitrary coverings of algebraic varieties (not necessarily nite). Corollary 2 Let L=K be an extension of discretely valued elds and a residue eld of K be a functional eld. Let a corresponding extension of residue elds be an extension of nite type. Then there is a solvable totally ramied extension K 0 =K in algebraic closure of K with the property that the extension LK 0 =K 0 is a tamely unramied extension for any continuation of the valuation of L to LK 0. Let us note that the statement of theorems above can be treated as some renement of Epp theorem (see [4]). 1 Purely unseparable extensions. We use approach of the paper [1] (see x3) and give a description of types of purely unseparable extensions of discretely valued elds. Let us remember that a valuation of K can be extended to a purely unseparable extension by unique way. We denote R(K) valuation ring, k residue eld, and? : R(K)! k natural projection of this eld. Let U(K) be a group of units and be a uniformizing parameter of K. As any purely unseparable extension can be presented as a tower of extensions of a prime degree we consider an extension L=K of a prime degree. The extension L=K is a eld of decomposition of some polynomial of the form X p? a for some a 2 KnK p. This extension can be dened by any element from the set ak p + K p = (a + K p )K p. So we can choose a from the ring R(K). Let us call these elements as representative elements of the extension. Lemma 1 Let L=K be a purely unseparable extension of discretely valued elds of a prime degree p. Then this extension is or an immediate extension or a wild purely ramied extension. In the second case there are two possibilities. (a) There is a representative element a 2 R(K) and ord(a) = 1. Then the extension is a totally ramied extension. (b) There is a representative element a 2 U(K) and a 62 k p. extension is a tamely unramied extension. Then the 3

4 Let us prove our lemma. Any immediate extension of K can be embedded in a maximal immediate extension, in the completion K. As K p is an open subset in K and K is a dense subset of K then a purely unseparable immediate extension of a prime degree is an extension that can be presented by some representative element from the eld K p \ K. Let us assume that the extension is not immediate extension. Then a 62 K p \ K. The extension is nite and we have that the extension L = L K is a wild extension of a prime degree p over K. Let us choose some eld of representatives in K. Let us identify this eld with a residue eld k by natural isomorphism. We can perform an element a from the complete eld K by power series: a = a 0 + a 1 + :::; a i 2 k. As the extension is a purely ramied wild extension we have a 2 K; a 62 K p. So there is an index n that or a n 6= 0 and (n; p) = 1 or a n 62 k p. Let us choose a minimal number n for these properties. Field K is dense in K so there is b 2 K and a = b p + a 0 n n + ::: = b p + c, c = a 0 n n + :::. Let us choose c instead of a. Case 1: (n; p) = 1. Let s; m 2 Z and ns + pm = 1. Let us prove that we can choose c s pm as representative element. This element has order 1. Indeed, let x be a root of a polynomial X p?c, so L = K(x). Then x ps = c s and element y = x s m is a root of irreducible polynomial Y p? c s pm. We have K(y) K(x) so K(x) = K(y). Case 2: pjn but a n 62 k p. Let n = pm then we can choose b = c?mp = c (?m ) p as a representative element. This element belongs to U(R) and b = a n 62 k p. We have proved the lemma. Let us remember lemma 7 from [1]: Lemma 2 Let L=K be some tamely unramied extension and K 1 =K be a totally ramied extension. Let these extensions are contained in some general discretely valued eld. Then the extension LK 1 =K 1 is a tamely unramied extension and the extension LK 1 =L is a totally ramied extension. Proposition 1 Let L=L 0 be a purely unseparable extension of a degree p n and L 0 =K be some tamely unramied extension of discretely valued elds. Let K 0 =K be a totally ramied purely unseparable extension of a degree p m. Then if m n then the extension LK 0 =K 0 is a tamely unramied extension for any continuation of valuation of K. 4

5 Let us prove the proposition. Let us assume L 0 be a maximal tamely unramied subextension of L=K. If L 0 = L then the extension is a tamely unramied extension. We can use lemma 2. So let us assume L 6= L 0. In the purely unseparable extension K 0 =K there is a subextension K n =K of a degree p n. By lemma 2 we can prove the proposition for K n =K instead of K 0 =K. So let us assume n = m. Let us present our two extensions by towers of extensions of a prime degree p: L = L n =:::=L 0 = K K 0 = K n =:::=K 0 = K. We will prove that a composite L i K i is a tamely unramied extension over K i by induction on i = 0; :::; n. If i = 0 then nothing to prove. This gives us a rst step of induction. Let we have proved that extension L i?1 K i?1 =K i?1 is a tamely unramied extension. Extension L i K i?1 =L i?1 K i?1 is a purely unseparable extension of a prime degree p and extension L i?1 K i?1 =K i?1 is a tamely unramied extension. The extension K i =K i?1 is a totally ramied purely unseparable extension of a prime degree. By lemma 2 we can restrict ourselves to a case jk 0 : Kj = jl : L 0 j = p; n = m = 1. As L 0 =K is a maximal tamely ramied extension then the extension L=L 0 is a totally ramied extension. As far as a completion of an extension preserves ramication index we can take completions of a elds. Let us assume our elds are complete elds. We choose some eld of representatives in K. We can identify this eld with residue eld by some canonical isomorphism. We use lemma 1 and present extension K 0 =K as a decomposition eld of a polynomial X p? for some uniformizing parameter. Then for some 2 U(K) the extension L=L 0 is a decomposition eld of a polynomial Y p?. So LK 0 =L 0 K 0 is a prime extension of L 0 K 0 =K 0 by some root of a polynomial Z p?. By lemma 2 it is a tame ramied extension. Let us take some representation of by power series : = :::; i 2 k. Let 2 K 0 and p = then is an uniformizing parameter of K 0. So the element in the eld LK 0 can be represented as = p +:::; i 2 k. If for any i we have i 2 k p then 2 K 0p and LK 0 = L 0 K 0. Theorem is proved. Let us consider second case i 62 k p. The extension LK 0 =L 0 K 0 is a tamely unramied extension. This extension denes a purely unseparable extension of the residue eld k. This extension is a primary extension by p - root of the element i with smallest index i. We have proved the theorem. 5

6 2 Abhyankar lemma. In the 50th S.S.Abhyankar have proved the following proposition (see [8], Exp.10, Lemme 3.6, P.279). Lemma 3 (Abhyankar lemma) Let L=K and K 0 =K be nite Galois extensions of a discretely valued eld K in some separable closure of K. Let these extensions are tame ramied extensions with ramication indices n and m. Then if a natural number n is a divisor of a natural number m then extension LK 0 =K 0 is an unramied extension for any continuation of the valuation of K to LK 0. Proof see in [8]. We can generalize this statement. Proposition 2 Let L 0 =L 0 be a nite separable extension of a tamely unramied extension of discretely valued elds L 0 =K. Then there is prime to p natural number n = n(l=l 0 ) 2 N with the following property. For any totally ramied and tamely ramied extension K 0 =K such that nje r (K 0 =K) the extension LK 0 =L 0 K 0 is a wild extension. Besides e r (LK 0 =L 0 K 0 ) is a divisor of e r (L=K). Let us prove proposition. We use lemma 2 and reduce the proof of proposition to a particular case L 0 = K. Indeed any totally ramied and tame ramied extension K 0 =K gives us a totally ramied and a tamely ramied extension K 0 L 0 =L 0. It follows from multiplicative property of ramication index that e r (K 0 =K) = e r (K 0 L 0 =L 0 ). From now on let us assume K = L 0. As subextension of a wild extension is a wild extension too we can consider L=K as Galois extension. The extension K 0 =K is a totally ramied extension so the valuation of K can be continued to K 0 by a unique way. But any continuation of a valuation K to LK 0 is dened by continuation of the valuation K 0 to LK 0. As extension LK 0 =K 0 is Galois extension the whole set of continuations is conjugated by Galois group. These continuations have equals ramication indexes. So we have to prove proposition for only one continuation of valuation of K to LK 0. From now on let us choose some continuation of valuation of K to LK 0. Let K V be a ramication eld of Galois extension L=K. This is a maximal tame ramied extension, this is Galois extension and (e r (K V =K); p) = 1. 6

7 The extension K=K V is purely wild Galois p - extension, this is a solvable extension. Let n = e r (K V =K). We take any totally ramied and tamely ramied separable extension K 0 =K of a degree n, nje(k 0 =K). By Abhyankar lemma extension K V K 0 =K 0 is tamely unramied extension and extension LK 0 =K V K 0 is a wild extension by marked property of the class of wild solvable extensions (see proposition 2 in [1]). Proof of the proposition follows from a calculation of ramication indexes. Remark. Totally ramied and tamely ramied extension of index m can be obtained as a decomposition eld of a polynomial X m? for any uniformizing parameter of K ([2].1 x8,proposition 8.1). Such extensions are cyclic extensions. 3 Proof of theorem 1. Let us prove theorem 1. Let us assume that the conditions of theorem are satised. Let L n be a normal closure of the extension L=L 0. If the statement of theorem is true for extension L n =K then this statement is valid for any subextension. In particular for extension L=K this statement is true. So we will prove theorem for normal extension L=L 0. Let us present a normal extension L=L 0 as a tower L=L us =L 0. Here L us =L 0 is a purely unseparable extension and L=L us is Galois extension. The valuation of L 0 can be extended to valuation of purely unseparable extension L us =L 0 by a unique way. We use proposition 1 for L us =K and nd such totally ramied purely unseparable extension K 1 =K that the extension L us K 1 =K 1 is a tamely unramied extension for any continuation of valuation. As far as the extension L us K 1 =L 0 is purely unseparable such continuation is a uniquely dened continuation. Remark. As an example of an extension from above one can choose a decomposition eld of a polynomial X pm? where is a uniformizing parameter, K 1 = K( p?m ). Now the extension LK 1 =L us K 1 is nite Galois extension. Let us use proposition 2 for the tower LK 1 =L us K 1 =K 1 and nd totally ramied and tamely ramied cyclic Galois extension K 2 =K 1 that the extension LK 2 =L us K 2 is wild Galois extension for any continuation of the valuation of L 0 to LK 2. By lemma 2 the extension L us K 2 =K 2 is a tamely unramied extension. 7

8 Remark. As an example of the extension above we can choose a primary extension by a root of any uniformizing parameter of K 1, K 2 = K 1 ( 1=n ). The extension K 2 =K is a totally ramied extension and by lemma 2 the extension K 2 L 0 =L 0 is a totally ramied extension too and there are trivial extensions of residue elds. So we can use the proof of theorem 1 from [1] for tower LK 2 =L us K 2 =K 2 and nd a totally ramied wild and solvable extension K 0 =K 2 that the extension LK 0 =K 0 is a tamely unramied extension for any continuation of valuation K 2 L us to K 2 L. Because the valuation of L 0 can be extended to K 2 L us by a unique way we have that the extension LK 0 =K 0 is a tamely unramied extension for any continuation of the valuation L 0 to LK 0. This proves theorem. Corollary 1 follows from statement of theorem 1. We have L 0 = K and we use remarks from x4 in [1]. From proposition 3 of this paper follows the inclusion l ea k. 4 Proof of theorem 2. Let us prove theorem 2. At rst, we prove it for complete discrete valued elds. At second, we prove it for arbitrary elds. Let L=K be an extension of complete discretely valued elds and l=k be an extension of residue elds. By Cohen theorem [5] we choose in L some eld of representatives l 0. Let be a uniformizing parameter of K. We denote by k 0 1 = l 0 \ K a proper subeld of the valuation ring of K. We dene in K a complete subeld generated by completion of the eld of fractions of a prime extension k 0 (), F = 1 k0 1 (()). This eld is a complete eld, it contains uniformizing parameter and it contains a eld of representatives k1. 0 So extension K=F is a tamely unramied extension. We note that the condition on the extension of the residue elds is satised already in extension L=F. Moreover the condition is satised already for elds of representatives. This follows from two lemmata. Lemma 4 Let l=k be an extension of elds. equivalent to the conditions 1) l e = k e, 2) k ea k. Then the inclusion l ea k is Indeed, from the inclusion k l it follows k e l e. Because l ea k then l e k. So for any natural n we have l e = (l e ) pn = k pn and l e k e. It 8

9 follows l e = k e, l ea = k ea and k ea k. Conversely, these two conditions give us the inclusions l ea = k ea k. Lemma is proved. Lemma 5 Let L be a complete discretely valued eld and l be a eld of representatives. Then L e = l e and l ea \ L l. Indeed, rst equality follows from the structure of complete discretely valued elds. Second equality follows from maximality of a eld of representatives in a valuation ring of L. Field l e is an perfect eld and extension l ea =l e is separable. So the extension (l ea \ L) l=l is a separable extension in L. This extension gives us a greater eld of representatives. That is impossible. From these remarks we see that under conditions above we have l 0e = L e K e = k 0e. From the condition l ea k of theorem 2 we have equality l e = k e. In particular we have l 0e = k 0e. So l 0ea = k 0ea k 0 K L. It follows l 0ea = l 0ea \ L l 0. So l 0ea K \ l 0 = k1. 0 We see that for extension L=F all the conditions of theorem 2 are satised. Let us note that we have to prove theorem 2 for extension L=F only. Indeed, it follows from lemma 2 that any totally ramied and solvable extension denes a totally ramied and solvable extension F 0 K=K. If the extension LF 0 =F 0 is a tamely unramied extension then the subextension LF 0 = LKF 0 =KF 0 is a tamely unramied extension too. Then for the extension K 0 = KF 0 all the conditions of the theorem are satised. From now on we suppose K = F = k1(()). 0 We denote a completion of a eld of fractions of a prime extension l 0 () by L 0. By structure theorem we have L 0 = l 0 (()) and a residue eld of L 0 is equals to l. Because the extension of the residue elds in the extension L=L 0 is trivial so e r (L=L 0 ) = jl : L 0 j and extension L=L 0 is a totally ramied and a nite extension. We use theorem 1 for tower L=L 0 =K and nd an extension required K 0 =K. This proves theorem 2 for complete elds. Now let us consider arbitrary elds. Let L=K be some extension of discretely valued elds. We take a completion of L and dene extension of complete elds L=K. We use the above statement for this extension and nd a totally ramied solvable extension K 0 =K that the extension LK 0 =K 0 is a tamely unramied extension. 9

10 We use theorem 1 and nd totally ramied solvable extension K 0 =K in the eld K 0 that K 0 = K 0 K. Then extension LK 0 =K 0 is contained in the tower of tamely unramied extensions LK 0 =K 0 =K 0, it is a tamely unramied extension. The statements of theorem 2 are satised for this extension and for continuous extension of valuation v of the eld L to the eld LK 0. Indeed, from the proof of theorem 1 it follows that the extension K 0 =K can be presented as a tower K 0 = K 3 =K 2 =K 1 =K. Where the extension K 1 =K is a decomposition eld of a polynomial X pn? for some uniformizing parameter 2 K. We can choose uniformizing parameter from the eld K. Then we have a totally ramied solvable extension K 1 =K and K 1 K = K 1. The extension K 1 =K 1 is an unramied extension and the eld K 1 is dense in K 1. The extension K 2 =K 1 is a decomposition eld of polynomial X n? for some uniformizing parameter 2 K 1. We can choose 2 K 1 and dene such a eld K 2 that K 2 K = K 2. The extension K 2 =K 2 is an unramied and immediate extension. At last, the extension K 3 =K 2 is a separable totally ramied extension. By proposition 1 from [1] there is a totally ramied solvable extension K 3 =K 2 and K 3 K = K 3 = K 0. So for a given extension of the valuation we can take K 0 = K 3. For any continuation of valuation the statement follows from the remark. Lemma 6 Let L=K be some extension of valued elds and K 0 =K be some nite extension. Then a reduced ramication index of the extension e r (LK 0 =K 0 ) does not depend on the continuation of valuation of L to LK 0. Let L be a completion of L. Then the completion of K in L gives us a completion K of K. We have the extension of complete elds L=K. We choose some extension of valuation of L to LK 0. Completion of a eld preserves a group of valuation so it preserves reduced ramication index. So e r (LK 0 =K 0 ) = e r (LK 0 =K 0 ). But the second extension is a nite extension of a complete eld L. Any complete valuation can be extended to any nite extension by a unique way. So completions LK 0 (and K 0 ) does not depend on extension of valuation. This proves our lemma and theorem. The statement of corollary 2 follows from the statement of theorem. We have use x4 from [1]. 10

11 References [1] K.N.Ponomaryov. Solvable elimination of ramication in extensions of discretely valued elds. Algebra and logic, 37N1 (1998), [2] Algebraic number theory. Edited by J.Cassels, A.Frohlich. Academic Press, [3] O.Zariski, P.Samuel. Commutative algebra, v.1. D.van Nostrand Comp., [4] H.Epp,Eliminating wild ramication. Inv.math., 19, , [5] O.Zariski, P.Samuel. Commutative algebra, v.2. D.van Nostrand Comp., [6] S.Lang.Algebra. Add.-Wes.Publ.Comp., [7] S.Lang.Algebraic numbers. Add.-Wes.Publ.Comp., [8] Revetements etales et Groupe Fondamental, Lect. Notes in Math Spr.-Verl.-Heidelb.,

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