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Tame Fields and Tame Extensions
201, 647]655 Ž1998. JA977298
JOURNAL OF ALGEBRA ARTICLE NO.
Tame Fields and Tame ExtensionsU Sudesh K. Khanduja Department of Mathematics, Panjab Uni¨ ersity, Chandigarh-160014, India Communicated by Walter Feit Received August 16, 1997
Let V be a henselian valuation of any rank of a field K and let V be the extension of V to a fixed algebraic closure K of K. In this paper, it is proved that Ž K, V . is a tame field, i.e., every finite extension of Ž K, V . is tamely ramified, if and only if, to each a g K R K, there corresponds a g K for which V Ž a y a. G D K Ž a ., X X where D K Ž a . s minV Ž a y a .< a runs over all K-conjugates of a 4. A special case of the previous result, when K is a perfect field of nonzero characteristic was proved in 1995, with the purpose of completing a result of James Ax wS. K. Khanduja, J. Algebra 172 Ž1995., 147]151x. Q 1998 Academic Press Key Words: valued fields; valuations and their generalizations
1. INTRODUCTION Throughout V is a henselian valuation of any rank of a field K and V is a Žunique. prolongation of V to a fixed algebraic closure K of K. In 1970, James Ax w1, Sect. 2, Proposition 2X x pointed out that if K is a perfect field of nonzero characteristic and V is of rank one, then to each a g K, there corresponds a g K for which V Ž a y a. G D K Ž a ., where, D K s min V Ž a y a X . : a X runs over K-conjugates of a 4 .
Ž 1.
In 1991, a counterexample was given to show that this result is false Žsee w5x.. In 1995, we gave a necessary and sufficient condition for the foregoing result to be true Žcf. w6, Theorem 1.1x.. In fact we proved THEOREM A. Let Ž K, V . be a henselian ¨ alued field of any rank which is a perfect field of nonzero characteristic and let Ž K, V . be as in the pre¨ ious text. Then to each a g K, there corresponds an element a g K satisfying V Ž a y a. G D K Ž a . if and only if Ž K, V . is a defectless ¨ alued field. * The research is partially supported by CSIR vide Grant No. 25Ž0072.r93rEMR-II. 647 0021-8693r98 $25.00 Copyright Q 1998 by Academic Press All rights of reproduction in any form reserved.
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SUDESH K. KHANDUJA
In this paper our main aim is to formulate and to prove an equivalent form of Theorem A which holds even when characteristic Žto be abbreviated as char. of K is zero. Recall that a finite extension Ž K X , V X . of Ž K, V . is said to be defectless if w K X : K x s ef, where e, f are, respectively, the index of ramification and the residual degree of V XrV. A finite extension Ž K X , V X . of Ž K, V . is said to be tame or tamely ramified if Ža. it is defectless; Žb. the residue field of V X is a separable extension of the residue field of V; Žc. the ramification index of V XrV is not divisible by the char of the residue field of V. Further a henselian valued field is said to be a defectless field Žresp., a tame field. if each finite extension of Ž K, V . is defectless Žresp., tame.. Observe that when K is a perfect field of char p ) 0, then the residue field of V is also perfect and its value group is p-divisible; consequently every finite defectless extension of Ž K, V . is tame in this instance. Therefore Theorem A is a particular case of the following theorem which we prove in the second section. THEOREM 1.1. Let V be a henselian ¨ aluation of any rank of a field K and let K, V be as in the foregoing text. Then to each a g K, there corresponds an element a g K satisfying V Ž a y a. G D K Ž a ., if and only if Ž K, V . is a tame field. In the course of proof of the preceding theorem, we show that if Ž K X , V X . is any finite Galois tame extension of a henselian valued field Ž K, V ., then to each a g K X there corresponds a g K satisfying V X Ž a y a. G D K Ž a . Žsee Theorem 2.3.. The following question regarding its converse naturally arises: If Ž K X , V X . is a finite Galois extension of a henselian field Ž K, V . such that to each a g K X , there corresponds a g K satisfying V X Ž a y a. G D K Ž a ., then does it follow that Ž K X , V X . is a tame extension of Ž K, V .? We show by means of an example Žsee Example 2.5. that the answer to the previous question is ‘‘no’’ in general even if K is perfect. We prove that the answer is ‘‘yes’’ if we have the additional hypothesis D K Ž a . s v K Ž a . for some generator a of the extension K XrK, where v K Ž a . s ` if a g K and
v K Ž a . s max V Ž a y a X . < a X / a runs over K-conjugates of a 4 , otherwise. In this direction, the following two theorems are proved. THEOREM 1.2. Let V be a henselian ¨ aluation of a field K with prolongation V to an algebraic closure K of K and let a g K R K be separable o¨ er K. If there exists a g K such that V Ž a y a. G v K Ž a ., then K Ž a . is a tame extension of the ¨ alued field Ž K, V ..
TAME FIELDS AND TAME EXTENSIONS
649
THEOREM 1.3. Let V be a henselian ¨ aluation of a field K with residue field k and let V X be the prolongation of V to a finite Galois extension K X of K of degree n with residue field kX . Then the following statements are equi¨ alent. Ži. kX / k and to each a g K X R K, there corresponds a g K such that V Ž a y a. G v K Ž a .. X
Žii.
n is prime and kXrk is a Galois extension of degree n.
2. PROOF OF THEOREM 1.1 The proof of the theorem is on the same lines as that of Theorem 1.1 of w6x, however there are some significant changes. We retain the notations of w6x. For any b in the valuation ring of a valuation V X extending the given valuation V, b U will stand for its V X-residue, i.e., the image of b under the canonical homomorphism from the valuation ring of V X onto the residue field of V X . When there is no chance of confusion, we shall write DŽ a . for D K Ž a . defined by Ž1.. Note that D K Ž a . s ` if a is purely inseparable over K. We first prove some preliminary results. LEMMA 2.1. Let V be a henselian ¨ aluation of a field K whose residue field has characteristic p G 2. Let Ž K X , V X . be a tame Galois extension of Ž K, V . of degree p. Then to each a g K X , there corresponds a g K such that V X Ž a y a. G D K Ž a .. Proof. The index of ramification of V XrV being a divisor of p must be one, as it should be coprime to p. So if k, kX are the residue fields of V, V X , respectively, then in view of Ž K X , V X .rŽ K, V . being defectless, we conclude that w kX :k x s p. The rest of the proof of the lemma is exactly the same as that of Lemma 2.1 of w6x and is omitted. LEMMA 2.2. Let Ž K, V . be a henselian ¨ alued field with the residue field of V ha¨ ing characteristic p G 0. Let Ž K X , V X . be a finite extension of Ž K, V . of degree not di¨ isible by p. Then for any a g K X , one can choose a g K with V X Ž a y a. G D K Ž a .. Proof. It is clear from the hypothesis of the lemma that K X is a separable extension of K. Let a be any element of K X R K. Denote K Ž a . by L and denote the set of K-monomorphisms of L into the algebraic closure K of K by A. By hypothesis p does not divide w L: K x s < A <, i.e.,
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SUDESH K. KHANDUJA
V X Žw L: K x. s 0. So we have V X a y w L: K x
ž
y1
tr L r K Ž a . s V X Ž w L: K x a y tr L r K Ž a . .
/
s VX
ž Ý Ž a y sa . / G D sgA
K
Ža.,
and the lemma is proved. THEOREM 2.3. Let Ž K X , V X . be a finite Galois tame extension of a henselian ¨ alued field Ž K, V .. Then for any a g K X , there exists a g K such that V X Ž a y a. G D K Ž a .. Proof. Let p denote the char of the residue field of V. If p s 0, then the theorem is already proved by virtue of Lemma 2.2. So we may assume that p / 0. We prove the theorem by induction on n where p n is the exact power of p dividing w K X : K x. If n s 0, then the result is true in view of Lemma 2.2. In the general case, let L be the fixed field of a p-Sylow subgroup of the Galois group of K XrK. Then p does not divide w L: K x and there exists a tower of fields L s L0 : L1 : ??? : L n s K X such that L iq1 is a normal extension of L i of degree p for i s 0, . . . , n y 1. Since Ž K X , V X .rŽ K, V . is tame, so is L iq1rL i for each i. By Lemma 2.1, there exists g g L ny 1 such that V X Ž a y g . G D L Ž ny 1.Ž a . G D K Ž a . .
Ž 2.
Proceeding exactly as in the proof of w6, Lemma 2.3x, we can show that there exists an element a g K for which V X Ž g y a. G D K Ž a . .
Ž 3.
It follows from Ž2. and Ž3. that V X Ž a y a. G D K Ž a . which completes the proof of the theorem. Remark. At the end of this section, we shall give an example to show that the converse of Theorem 2.3 is false. Recall that an extension Ž K X , V X .rŽ K, V . of valued fields is said to be immediate if V and V X have the same value group and the same residue field. The following lemma is already known Žsee w6, Lemma 2.4x. and we omit its proof. LEMMA 2.4. Let Ž K X , V X . be an immediate algebraic extension of a perfect henselian ¨ alued field Ž K, V .. Assume that for each a g K X , there exists an element a g K with V Ž a y a. G D K Ž a .. Then K s K X .
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651
Proof of Theorem 1.1. Assume that Ž K, V . is a tame field. Then it can be easily seen that K is a perfect field. Hence each a g K lies in a Galois extension of K and by Theorem 2.3 there corresponds a g K with V Ž a y a. G D K Ž a . .
Ž 4.
To prove the converse, assume that for every a g K, there exists a g K satisfying Ž4.. Therefore K must be perfect, for if K is of char p ) 0 and a g K 1r p , then D K Ž a . s ` by definition and Ž4. implies that a s a which is in K. Now arguing exactly as in the concluding lines of proof of Theorem 1.1 of w6, p. 151x, one can show that any finite extension Ž K X , V X . of Ž K, V . is tame. The following is an example of a finite Galois extension Ž K X , V X .rŽ K, V . such that for each a in K X there exists a g K satisfying Ž4. but Ž K X , V X . is not a tame extension of Ž K, V .. EXAMPLE 2.5. Let Q denote the field of rational numbers with the 2-adic valuation u 2 characterized by u 2 Ž2. s 1. Let u be a primitive third root of unity. However, as is well known, u 2 has a unique prolongation to QŽ u . and is unramified in QŽ u . Žcf. w2, Chapter 5, Section 2.2x.. So the value group of the prolongation V2 of u 2 to QŽ u . is Z and its residue field is the field D of 4 elements. Let t be an indeterminate and let V2t be the Gaussian valuation of QŽ u , t . which extends V2 and is defined by V2t
ž Ý a t / s min V Ž a . , i
i
i
i
2
i
ai g Q Ž u . .
Let Ž K, V . be the henselisation of Ž QŽ u , t ., V2t . and let K X s K Ž t 1r6 . be the Galois extension of K of degree 6 with valuation V X which is the unique prolongation of V to it. However, as in w3, Section 10.1, Proposition 2x, it can be easily seen that the residue field k of V is DŽ tU . and the residue field kX of V X is DŽ tU 1r6 ., where the V-residue tU of t is transcendental over D. Since kXrk is not a separable extension, it follows that Ž K X , V X .rŽ K, V . is not tame. We show that for any a g K X R K, there corresponds a g K satisfying Ž4.. We fix a sixth root of t and denote it by x. Let z be a primitive sixth root of unity and let s 1 , s 2 , s 3 , s4 , s5 , s6 be the automorphisms of K XrK with si characterized by
si Ž x . s z i x. Because z 3 is a primitive second root of unity, we have V Ž 1 y z 3 . s V2 Ž 2 . s 1.
Ž 5.
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SUDESH K. KHANDUJA
Because Ł 5is1Ž1 y z i . s 6 and the value group of V2 is Z, we have by virtue of Ž5., 5
Ý V2 Ž 1 y z i . s V2 Ž 6 . s 1 s V2 Ž 1 y z 3 . . is1
Therefore, as V2 Ž1 y z i . G 0 for each i, it follows that V2 Ž 1 y z
i
. s 0,
1 F i F 5,
i / 3.
Ž 6.
Keeping in view that the V X-residues of 1, x, x 2 , x 3 , x 4 , x 5 form a linearly independent set over k, we can easily check Žcf. w7, Cor. 3.10x. that if a i g K, then VX
5
a i x i s min V X Ž a i . .
žÝ / is0
i
Ž 7.
Using Ž5., Ž6., and Ž7., a simple calculation shows that for any a s Ý5is0 a i x i g K X R K with a i g K, we have V X Ž a y s 1 a . s min V Ž a1 . , V Ž a2 . , V Ž a3 . q 1, V Ž a4 . , V Ž a5 . 4 ; V X Ž a y s 2 a . s min V Ž a1 . , V Ž a2 . , V Ž a4 . , V Ž a5 . 4 ; V X Ž a y s 3 a . s min V Ž a1 . q 1, V Ž a3 . q 1, V Ž a5 . q 1 4 ; V X Ž a y s4 a . s min V Ž a1 . , V Ž a2 . , V Ž a4 . , V Ž a5 . 4 ; V X Ž a y s5 a . s min V Ž a1 . , V Ž a2 . , V Ž a3 . q 1, V Ž a4 . , V Ž a5 . 4 . Consequently, D K Ž a . s min V Ž a1 . , V Ž a2 . , V Ž a3 . q 1, V Ž a4 . , V Ž a5 . 4 . Two cases are distinguished. If D K Ž a . s V Ž a3 . q 1, then it is clear that a s a0 q 2 a3 satisfies V X Ž a y a. s D K Ž a .. If D K Ž a . - V Ž a3 . q 1 say D K Ž a . s V Ž a j ., 1 F j F 5, j / 3. Because V Ž a i . are integers, we have V Ž a j . F V Ž a3 .. Therefore a s a0 q a j works in this case.
TAME FIELDS AND TAME EXTENSIONS
653
3. PROOF OF THEOREMS 1.2 AND 1.3 Proof of Theorem 1.2. Let K X be the smallest Galois extension of K containing a and let K V be the maximal tame extension of Ž K, V . contained in K X . By hypothesis, there exists a g K satisfying V Ž a y a. G v K Ž a . .
Ž 8.
Suppose that the theorem is false, i.e., a f K V . So there exists s g GalŽ K XrK V . such that s Ž a . / a . By definition of the ramification group GalŽ K XrK V . Žsee w4, Section 20x., we have V Ž s Ž a y a. y Ž a y a. . ) V Ž a y a. .
Ž 9.
Combining Ž8. and Ž9., we conclude that V Ž s Ž a . y a . ) vK Ž a . which contradicts the definition of v K Ž a .. Hence the theorem. Proof of Theorem 1.3. By an elementary result w4, 14.5x kX is a normal extension of k. For any s g GalŽ K XrK ., we shall denote by s the image of s under the canonical homomorphism from GalŽ K XrK . onto the group of automorphisms of kXrk, characterized by s Ž j U . s s Ž j .U , where for any j the valuation ring of V X , j U stands for its V X-residue. Assume first that Ži. holds. Then by Theorem 1.2 Ž K X , V X . is a tame extension of Ž K, V ., in particular kX is a separable and hence Galois extension of k. So assertion Žii. is proved as soon as we show that for each b U g kX R k, we have k Ž b U . :k s w K X : K x s w kX :k x .
Ž 10 .
Let b U be any element of kX R k with b g K X and V X Ž b . s 0. By assertion Ži., there exists b g K such that V X Ž b y b . G vK Ž b . .
Ž 11 .
If v K Ž b . ) 0, then by Ž11., b U s bU g k which is not so. It follows that
v K Ž b . s 0.
Ž 12 .
Observe that if s is any element of GalŽ K XrK . with s Ž b . / b , then s Ž b U . / b U , for otherwise V X Ž s Ž b . y b . ) 0 which leads to v K Ž b . ) 0 contrary to Ž12.. It follows that K Ž b . : K s k Ž b U . :k .
Ž 13 .
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SUDESH K. KHANDUJA
We next show that K Ž b . s KX.
Ž 14 .
Suppose that Ž14. is false. Let a be a generator of the extension K XrK and let d be an element of K such that V X Ž d a . ) 0. Because K Ž b . is assumed to be a proper subfield of K Ž a . s K X , there exists t g GalŽ K XrK . such that t Ž a . / a , t Ž b . s b . Therefore keeping in view the choice of d, we have V X Ž t Ž d a q b . y Ž d a q b . . s V X Ž d Ž t Ž a . y a . . ) 0; consequently v K Ž d a q b . ) 0. By assertion Ži. there exists c g K satisfying V X Ž d a q b y c . G v K Ž d a q b . ) 0; Because V X Ž d a . ) 0, we see that V X Ž b y c . ) 0, i.e., b U s cU which is impossible. This contradiction proves Ž14.. Combining Ž13. and Ž14., we see that k Ž b U . :k s K Ž b . : K s w K X : K x . The desired equation Ž10. is now obvious, because by fundamental inequality w4, 17.5x, k Ž b U . :k F w kX :k x F w K X : K x . Conversely suppose that Žii. holds. Then K X is an unramified and hence tame extension of K. Let a be any element of K X R K. By Theorem 2.3, there exists a g K satisfying V X Ž a y a. G D K Ž a .. Assertion Ži. is proved once we show that D K Ž a . s vK Ž a . .
Ž 15 .
V X Ž a y a . s max V X Ž a y b . : b g K 4 s D K Ž a . .
Ž 16 .
We first show that
Because Ž K, V . is a henselian valuation, for any K-conjugate a X of a and b g K, we have V X Ž a y b . s V X Ž a X y b .; consequently, V X Ž a y a X . G min V X Ž a y b . ; V X Ž a X y b . 4 s V X Ž a y b . . Therefore D K Ž a . G V X Ž a y b . for all b g K. On recalling that V X Ž a y a. G D K Ž a ., we quickly obtain Ž16.. Suppose, if possible, Ž15. is false. Keeping in view that Ž K X , V X .rŽ K, V . is unramified, we can choose c g K such that V X Ž a y a. s D K Ž a . s V Ž c . .
TAME FIELDS AND TAME EXTENSIONS
655
Set b s Ž a y a.rc. The claim is that b U f k. If b U g k, say b U s dU , d g K, V Ž d . s 0; this leads to V X Ž Ž a y a . y cd . ) V Ž c . s D K Ž a . , which contradicts the last equality of Ž16. and proves the claim. By hypothesis w K X : K x s w kX :k x is prime, it follows that kX s k Ž b U ., K X s K Ž b .. A simple calculation shows that DK Ž b . s DK Ž a . y V Ž c. ,
vK Ž b . s vK Ž a . y V Ž c . . So the supposition D K Ž a . - v K Ž a . implies that D K Ž b . - v K Ž b .. Because V X Ž b . s 0, we have D K Ž b . G 0. Thus we are led to v K Ž b . ) 0. So by definition of v K Ž b ., there exists an automorphism t of K XrK, such that t Ž b . / b and V X Žt Ž b . y b . ) 0. So t Ž b U . s b U . Keeping in view that kX s k Ž b U ., we see that t is trivial on kX , which is impossible by virtue of the fact that Ž K X , V X .rŽ K, V . is an unramified extension. The proof of the theorem is now complete.
REFERENCES 1. J. Ax, Zeros of polynomials over local fields}The Galois action, J. Algebra 15 Ž1970., 417]428. 2. Z. I. Borevich and I. R. Shafarevich, ‘‘Number Theory,’’ Academic Press, New York, London, 1966. 3. N. Bourbaki, ‘‘Commutative Algebra Chapter 6 Valuations,’’ Hermann, Paris, 1972. 4. O. Endler, ‘‘Valuation Theory,’’ Springer-Verlag, BerlinrNew York, 1972. 5. S. K. Khanduja, A note on a result of J. Ax, J. Algebra 140 Ž1990., 360]361. 6. S. K. Khanduja, On a result of James Ax, J. Algebra 172 Ž1995., 147]151. 7. J. Ohm, Simple transcendental extensions of valued fields II: A fundamental inequality, J. Math. Kyoto Uni¨ . 25 Ž1985., 583]596.
JOURNAL OF ALGEBRA ARTICLE NO.
Tame Fields and Tame ExtensionsU Sudesh K. Khanduja Department of Mathematics, Panjab Uni¨ ersity, Chandigarh-160014, India Communicated by Walter Feit Received August 16, 1997
Let V be a henselian valuation of any rank of a field K and let V be the extension of V to a fixed algebraic closure K of K. In this paper, it is proved that Ž K, V . is a tame field, i.e., every finite extension of Ž K, V . is tamely ramified, if and only if, to each a g K R K, there corresponds a g K for which V Ž a y a. G D K Ž a ., X X where D K Ž a . s minV Ž a y a .< a runs over all K-conjugates of a 4. A special case of the previous result, when K is a perfect field of nonzero characteristic was proved in 1995, with the purpose of completing a result of James Ax wS. K. Khanduja, J. Algebra 172 Ž1995., 147]151x. Q 1998 Academic Press Key Words: valued fields; valuations and their generalizations
1. INTRODUCTION Throughout V is a henselian valuation of any rank of a field K and V is a Žunique. prolongation of V to a fixed algebraic closure K of K. In 1970, James Ax w1, Sect. 2, Proposition 2X x pointed out that if K is a perfect field of nonzero characteristic and V is of rank one, then to each a g K, there corresponds a g K for which V Ž a y a. G D K Ž a ., where, D K s min V Ž a y a X . : a X runs over K-conjugates of a 4 .
Ž 1.
In 1991, a counterexample was given to show that this result is false Žsee w5x.. In 1995, we gave a necessary and sufficient condition for the foregoing result to be true Žcf. w6, Theorem 1.1x.. In fact we proved THEOREM A. Let Ž K, V . be a henselian ¨ alued field of any rank which is a perfect field of nonzero characteristic and let Ž K, V . be as in the pre¨ ious text. Then to each a g K, there corresponds an element a g K satisfying V Ž a y a. G D K Ž a . if and only if Ž K, V . is a defectless ¨ alued field. * The research is partially supported by CSIR vide Grant No. 25Ž0072.r93rEMR-II. 647 0021-8693r98 $25.00 Copyright Q 1998 by Academic Press All rights of reproduction in any form reserved.
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SUDESH K. KHANDUJA
In this paper our main aim is to formulate and to prove an equivalent form of Theorem A which holds even when characteristic Žto be abbreviated as char. of K is zero. Recall that a finite extension Ž K X , V X . of Ž K, V . is said to be defectless if w K X : K x s ef, where e, f are, respectively, the index of ramification and the residual degree of V XrV. A finite extension Ž K X , V X . of Ž K, V . is said to be tame or tamely ramified if Ža. it is defectless; Žb. the residue field of V X is a separable extension of the residue field of V; Žc. the ramification index of V XrV is not divisible by the char of the residue field of V. Further a henselian valued field is said to be a defectless field Žresp., a tame field. if each finite extension of Ž K, V . is defectless Žresp., tame.. Observe that when K is a perfect field of char p ) 0, then the residue field of V is also perfect and its value group is p-divisible; consequently every finite defectless extension of Ž K, V . is tame in this instance. Therefore Theorem A is a particular case of the following theorem which we prove in the second section. THEOREM 1.1. Let V be a henselian ¨ aluation of any rank of a field K and let K, V be as in the foregoing text. Then to each a g K, there corresponds an element a g K satisfying V Ž a y a. G D K Ž a ., if and only if Ž K, V . is a tame field. In the course of proof of the preceding theorem, we show that if Ž K X , V X . is any finite Galois tame extension of a henselian valued field Ž K, V ., then to each a g K X there corresponds a g K satisfying V X Ž a y a. G D K Ž a . Žsee Theorem 2.3.. The following question regarding its converse naturally arises: If Ž K X , V X . is a finite Galois extension of a henselian field Ž K, V . such that to each a g K X , there corresponds a g K satisfying V X Ž a y a. G D K Ž a ., then does it follow that Ž K X , V X . is a tame extension of Ž K, V .? We show by means of an example Žsee Example 2.5. that the answer to the previous question is ‘‘no’’ in general even if K is perfect. We prove that the answer is ‘‘yes’’ if we have the additional hypothesis D K Ž a . s v K Ž a . for some generator a of the extension K XrK, where v K Ž a . s ` if a g K and
v K Ž a . s max V Ž a y a X . < a X / a runs over K-conjugates of a 4 , otherwise. In this direction, the following two theorems are proved. THEOREM 1.2. Let V be a henselian ¨ aluation of a field K with prolongation V to an algebraic closure K of K and let a g K R K be separable o¨ er K. If there exists a g K such that V Ž a y a. G v K Ž a ., then K Ž a . is a tame extension of the ¨ alued field Ž K, V ..
TAME FIELDS AND TAME EXTENSIONS
649
THEOREM 1.3. Let V be a henselian ¨ aluation of a field K with residue field k and let V X be the prolongation of V to a finite Galois extension K X of K of degree n with residue field kX . Then the following statements are equi¨ alent. Ži. kX / k and to each a g K X R K, there corresponds a g K such that V Ž a y a. G v K Ž a .. X
Žii.
n is prime and kXrk is a Galois extension of degree n.
2. PROOF OF THEOREM 1.1 The proof of the theorem is on the same lines as that of Theorem 1.1 of w6x, however there are some significant changes. We retain the notations of w6x. For any b in the valuation ring of a valuation V X extending the given valuation V, b U will stand for its V X-residue, i.e., the image of b under the canonical homomorphism from the valuation ring of V X onto the residue field of V X . When there is no chance of confusion, we shall write DŽ a . for D K Ž a . defined by Ž1.. Note that D K Ž a . s ` if a is purely inseparable over K. We first prove some preliminary results. LEMMA 2.1. Let V be a henselian ¨ aluation of a field K whose residue field has characteristic p G 2. Let Ž K X , V X . be a tame Galois extension of Ž K, V . of degree p. Then to each a g K X , there corresponds a g K such that V X Ž a y a. G D K Ž a .. Proof. The index of ramification of V XrV being a divisor of p must be one, as it should be coprime to p. So if k, kX are the residue fields of V, V X , respectively, then in view of Ž K X , V X .rŽ K, V . being defectless, we conclude that w kX :k x s p. The rest of the proof of the lemma is exactly the same as that of Lemma 2.1 of w6x and is omitted. LEMMA 2.2. Let Ž K, V . be a henselian ¨ alued field with the residue field of V ha¨ ing characteristic p G 0. Let Ž K X , V X . be a finite extension of Ž K, V . of degree not di¨ isible by p. Then for any a g K X , one can choose a g K with V X Ž a y a. G D K Ž a .. Proof. It is clear from the hypothesis of the lemma that K X is a separable extension of K. Let a be any element of K X R K. Denote K Ž a . by L and denote the set of K-monomorphisms of L into the algebraic closure K of K by A. By hypothesis p does not divide w L: K x s < A <, i.e.,
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SUDESH K. KHANDUJA
V X Žw L: K x. s 0. So we have V X a y w L: K x
ž
y1
tr L r K Ž a . s V X Ž w L: K x a y tr L r K Ž a . .
/
s VX
ž Ý Ž a y sa . / G D sgA
K
Ža.,
and the lemma is proved. THEOREM 2.3. Let Ž K X , V X . be a finite Galois tame extension of a henselian ¨ alued field Ž K, V .. Then for any a g K X , there exists a g K such that V X Ž a y a. G D K Ž a .. Proof. Let p denote the char of the residue field of V. If p s 0, then the theorem is already proved by virtue of Lemma 2.2. So we may assume that p / 0. We prove the theorem by induction on n where p n is the exact power of p dividing w K X : K x. If n s 0, then the result is true in view of Lemma 2.2. In the general case, let L be the fixed field of a p-Sylow subgroup of the Galois group of K XrK. Then p does not divide w L: K x and there exists a tower of fields L s L0 : L1 : ??? : L n s K X such that L iq1 is a normal extension of L i of degree p for i s 0, . . . , n y 1. Since Ž K X , V X .rŽ K, V . is tame, so is L iq1rL i for each i. By Lemma 2.1, there exists g g L ny 1 such that V X Ž a y g . G D L Ž ny 1.Ž a . G D K Ž a . .
Ž 2.
Proceeding exactly as in the proof of w6, Lemma 2.3x, we can show that there exists an element a g K for which V X Ž g y a. G D K Ž a . .
Ž 3.
It follows from Ž2. and Ž3. that V X Ž a y a. G D K Ž a . which completes the proof of the theorem. Remark. At the end of this section, we shall give an example to show that the converse of Theorem 2.3 is false. Recall that an extension Ž K X , V X .rŽ K, V . of valued fields is said to be immediate if V and V X have the same value group and the same residue field. The following lemma is already known Žsee w6, Lemma 2.4x. and we omit its proof. LEMMA 2.4. Let Ž K X , V X . be an immediate algebraic extension of a perfect henselian ¨ alued field Ž K, V .. Assume that for each a g K X , there exists an element a g K with V Ž a y a. G D K Ž a .. Then K s K X .
TAME FIELDS AND TAME EXTENSIONS
651
Proof of Theorem 1.1. Assume that Ž K, V . is a tame field. Then it can be easily seen that K is a perfect field. Hence each a g K lies in a Galois extension of K and by Theorem 2.3 there corresponds a g K with V Ž a y a. G D K Ž a . .
Ž 4.
To prove the converse, assume that for every a g K, there exists a g K satisfying Ž4.. Therefore K must be perfect, for if K is of char p ) 0 and a g K 1r p , then D K Ž a . s ` by definition and Ž4. implies that a s a which is in K. Now arguing exactly as in the concluding lines of proof of Theorem 1.1 of w6, p. 151x, one can show that any finite extension Ž K X , V X . of Ž K, V . is tame. The following is an example of a finite Galois extension Ž K X , V X .rŽ K, V . such that for each a in K X there exists a g K satisfying Ž4. but Ž K X , V X . is not a tame extension of Ž K, V .. EXAMPLE 2.5. Let Q denote the field of rational numbers with the 2-adic valuation u 2 characterized by u 2 Ž2. s 1. Let u be a primitive third root of unity. However, as is well known, u 2 has a unique prolongation to QŽ u . and is unramified in QŽ u . Žcf. w2, Chapter 5, Section 2.2x.. So the value group of the prolongation V2 of u 2 to QŽ u . is Z and its residue field is the field D of 4 elements. Let t be an indeterminate and let V2t be the Gaussian valuation of QŽ u , t . which extends V2 and is defined by V2t
ž Ý a t / s min V Ž a . , i
i
i
i
2
i
ai g Q Ž u . .
Let Ž K, V . be the henselisation of Ž QŽ u , t ., V2t . and let K X s K Ž t 1r6 . be the Galois extension of K of degree 6 with valuation V X which is the unique prolongation of V to it. However, as in w3, Section 10.1, Proposition 2x, it can be easily seen that the residue field k of V is DŽ tU . and the residue field kX of V X is DŽ tU 1r6 ., where the V-residue tU of t is transcendental over D. Since kXrk is not a separable extension, it follows that Ž K X , V X .rŽ K, V . is not tame. We show that for any a g K X R K, there corresponds a g K satisfying Ž4.. We fix a sixth root of t and denote it by x. Let z be a primitive sixth root of unity and let s 1 , s 2 , s 3 , s4 , s5 , s6 be the automorphisms of K XrK with si characterized by
si Ž x . s z i x. Because z 3 is a primitive second root of unity, we have V Ž 1 y z 3 . s V2 Ž 2 . s 1.
Ž 5.
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SUDESH K. KHANDUJA
Because Ł 5is1Ž1 y z i . s 6 and the value group of V2 is Z, we have by virtue of Ž5., 5
Ý V2 Ž 1 y z i . s V2 Ž 6 . s 1 s V2 Ž 1 y z 3 . . is1
Therefore, as V2 Ž1 y z i . G 0 for each i, it follows that V2 Ž 1 y z
i
. s 0,
1 F i F 5,
i / 3.
Ž 6.
Keeping in view that the V X-residues of 1, x, x 2 , x 3 , x 4 , x 5 form a linearly independent set over k, we can easily check Žcf. w7, Cor. 3.10x. that if a i g K, then VX
5
a i x i s min V X Ž a i . .
žÝ / is0
i
Ž 7.
Using Ž5., Ž6., and Ž7., a simple calculation shows that for any a s Ý5is0 a i x i g K X R K with a i g K, we have V X Ž a y s 1 a . s min V Ž a1 . , V Ž a2 . , V Ž a3 . q 1, V Ž a4 . , V Ž a5 . 4 ; V X Ž a y s 2 a . s min V Ž a1 . , V Ž a2 . , V Ž a4 . , V Ž a5 . 4 ; V X Ž a y s 3 a . s min V Ž a1 . q 1, V Ž a3 . q 1, V Ž a5 . q 1 4 ; V X Ž a y s4 a . s min V Ž a1 . , V Ž a2 . , V Ž a4 . , V Ž a5 . 4 ; V X Ž a y s5 a . s min V Ž a1 . , V Ž a2 . , V Ž a3 . q 1, V Ž a4 . , V Ž a5 . 4 . Consequently, D K Ž a . s min V Ž a1 . , V Ž a2 . , V Ž a3 . q 1, V Ž a4 . , V Ž a5 . 4 . Two cases are distinguished. If D K Ž a . s V Ž a3 . q 1, then it is clear that a s a0 q 2 a3 satisfies V X Ž a y a. s D K Ž a .. If D K Ž a . - V Ž a3 . q 1 say D K Ž a . s V Ž a j ., 1 F j F 5, j / 3. Because V Ž a i . are integers, we have V Ž a j . F V Ž a3 .. Therefore a s a0 q a j works in this case.
TAME FIELDS AND TAME EXTENSIONS
653
3. PROOF OF THEOREMS 1.2 AND 1.3 Proof of Theorem 1.2. Let K X be the smallest Galois extension of K containing a and let K V be the maximal tame extension of Ž K, V . contained in K X . By hypothesis, there exists a g K satisfying V Ž a y a. G v K Ž a . .
Ž 8.
Suppose that the theorem is false, i.e., a f K V . So there exists s g GalŽ K XrK V . such that s Ž a . / a . By definition of the ramification group GalŽ K XrK V . Žsee w4, Section 20x., we have V Ž s Ž a y a. y Ž a y a. . ) V Ž a y a. .
Ž 9.
Combining Ž8. and Ž9., we conclude that V Ž s Ž a . y a . ) vK Ž a . which contradicts the definition of v K Ž a .. Hence the theorem. Proof of Theorem 1.3. By an elementary result w4, 14.5x kX is a normal extension of k. For any s g GalŽ K XrK ., we shall denote by s the image of s under the canonical homomorphism from GalŽ K XrK . onto the group of automorphisms of kXrk, characterized by s Ž j U . s s Ž j .U , where for any j the valuation ring of V X , j U stands for its V X-residue. Assume first that Ži. holds. Then by Theorem 1.2 Ž K X , V X . is a tame extension of Ž K, V ., in particular kX is a separable and hence Galois extension of k. So assertion Žii. is proved as soon as we show that for each b U g kX R k, we have k Ž b U . :k s w K X : K x s w kX :k x .
Ž 10 .
Let b U be any element of kX R k with b g K X and V X Ž b . s 0. By assertion Ži., there exists b g K such that V X Ž b y b . G vK Ž b . .
Ž 11 .
If v K Ž b . ) 0, then by Ž11., b U s bU g k which is not so. It follows that
v K Ž b . s 0.
Ž 12 .
Observe that if s is any element of GalŽ K XrK . with s Ž b . / b , then s Ž b U . / b U , for otherwise V X Ž s Ž b . y b . ) 0 which leads to v K Ž b . ) 0 contrary to Ž12.. It follows that K Ž b . : K s k Ž b U . :k .
Ž 13 .
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SUDESH K. KHANDUJA
We next show that K Ž b . s KX.
Ž 14 .
Suppose that Ž14. is false. Let a be a generator of the extension K XrK and let d be an element of K such that V X Ž d a . ) 0. Because K Ž b . is assumed to be a proper subfield of K Ž a . s K X , there exists t g GalŽ K XrK . such that t Ž a . / a , t Ž b . s b . Therefore keeping in view the choice of d, we have V X Ž t Ž d a q b . y Ž d a q b . . s V X Ž d Ž t Ž a . y a . . ) 0; consequently v K Ž d a q b . ) 0. By assertion Ži. there exists c g K satisfying V X Ž d a q b y c . G v K Ž d a q b . ) 0; Because V X Ž d a . ) 0, we see that V X Ž b y c . ) 0, i.e., b U s cU which is impossible. This contradiction proves Ž14.. Combining Ž13. and Ž14., we see that k Ž b U . :k s K Ž b . : K s w K X : K x . The desired equation Ž10. is now obvious, because by fundamental inequality w4, 17.5x, k Ž b U . :k F w kX :k x F w K X : K x . Conversely suppose that Žii. holds. Then K X is an unramified and hence tame extension of K. Let a be any element of K X R K. By Theorem 2.3, there exists a g K satisfying V X Ž a y a. G D K Ž a .. Assertion Ži. is proved once we show that D K Ž a . s vK Ž a . .
Ž 15 .
V X Ž a y a . s max V X Ž a y b . : b g K 4 s D K Ž a . .
Ž 16 .
We first show that
Because Ž K, V . is a henselian valuation, for any K-conjugate a X of a and b g K, we have V X Ž a y b . s V X Ž a X y b .; consequently, V X Ž a y a X . G min V X Ž a y b . ; V X Ž a X y b . 4 s V X Ž a y b . . Therefore D K Ž a . G V X Ž a y b . for all b g K. On recalling that V X Ž a y a. G D K Ž a ., we quickly obtain Ž16.. Suppose, if possible, Ž15. is false. Keeping in view that Ž K X , V X .rŽ K, V . is unramified, we can choose c g K such that V X Ž a y a. s D K Ž a . s V Ž c . .
TAME FIELDS AND TAME EXTENSIONS
655
Set b s Ž a y a.rc. The claim is that b U f k. If b U g k, say b U s dU , d g K, V Ž d . s 0; this leads to V X Ž Ž a y a . y cd . ) V Ž c . s D K Ž a . , which contradicts the last equality of Ž16. and proves the claim. By hypothesis w K X : K x s w kX :k x is prime, it follows that kX s k Ž b U ., K X s K Ž b .. A simple calculation shows that DK Ž b . s DK Ž a . y V Ž c. ,
vK Ž b . s vK Ž a . y V Ž c . . So the supposition D K Ž a . - v K Ž a . implies that D K Ž b . - v K Ž b .. Because V X Ž b . s 0, we have D K Ž b . G 0. Thus we are led to v K Ž b . ) 0. So by definition of v K Ž b ., there exists an automorphism t of K XrK, such that t Ž b . / b and V X Žt Ž b . y b . ) 0. So t Ž b U . s b U . Keeping in view that kX s k Ž b U ., we see that t is trivial on kX , which is impossible by virtue of the fact that Ž K X , V X .rŽ K, V . is an unramified extension. The proof of the theorem is now complete.
REFERENCES 1. J. Ax, Zeros of polynomials over local fields}The Galois action, J. Algebra 15 Ž1970., 417]428. 2. Z. I. Borevich and I. R. Shafarevich, ‘‘Number Theory,’’ Academic Press, New York, London, 1966. 3. N. Bourbaki, ‘‘Commutative Algebra Chapter 6 Valuations,’’ Hermann, Paris, 1972. 4. O. Endler, ‘‘Valuation Theory,’’ Springer-Verlag, BerlinrNew York, 1972. 5. S. K. Khanduja, A note on a result of J. Ax, J. Algebra 140 Ž1990., 360]361. 6. S. K. Khanduja, On a result of James Ax, J. Algebra 172 Ž1995., 147]151. 7. J. Ohm, Simple transcendental extensions of valued fields II: A fundamental inequality, J. Math. Kyoto Uni¨ . 25 Ž1985., 583]596.