QUADRATIC SUBFIELDS OF QUARTIC EXTENSIONS OF LOCAL FIELDS

We show that any quartic extension of a local field of odd residue characteristic must contain an intermediate field. A consequence of this is that local fields of odd residue characteristic do not have extensions with Galois group A4 or S4. Counterexamples are given for even residue characteristic.


INTRODUCTION.
In Section 2, a simple application of local class field theory proves the existence of intermediate fields for quartic extensions of local fields with odd residue characteristic.This immediately implies the non-existence of Galois extensions of type A h or S h over such fields.
In Section 3, examples are given of A h and S h extensions of fields with even residue characteristic, and of a quartic extension with no intermediate field.
In Section h, the results of Section 2 are used to show that the splitting field of an irreducible quartic polynomial over a local field must have degree or 8, provided the residue characteristic is odd.The implications of the results of Section 2 and Section 3 for the theory of endoscopic groups are also discussed.
I wish to thank Noriko Yui for helpful conversations about this work.

EXISTENCE OF INTERMEDIATE EXTENSIONS.
Let F be a non-archimedean local field.Let o o F and P PF respectively, be the ring of integers of F and its prime ideal.
THEOREM 2.1.Suppose the residue characteristic of F is odd, and E/F is a quartic extension (i.e.[E:F] 4).Then there must be an inter- N6w suppose e h so f 1 Any unit in E is of the form u+p and P PE The norm of such an element is u + p' with with u e o F P' e PE 0 F PF So by Hensel's Lemma the only units contained in the image of NE/F are fourth powers.In particular, NE/F is not surjective, so Corollary 1 to Theorem h of chapter XII, Section 3 of Well [1] proves the theorem.
Translating this into the corresponding result on Galois groups, we obtain the following equivalent formulation THEOREM 2.2.If F has odd residue characteristic, there cannot be a Galois extension E/F whose Galois group is isomorphic to A h or S h PROOF: A h contains subgroups of index h (the cyclic group generated by any 3-cycle), none of which is properly contained in any proper subgroup (such a proper subgroup, if it existed, would be of order 6 and index 2, hence normal, hence would contain all 3-cycles, of which there are 8).
An Sh-extension of F would be an Ah-extension of a quadratic exten- Let F 2 and consider the Eisenstein polynomial (X) Xh-2x-2 eF[X].
Let E be the splitting field of (X) we shall show that Gal(E/F) S h and Gal(E/K) A h where K 2 () In the process we shall find a quartic extension L/F with no intermediate field.
Let e be a root of (X) and let L F(a) LEMMA 3.1.The norm NL/F is surjective.
PROOF: Notice that N(+l) =.(-1)I N(e-1) Also the characteristic polynomial of 3 is 3(X) X h 6X 3 + 12X 2 8X-8 so N(a3+l) 3(-1) 19 If N NL/F were not surJective, its image would be contained in the image of the norm map from some ramified quadratic extension of F Such an image contains exactly two of the four cosets of oX modulo (oX) 2 We have Just shown NL/F contains the three cosets con- taining l, -3, and 19.
In particular (by Corollary 1 to Theorem h of chapter XII, Section 3 of Well [1]), L/F is a quartic extension with no intermediate field.
Factoring the polynomial (X) over L we see that (X) (X-G)(X) where (X) X 3 + aX 2 + 2X + (3-2) PROPOSITION 3.2.(X) is irreducible over L PROOF: If all roots of (X) were in L then L E would be Galois, in contradiction of Lemma 3.1.The only other way for (X) to be reducible would be for exactly one root, ' say, to be in L In this case, F((' would be a quartic extension of F contained in L hence F(a' F() L Let e GaI(E/F) be such that (c) ' Then g(F()) F(') and a' F(a) implies that g(a') g F(') F(c) L Since g(a') ' g(') must equal the only other conjugate of ' in L i.e. g(') Hence the fixed field L contains + ' and ' so (X-a)(X-(') So E is the splitting field of T(X) over L and GaI(E/L) is either A 3 or S 3 Now T(X) where X' X + 2/3 Hence the discriminant of W(X) is 27((20/27)C3-2)2-4((2/3)(2) 3 4.27 + (368/27)c6-80 3 4.9.3 mod*(l+4PL) Since 4.9(i+4PL (LX) 2 the discriminant of (X) is a square in L if and only if 3 is.
LEMMA 3.3.The element 3 is not a square in L PROOF: If 3 were a square, truncation of its square root would give 'an element of the form x i +as+b(2+ca 3 with a b and c each equal to 0 or 1 and so that 3 x 2 4pL A trivial computation shows that this is impossible. Accordingly GaI(E/L) S 3 GaI(E/F) S 4 and GaI(E/K) A 4 where K F(/) 4. APPLICATIONS.
i.The splitting field of a quartic polynomial over a local field is severely constrained by the results of Section 2.
THEOREM 4.1.Let F be a local field with odd residue characteristic.Let f(X) e FIx] be an irreducible polynomial with deg f(X) 4 Let E be the splitting field of f(X) over F Then [E:F] 4 or 8 PROOF: GaI(E/F) is a subgroup of S 4 But by Theorem 2.2 it cannot be S 4 or A 4 Since 4 [E:F] the only possibilities are 4 or 8 The polynomial @(X) of Section 3 gives a counterexample to this result when the residue characteristic is 2 Theorem 4.1 is clearly equivalent to Theorem 2.2 (and henc to Theorem 2.1).
2. If F is a local field, let G SL(h,F) and let T be an elliptic torus in G To T is associated a quartic extension E/F so that the centralizer of T in GL(4,F) is isomorphic to E x and T it- : { , (x) } self is isomorphic to E 1

NE/F
The theory of endoscopic groups (cf.Langlands [2], Shelstad [3]) associates to G and T some other groups, among which the most interesting are constructed as follows: let E K m F and let G' {g GL(2,K) NK/F(detg) l} In G' it is possible to find an elliptic torus T' associated to the quadratic extension E/K and there is an isomorphism between T and T' The hope is to simplify calculations with orbital integrals over the G-conjugacy class of t T by comparing them with orbital integrals over the G'-conjugacy class of the corresponding t' T' The example of Section B shows that this approach will not apply for certain tori when the residue characteristic is 2 happily, for these tori the ordinary orbital integrals are invariant under stable conjugacy, so the problem does not arise.The results of Sections 2 encourage optimism in the case of odd residue characteristic.
mediate field K i.e.E o K F [E:K] [K:F] 2 J. REPKA PROOF: If E/F is unramified, the result is obvious.If the ramifica- tion index of E/F is e 2 then we must have f 2 and, by Corollary 4 to Theorem 7 of chapter I, Section h of Weil [1], there is an unramified quadratic intermediate field.