Totally ramified splitting fields of central simple algebras over Henselian fields

JOURNAL

OF ALGEBRA

98, 95-101 (1986)

Totally Ramified Splitting Fields of Central Simple Algebras over Henselian Fields J.-P. TIGNOL* UniurrsitP Calho[ique de Louvain. B-1348 Louuain-la-Neuur,

Belgium

AND

S. A. AMMITSUR Hebrew University at Givut Ram, 91904 Jerusalem, Israel Communicated hy I. N. Herstein Received October

15. 1984

Let D be a central division algebra over a Henselian field f with algebraically closed residue field I? It is proved that if the degree of D is relatively prime to the characteristic of F, then every finite extension of F which splits D contains a maximal subfield of D. ‘7’ 1986 Academic Press. Inc

1. Let D be a division algebra, finite dimensional over its center F. It is well known that every maximal subfield of D splits D [ 1, Theorem 4.271 but in general splitting fields of D need not contain any maximal subfield of D (up to isomorphism), even when they are algebraic over F; see [2, p. 2421 for an example where F is the field of rational numbers. However, we prove here a positive result of this type under some restrictions on the center F. Let F have a Henselian valuation. Assume that the residue field F is algebraically closed and that its characteristic does not divide the degree of D (this implies in particular that every maximal subfield of D is Galois over F with abelian Galois group; see [3, p. 661); then we will prove: THEOREM. Every finite extension of F which splits D contains (an isomorphic image of) a maximal subfield of D; moreover if G, G’ are the * The main results of this paper were obtained while the first author was visiting Hebrew University of Jerusalem, whose hospitality is gratefully acknowledged.

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96

TIGNOL AND AMITSUR

Galois groups over F of two such maximal subfields, then G/H N G’IH’ for

some isomorphic subgroups H 2: H’. 2. From now on, let F be a field with a Henselian valuation v. If M is a finite extension of F, we let also v denote the (unique) extension of v to A4 and we let AM, U(M) and ii;i denote the maximal ideal of the valuation ring of v, the multiplicative group of units in A4 and the residue held of M, respectively. We then have exact sequencesof natural maps l+ U(M)+M”+v(M)+O

(2.1)

l+l+J#~+U(M)+li;i-“+l.

(2.2)

Recall that when the characteristic of F does not divide [M : F], the extension M/F is said to be tame. Now, let G be a finite group acting on M by field automorphisms which leave F elementwise invariant. For any o E G, the composite vog is an extension of v from F to M; since this extension is unique, we have for all rsE G

lNo=v

whence U(M) and 1 + A,,,, are stable under G. The action of G on M thus induces actions of G on v(M) and li;r” via the exact sequences (2.1) and (2.2) and the induced action on v(M) is easily seen to be trivial. In the next three lemmas, we keep the following hypothesis: the characteristic of li;r does not divide the order of G.

Letting it denote the order of G, this hypothesis ensures that for every UEl+JH,$f, the equation Y-u=0 has a unique solution in 1 + A,,,,, by [3, Chap. 3, Lemma 11, whence 1 + AM is uniquely divisible by n. Therefore, H’(G, 1 +AM)=

1

for all i> 1,

so that the exact cohomology sequence associated to (2.2) yields natural isomorphisms H’(G, U(M)) 2: H’(G, M”)

for all i> 1.

Moreover, since G acts trivially on v(M) and v(M) is torsion free, H’(G, v(M)) = Hom(G, v(M)) = 0

and the exact cohomology sequence associated to (2.1) yields the following result:

TOTALLY

3.

LEMMA.

RAMIFIED

SPLITTING

97

FIELDS

There is a natural exact sequence:

1 + H2(G, A’) -+ H2(G, M”)

‘* +H*(G, u(M)),

Moreover, if N is an extension qf F contained in M and stable under the action qf G. then the inclusion i: N + A4 induces a commutative diagram:

1 + H2(G, NY) -+ H2(G, N’) A

H’(G, 4N) I* I* f* 1 1 I v(M)) 1 + H2(G, &?‘) + H’(G, M”) AH*(G,

(3.1)

The commutativity of diagram (3.1) easily checked, using the naturality of the maps. 4. LEMMA. Keep the same hypotheses and notations as in Lemma 3 and assume moreover that M is totally ramified over N. If CIE H2(G, N”) is such that u,(a) =0 and i,(a) = 1, then u = 1.

x=&f Proof: If M/N is totally ramified, then and i,: H*(G, 3.‘) + H2(G, ti”) is the identity map. The lemma then follows from an easy chase around diagram (3.1). 5. Assume now that M/N is a totally ramified Galois extension, that the characteristic of iV does not divide [M : N] (i.e., M/N is tame) and that N is stable under the action of G. Under these restrictions we prove the following important lemma: LEMMA. For every a E Ker[i,: H*(G, N”) -+ H’(G, A4”)], there exists a Galois extension L of N contained in M and stable under the action of G such that:

(a) the Galois group Gal(L/N) is a homomorphic image of G, and (b) CIEKer[ii: H2(G, N”) + H2(G, L’)]. Proof: Let CIE H’(G, N”) be such that i,(a) = 1 in H*(G, M”). Since diagram (3.1) is commutative, we have

u*(g) E Ker[i,: H’(G, o(N)) -+ H2(G, u(M))]. From the exact sequence 0 -+ v(N) -+ u(M) + v(M)/u(N)

--. 0

(5.1)

98

TIGNOL

AND

AMITSUR

we obtain the associated cohomology sequence: H’(G, u(M)/u(N)) AH*(G,

u(N)) AH2(G,

u(M))

(5.2)

and hence u*(c()E Im[a,:

H’(G, u(M)/u(N)) + H’(G, v(N))];

u*(a) = d,B

for some BE H’(G, u(M)/u(N)).

so, let (5.3)

Since u(M)/u(N) is a trivial G-module, p may be regarded as a homomorphism from G to u(M)/u(N) and /?(G) is thus a subgroup of u(M)/u(N). The extension M/N being totally ramified and tame, the subgroups of u(M)/u(N) are in l-l correspondence with the extensions of N contained in M, in such a way that an extension K corresponds to the subgroup u(K)/u(N): see [3, Chap. 3, Sect. 21. We then define L as the (unique) extension of N contained in M such that 4L)lQN) = P(G).

(5.4)

Being unique, this extension L is stable under the action of G. Moreover, the Galois group of M/N is abelian, by [3, Chap. 3, Theorem 41, whence L/N is Galois and it remains to show that L satisfies conditions (a) and lb). Indeed, L/N satisfies (a): since L/N is totally ramified and tame, the Galois group Gal(L/N) is isomorphic (although not canonically, in general) to u(L)/u(N), by [3, Chap. 3, Theorem 51. Equation (5.4) then shows that Gal(L/N) is a homomorphic image of G. To prove that L satisfies (b), we consider the inclusion maps N+“L+ i” A4 of fields. Denote by ii ii the induced maps of cohomology groups. From diagram (3.1) and sequence (5.2) as applied to both i’ and i” we obtain a commutative diagram

H*(G, N”) L i; I

H*(G, L-') A

H*(G, M”)

H’(G u(L)lu(W) AH’(G, dl. I H’(G, u(N)) = r’* I

H'(G, u(L))

u(M)/u(N)) dnr I H*(G, u(N))

TOTALLY RAMIFIED SPLITTING FIELDS

99

The lower part of this diagram is (3.1) with L replacing M and i’: N-+ L replacing i: N + M. The second vertical sequence is exact: it is (5.2) with L replacing M. Finally, the upper commutative square is part of the commutative diagram of cohomology groups obtained from the following diagram: 0 + u(N) -+ u(L) + u(L)/u(N) + 0

II

ii’

Ii”

0 + u(N) + u(M) -+ u(M)/u(N)

+ 0.

Our aim is to prove that if txEH*(G,N’) and i,(r)= 1, then also i’,(~)=0, where il,: H*(G, N’) + H*(G, L’). This will be obtained by applying Lemma 4 to the lower two legs of the big diagram: by showing that ii = 0 and u,(Q) = 0, it will follow from the lemma that &cc = 0. The first part is immediate since i, = ii ii and i,a = 0 assumption. To prove that u,i’,x = 0, we note that from the comutativity of the square it suffices to show that lower iiu,a=O, i.e., that z;,a E Ker[i’,: H’(G, u(N)) + H*(G, c(L))]. Since the second vertical sequence is exact, this will follow if U*CIE Im 8L. But now, the choice of L yields that fi E Im[i;: H1(G, u(L)/u(N)) -+ H’(G, u(M)/u(N))], since by (5.4), /?E Hom(G, u(L)/u(N)) = H’(G, u(L)/u(N)). Hence, by (5.3) tl,x=a,BEL7MImi’;=Ima,, and the proofis complete. 6. The preceding lemma can be translated from cocycles to central simple algebras as follows: let N, T be two finite Galois tame extensions of a Henselian field F, both contained in some algebraic closure of F, and let M = N. T be the composite field of N and T. M E NT

PROPOSITION. If M/N is totally ramified, then for every central simple algebra ouer F split both by T and by N there exists a splitting field Kc T, Galois over Nn T, such that Gal(K/Nn T) is a homomorphic image of Gal( N/N n T).

100

TIGNOL AND AMITSUR

Proof Let G = Gal(N/N n T) = Gal(M/T). The field N is thus obviously stable under G. Let A be a central simple algebra over F, split both by N and by T; then the algebra A OF (N n T) is split by N and is therefore similar to a crossed product A OI;(Nn

T)-(N,

G, CZ)

for some a E H2(G, N’). By [l, Theorem 5.81, we have (N G, ~1 0 ,wnT T-(M)

G, a).

Since A @ (Nn T) is split by T, this relation implies that c( is in the kernel of the natural map H2(G, N’) - H2(G, M”). It then follows from Lemma 5 that there exists a Galois extension L of N contained in M such that: (a)

Gal(L/N)

is a homomorphic

image of G

(b)

the crossed product (L, G, a) splits.

We then set K = T n L so that KN = L. The field K also splits A since by [ 1, Theorem 5.81 A O,K-(N, and condition

G, ~1 O,v,,r

K- (NKIK, G, a) = (L, G, a),

(b) implies that K splits A. Finally,

since N n K = N n T,

Gal( K/N n T) = Gal(K/N n K) = Gal( L/N) and the latter is, by (a), a homorphic

image of Gal(N/N n T).

7. This result has an immediate corollary (with the notations of the previous proposition, but not assuming that M/N is totally ramified): COROLLARY. [f A is a central simple algebra over F, split both by T and N, then T contains a splitting field K of A such that [K : F] divides [N : F] [lG:N].

Proof. Let U be the maximal inertial extension of N preceding proposition (with U instead of N) yields a splitting such that Gal(K/Un T) is a homomorphic image of Gal( Gal( N/N n T). Therefore, [K: Un T] divides [N: Nn [UnT: NnT]= [U: N]= [M:N], this shows that [K:fl [N:F] [li;i:m].

in M. The field Kc T U/Un T) = T]. Since divides

Remark. In fact, the ramification factor e( U n T/N n T) divides that [K : F] divides and generally it follows eW/Nn 0, [N : F] e(N/Nn T) f( T/N n T). In particular, if T is totally ramified, then [K : F] ) [N : F] e(N/Nn T) and the latter divides [N : F]‘.

TOTALLY

RAMIFIED

SPLITTING

FIELDS

101

Indeed, let K, be the maximal inertia subfield of N n T in Un T then Gal(UnT/K,)~u(UnT)/u(K,), but v(K,)=u(NnT), u(UnT)Gu(U)= u(N) and, therefore, order Gal( U n T/K,) = e( U n T/N n T) will divide the order of u(N)/v(N n T). Also [& : N n T] = f(K,/N n T) will divide Consequently [lii:m]=[UnT:NnT]=[UnT:K,] f(T/Nn T). [K,: Nn T] =e(Un T/Nn T)f(K,,/Nn T) divides e(N/Nn T)f(T/Nn T). Finally, if T/F is totally ramified then f (T/N n T) = 1 and the rest of the remark follows. 8. Proqf of the Theorem. Let p be the characteristic of i? (possibly, p=O)andlet Nb e a maximal subfield of the division algebra D. Since F is algebraically closed and since p does not divide the degree of D, the extension N/F is tame and totally ramified, and is thus Galois (with abelian Galois group: see [3, p. 661). Let now T be a finite extension of F which splits D and let T’ be the unique maximal tame extension of F in T (see [3, p. 651). The field T’ is Galois over F and it splits D, since the degree of D is relatively prime to the rank of T over T’. Corollary 7 then shows that T’ contains a splitting field K such that [K : F] divides [N : F]. Since N is a maximal subfield of D, it follows that [K : F] = [N : F], whence K is isomorphic to a maximal subfield of D. To complete the proof, we observe that if N and N’ are two subfields of some algebraic closure of F which are isomorphic to (Galois) maximal subfields of D, then Proposition 6 shows that Gal(N/Nn N’) = Gal(N’/N n N’); if we denote by G (resp. G’) the Galois group of N/F (resp. N’/F) and by H (resp. H’) the subgroup of G (resp. G’) corresponding to other hand Since on the means: H= H’. Nn N’, this G/H E Gal(N n N’/F) 21G’IH’, the proof is complete.

ACKNOWLEDGMENTS

The authors wish to express their thanks to M. Schacher and A. Wadsworth, out an error in the first version of this paper.

who pointed

REFERENCES

1. A. A. ALBERT, “Structure of Algebras,” Amer. Math. Sot. Colloq. Puhl. Vol. 24, Amer. Math. Sot., Providence, RI., 1961. 2. R. S. PIERCE, “Associative Algebras,” Graduate Texts in Math. Vol. 88, Springer-Verlag, New York/Heidelberg/Berlin, 1982. Math. Surveys Vol. 4, Amer. Math. Sot., 3. 0. F. G. SCHILLING, “The Theory of Valuations,” Providence, R.I., 1950.