Multidimensional local skew-fields (792481), страница 12

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Since j(1) − i = 0 mod p,r−3j(1)−i δj(n) (t) = 0 and j(1)−i δj(n)+i (t) = 0. By lemma 0.44, m δs |F (t) = c1 δ + . . . + cr−3 δfor s < (r − 2)j(n). So, j(1)−i δ(r−1)j(n)+i |F (t) = c1 δ + . . . + cr−2 δ r−2 .Hence by lemma 0.44, j(n)−j(1)+i δ̃rj(n)+i |F (t) = c1 δ + .

. . + cr δ r withcr = 1r j(n)−j(1)+i δ̃j(n)+i (t)(i δj(n) (t))r−1 = 1r i δ̃j(n)+i (t)(i δj(n) (t))r−1 = 0.The same arguments show that j(n)−j(1)+i δ̃ζ |F (t) = c1 δ + . . . + cr−1 δ r−1 for ζ <rj(n) + i.Therefore we havej(n)−j(1)+i δ(p−1)j(n)+i+ j(n)−j(1)+i δ̃(p−1)j(n)+i +(p−1)j(n)+i−1j(n)−j(1)+i δw· j(n)−j(1)+i δ̃(p−1)j(n)+i−w = 0,w=1where j(n)−j(1)+i δw · j(n)−j(1)+i δ̃(p−1)j(n)+i−w |F (t) = c1 δ + . .

. + cp−1 δ p−1 only if w = rj(n).1p−2In this case cp−1 = p−1−r.i δ̃j(n)+i (t)(i δj(n) (t))Since j(n)−j(1)+i δ̃j(n) (t) ∈ Z(D), we have j(n)−j(1)+i δi (j(n)−j(1)+i δ̃j(n) )p−1 = 0. So usinginduction, we get j(n)−j(1)+i δ(p−1)j(n)+i |F (t) = c1 δ + . . . + cp−1 δ p−1 withcp−1 = −(1 + . . . +1)i δ̃j(n)+i (t)(i δj(n) (t))p−2 −p−153(1 + . . . +1)i δ̃j(n)+i (t)(i δj(n) (t))p−2 − . . . − i δ̃j(n)+i (t)(i δj(n) (t))p−2 =p−2−i δ̃j(n)+i (t)(i δj(n) (t))p−2 = 0.Therefore, j(1) δ̃j(n+1)+i (tp ) = −j(1) δ̃j(n)+i (t)cp−1 ∈ Z(D̄).

This proves ii) and iii).The lemma is proved.2The theorem is proved.2Lemma 0.50 Let D be a splittable division algebra. Let n = |Gal(Z(D̄)/F̄ )|.There exists a parameter z such thatzaz −1 = α(a) + δn (a)z n + δ2n (a)z 2n + . . . ,a ∈ D̄So, δj = 0 if n |j.One can repeat the proof of lemma 0.29 to prove the lemma.Proposition 0.51 Let D be a good splittable division algebra.

Suppose Z(D̄)/F̄ is nota separable extension.Then p does not divide |Gal(Z(D̄)/F̄ )|.Proof. Suppose p divides |Gal(Z(D̄)/F̄ )|. By lemma 0.50 there exists a parameterz such thatzaz −1 = α(a) + δn (a)z n + δ2n (a)z 2n + . . . , a ∈ D̄,where n = |Gal(Z(D̄)/F̄ )|.Since Z(D̄)/F̄ is a compositum of a purely inseparable extension and Abelian Galoisextension, there exists an element u ∈ Z(D̄) such that α(u) = u, i.e.

u is a purelyinseparable element; so by theorem 0.43 up ∈ Z(D).In this case lemma 0.44 holds for l = 0 and we can repeat the arguments in theproof of lemma 0.45 to show that δpi (up ) = 0, which is a contradiction.2Proposition 0.52 Let D be a good splittable division algebra. Then we have D ∼=D1 ⊗F D2 , where D1 , D2 are division algebras such that D1 is an inertially split algebra,Z(D̄2 )/F̄ is a purely inseparable extension and D2 is a good splittable algebra (D1 orD2 may be trivial).So, D ∼ A ⊗F B ⊗F D2 , where A is a cyclic division algebra and B is an unramifieddivision algebra.54rProof.

By [25], p.261, D ∼= D1 ⊗F . . . ⊗F Dk , where [D : F ] = pr11 . . . pkk and[Di : F ] = pri i . Let p2 = p. By proposition 0.51, Z(D̄2 )/F̄ is a purely inseparableextension. Since Di are defectless over F , D1 , D3 , . . . Dk are inertially split. Therefore,by theorem 0.37 the algebra D1 ⊗ D3 ⊗ . . . ⊗ Dk is good splittable.Let L be an inertial lift of a Galois part of the extension Z(D̄)/Z(D). Consider thecentraliser D = CD (L). It’s clear that D ∼= D2 ⊗L B, where B is a division algebrasimilar to the algebra D1 ⊗ D3 ⊗ . . . ⊗ Dk ⊗ L. The algebra B is inertial, becauseZ(B̄)/Z(B) is trivial and B is inertially split. Since D̄ ∼= D̄2 ⊗ B̄ and D̄ → D is agood embedding, D̄ contain a subalgebra B̄ ⊂ B̄ ⊗L L ∼= B ⊂ D .

Now the centraliser∼CD (B) = D2 ⊗F L and it is good splittable, so D2 is good splittable.Decomposition theorems [9], Thm. 5.6-5.15 complete the proof.2Proposition 0.53 Let D2 be a good splittable division algebra such that Z(D̄2 )/Z(D2 )is a purely inseparable extension. Then D2 ∼= D3 ⊗Z(D2 ) D4 , where D3 is an unramifieddivision algebra and D4 is a good splittable division algebra such that D̄4 is a field,D̄4 /Z(D2 ) is a purely inseparable extension, [D̄4 : Z(D2 )] = [ΓD4 : ΓZ(D2 ) ].Proof.

For a good embedding there exists a subfield Z(D2 ) ⊂ K ⊂ Z(D̄2 ) suchthat the extension Z(D̄2 )/K has degree p. Then by theorem 0.37 and 0.43 there exists¯ = K, Γ = Γa lift K̃ of K in D2 , i.e. K̃Z(D2 ) , K ⊂ K̃.K̃Consider the centraliser C1 = CD2 (K̃). We have C̄1 = D̄2 , Z(C̄1 )/Z(C1 ) is a purelyinseparable extension of degree p, say Z(C̄1 ) = Z(C1 )(u). Using similar arguments asin the proof of theorem 0.43 one can show that there exists a parameter z such thatC1 ∼= C̄1 ((z)) as a vector space with the relationzaz −1 = a + δi (a)z i + c2i δi2 (a)z 2i + .

. . ,cki ∈ Fpand zuz −1 = u + xz i , where x ∈ Z(C1 ). Therefore, δip is a derivation trivial on thecentre Z(C̄1 ), hence by Scolem-Noether theorem it is an inner derivation.We claim that z p ∈ Z(C1 ). To prove it, consider a subalgebra W = C̄1 ((z i )) ⊂ C1(note that Z(W ) = Z(C1 )). It exists because of the type of the relation in C1 .We havez −i az i = a − iδi (a)z i , a ∈ C̄1in W .

Therefore,andz −pi az pi = a − ip δip (a)z pi ,a ∈ C̄12z pi az −pi = a + δ1 (a)z pi + δ1 (a)z 2pi + . . . ,where δ1 = ip δip . So,11 2z p az −p = a + δ1 (a)z pi + c2 2 δ1 (a)z 2pi + . . . ,ii55where ck are given by (1) in theorem 0.43. So, z p ∈ Z(C1 ) iff δip = 0. Suppose δip = 0.Consider an element Y ∈ Z(C1 ), w(Y ) > 0. LetY = a1 z p + . . .First note thatY = a1 z p + a2 z 2p + a3 z 3p + . . .

,ai ∈ C̄1Indeed, Y must satisfy [Y, u] = 0. Since u ∈ Z(C̄1 ), we then have [z ik , u] = 0 for everyk, where∞ak z ikY =k=1Therefore, p|ik .Then, Y must satisfy Y a = aY for any a ∈ C̄1 . Therefore, a1 , . . . ai ∈ Z(C̄1 ) andwe must haveaai+1 − ai+1 a = a1 δ1 (a)/iand2aa2i+1 − a2i+1 a = ai δ1 (a) + a1 c2 δ1 (a).Since ∆(a) = aa2i+1 − a2i+1 a is an inner derivation, we get δ1 2 = δ, where δ is aderivation, which is a contradiction. Therefore, δ1 2 = δ = 0 and δ1 = 0, and z p ∈ Z(C1 ).Consider the centraliser C2 = CC1 (K̃(z)).

It’s clear that [C̄2 : Z(C̄2 )] = [C̄2 :Z(C2 )] = indC̄1 and there exists a subalgebra C̄2 ⊂ C2 , C̄2 ⊂ D̄2 . Consider nowthe centraliser C3 = CD2 (Z(D2 )(z)). We have C̄2 ⊂ C3 , C̄3 ∼= C̄2 , because [Z(C̄3 ) :Z(D2 )] = [ΓC3 : ΓZ(D2 ) ] = [Z(C̄2 ) : Z(D2 )] = [K : Z(D2 )].

By induction on dimensionof D̄2 we get the existence of a subalgebra C̄4 ⊂ D̄2 such that [C̄4 : Z(C̄4 )] = [C̄2 :Z(C̄2 )], Z(C̄4 ) = Z(D2 ). Therefore there exists an unramified subalgebra C4 ⊂ D2 suchthat [C4 : Z(D2 )] = [C4 : Z(D2 )] = [C4 : Z(C4 )] = [D̄2 : Z(D̄2 )]. By Double CentraliserTheorem, D2 ∼= C4 ⊗Z(D2 ) D4 , where D4 is a division algebra with D̄4 = Z(D̄2 ). SinceD̄2 → D2 is a good embedding, [D̄4 : Z(D2 )] must be equal to [ΓD4 : ΓZ(D2 ) ]. It is easyto see that D4 is also a good splittable division algebra.The proposition is proved.2Proposition 0.54 Let D2 be a good splittable division algebra such that D̄2 is a field,D̄2 /Z(D2 ) is a purely inseparable extension and dD2 (uk ) ≤ 2i(uk ) or dD2 (uk ) = ∞ forall generators uk of the extension D̄2 /Z(D2 ).Then D2 ∼= A1 ⊗Z(D2 ) . .

. ⊗Z(D2 ) Am , where Ai are cyclic division algebras of degreep, [Āi : Z(D2 )] = [ΓAi : ΓZ(D2 ) ].56Proof. The proof immediately follows from theorem 0.43 and [26], Thm.3, §2.8.(see [1] for the proof of this theorem).2So, we get the following decomposition theorem.Theorem 0.55 Let D be a finite dimensional good splittable central division algebraover a field F with a discrete complete rank 1 valuation, char(F ) = p > 2, such thatdD2 (uk ) ≤ 2i(uk ) or dD2 (uk ) = ∞ for all generators uk of the extension Z(D̄)/Z(D).Then D ∼= D1 ⊗F D2 ⊗F A1 ⊗F . . . ⊗F Am , where Ai are cyclic division algebrasof degree p, [Āi : Z(D)] = [ΓAi : ΓZ(D) ], D1 is an inertially split division algebra,(ind(D2 ), p) = 1, D2 is an unramified division algebra (D1 , D2 , Ai may be trivial).Recall that a field F is called a Ci -field if any homogeneous form f (x1 , .

. . , xn ) ofdegree d in n > di variables with coefficients in F has a non-trivial zero.Corollary 8 The following conjecture: the exponent of A is equal to its index for anydivision algebra A (here we don’t demand that A is splittable) over a C2 -field F (see forexample [26], 3.4.5.) has the positive answer for F = F1 ((t)), where F1 is a C1 -field.Proof. 1) Let’s prove that A is splittable.

Since F̄ is a C1 -field, Ā is a field. Wecan assume Ā/F̄ is a purely inseparable extension. We claim that Ā = F̄ (u) for someu ∈ Ā, so by classical Cohen’s theorem, A is splittable.Indeed, suppose Ā = F̄ (u1 , . . . , ur ). Consider the field K = F̄ (up1 , . . . , upr ). By Tsen’s+xpp+1 u2 has a nontheorem, K and Ā are C1 -fields. So, the form xp1 +xp2 u1 +.

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