Multidimensional local skew-fields (792481), страница 13

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. .+xpp up−11trivial zero in Ā. But xpi ∈ K and elements 1, u1 , . . . , up−11 , u2 are linearly independentover K, a contradiction.2) Assume the corollary is known in the prime exponent case. We deduce the corollary by ascending induction on e = expA. If e is not a prime number, then write e = lm.By assumption A⊗m can be split by a field extension F ⊂ F of degree l.

This impliesthat AF has exponent dividing m. Note that F is also a Laurent series field. By theinduction hypothesis applied to the pair (F , AF ), there exists a field extension F ⊂ Lof degree dividing m splitting AF . Therefore A is split by the extension F ⊂ L ofdegree dividing lm and we conclude the corollary.3) So, let expA = l be a prime number. By the basic properties of the exponentand the index (see, e.g. [26]) we have then indA = lk for some natural k.Suppose (l, p = charF ) = 1.It is known that the conjecture is true for all division algebras of index indA = 2a 3b(see, e.g. [26]), so we can assume charF = 2, 3. Then we can assume F contains thegroup µl of l-roots of unity, because [F (µl ) : F ] < l and we can reduce the problemto the algebra A ⊗F F (µl ).

Then by the Merkuriev-Suslin theorem A is similar to thetensor product of symbol-algebras of index l.57Every symbol-algebra of index l over F is good splittable and cyclic and its residuefield is a cyclic Kummer extension of F̄ . To conclude the statement of the corollaryit is sufficient to prove that every two symbol algebras A1 , A2 contain F -isomorphicmaximal subfields.Since Ai , i = 1, 2 is cyclic, it contains an element zi , zil ∈ F . Since Ai is a goodsplittable algebra and by lemma 0.50 (which is true also if charF = 0), we can assumev(zil ) = 1 (v is the valuation on F ).To prove it we show that A1 contains any l-root of elements u in F with v(u) = 0.Since for any element 1 + b, v(b) > 0 there exists an element (1 + b)1/l ∈ F , it issufficient to prove that A1 contains any l-root of elements ct, c ∈ F̄ , where we fix someembedding i : F̄ → F .Indeed, since A1 is a good splittable algebra and by lemma 0.50 (which is true alsoif charF = 0) we can assume there exists an element z such that v(z l ) = 1, z l = ct,c ∈ F̄ , ad(z) acts on Ā1 , where Ā1 is embedded in A1 by a good embedding withrespect to i.

Note that for any element b ∈ Ā1 we have (bz)l = NA¯1 /F̄ (b)z l . But thenorm map NA¯1 /F̄ is surjective because F̄ is a C1 -field (see, e.g. [26], 3.4.2), so for anyc there exists b such that (bz)l = ct.4) Suppose now expA = p. Then indA = pk .1/p1/pBy Albert’s theorem (in [1]) there exists a field F = F (u1 , . .

. , uk ) which splits A.Using the same arguments as in 1) one can show that every such a field has maximum1/p1/ptwo generators, say F = F (u1 , u2 ). Therefore, indA ≤ p2 . If indA = p, there isnothing to prove, so we assume indA = p2 and F is a maximal subfield in A.5) Suppose F1 is a perfect field.By Albert’s theorem, A ∼= A1 ⊗F A2 , where A1 , A2 are cyclic algebras of degreep, A1 = (L1 /F, σ1 , u1 ), A2 = (L2 /F, σ2 , u2 ). Since F1 is perfect, Ā1 /F̄ , Ā2 /F̄ are Galois extensions.

So, A1 , A2 are good splittable. Let us show that A1 , A2 have commonsplitting field of degree p over F . This leads to a contradiction.By lemma 0.50 there exist parameters z1 ∈ A1 , z2 ∈ A2 such that they act on Ā1 ,Ā2 as Galois automorphisms. Note that then z1p , z2p ∈ F . Let us show that F (z1 ) splitsA2 .Consider the centralizer D = CA (F (z1 )). Consider the element t1 = z2 z1−1 . We havept1 ∈ F , w(t1 ) = 0, where w denote the unique extension of the valuation v on F . Since¯ is a Galois extension, there exists an element b1 ∈ F such that w(t1 − b1 ) > 0.D̄/Z(D)Since (t1 − b1 )p ∈ F , there exists natural k1 such that w((t1 − b1 )z1−k1 ) = 0. Denotet2 = (t1 − b1 )z1−k1 .

We have again tp2 ∈ F . Repeating this arguments and using thecompleteness of D ⊂ A we getz2 = t1 z1 = (t2 z1k1 + b1 )z1 = . . . = b1 z1 + b2 z1k1 +1 + . . .,so, z2 ∈ F (z1 ) = Z(D).6) Suppose F1 is not perfect.58Since F is generated by two elements over F , it contains all p-roots of F . Then,/ F (u1/p ), where z 1/p , u1/p ∈ F , alsoevery two elements u, z ∈ F such that z 1/p ∈generate F over F .

This follows from the same arguments as in 1), 4).Now take u ∈ F1 \F1p , z = u + t. It’s clear that p-roots of these elements generateF over F . Moreover, the fields F (u1/p ), F (z 1/p ) are ”unramified” over F , i.e. [F (u1/p ) :F̄ ] = p = [F (u1/p ) : F ], [F (z 1/p ) : F̄ ] = p. Denote u1 = u1/p , u2 = z 1/p in F . Thenby Albert’s theorem, A ∼= A1 ⊗F A2 , where A1 , A2 are cyclic algebras of degree p,A1 = (L1 /F, σ1 , u), A2 = (L2 /F, σ2 , z).Since the fields F (u1/p ) ⊂ A1 , F (z 1/p ) ⊂ A2 are ”unramified” and purely inseparableof degree p over F , the algebras A1 , A2 are good splittable.

Moreover, there existembeddings Ā1 → A1 , Ā2 → A2 such that u1 ∈ Ā1 , u2 ∈ Ā2 . Then by theorem0.43 there exist parameters z1 ∈ A1 , z2 ∈ A2 such that z1p , z2p ∈ F andz2 u2 z2−1 = u2 + cz2i ,where c ∈ F , v(c) = 0. So, for the element u2 = c−1 u2 we havez2 u2 z2−1 = u2 + z2i ,/ F , u p2 ∈ F .and u2 ∈Since F̄ is a C1 -field, we have Ā1 = Ā2 and therefore there exist an element b ∈F (u1 ) ⊂ A1 ⊂ A such that w(u2 − b) > 0, where w is the unique extension of v onA. Since b commutes with u2 , we have (u2 − b)p ∈ F . Therefore w(u2 − b) ∈ 1/pZ (weassume the value groups of w and v lie in a common divisible hull Γv ⊗Z ”).

Hence−pw(u2 −b)−pw(u2 −b)) = 0. Put u3 = (u2 − b)z2.w((u2 − b)z2Note thatz2 (u2 − b)z2−1 = (u2 − b) + z2i ,z2 u3 z2−1 = u3 + z2i1 ,i1 < iSo, the elements (u2 − b), z2 generate a division algebra C of degree p over F andu3 ∈ C. Then, up3 commutes with z2 if i1 > 0. Therefore, in this case up3 ∈ F and C is agood splittable division algebra. Note that u1 ∈ CA (C), so A ∼= A1 ⊗F C with u1 ∈ A1 .Using the same arguments we get that there exists an element u4 with w(u4 ) = 0 andz2 u4 z2−1 = u4 + z2i2 ,i2 ≤ 0So, i2 must be equal to 0 and therefore u4 , z2 generate a division algebra C of degreep over F such that C̄ /F̄ is a Galois extension and u1 ∈ CA (C ). So, A ∼= D ⊗F C withu1 ∈ D.Therefore, A contains the maximal subfield F (u1 )F (u4 ), which is a compositum ofa purely inseparable and Galois extension.

Moreover, this field is ”unramified” overF , so it is good splittable field and A is a good splittable algebra with p dividing59|Gal(Ā/F̄ )|. But this is a contradiction with proposition 0.51.2Corollary 9 Let A be a central division p-algebra over a C2 -field F = F1 ((t)), F1 isa C1 -field. Then A contains a maximal purely inseparable over F subfield, i.e. A is acyclic algebra.Moreover, A is a good splittable algebra.Proof. The proof of the first statement is by induction on degree of A.

If indA = p,then by Tignol’s theorem in [32] A is cyclic, so it contains such a maximal subfield.If indA = pk , k > 1, then by assumption a division algebra similar to A⊗p has theexponent and index pk−1 and so can be split by a field extension F ⊂ F of degree pk−1 .By corollary 8, the exponent and the index of AF is p, so there exists an extensionL/F of degree pk such that L splits A.To prove the second statement note that it is sufficient to prove it only for algebrasA with Ā/F̄ — purely inseparable. Now to prove the assertion we use lemma 0.24.Note that, using a similar induction, it is sufficient to prove the statement for algebrasA of degree p.Let z be a purely inseparable element in A, indA = p.

If F (z) is an ”unramified”over F , there is nothing to prove. So, we may assume F (z) is totally ramified over Fand z is a parameter of A.Choose an element a ∈ A such that ā generates Ā over F̄ . Suppose a ∈ Ā for someembedding Ā → A. Supposezaz −1 = a + δi (a)z i + δi+1 (a)z i+1 + . . .Then we havepz az−p=a+∞k=pi (i1 ,...ip )δi1 . . . δip (a)z k = a,(2)where ij = k and the second sum is taken over all such nonrepeating sets (i1 , . . .

, ip ).Therefore, δip (a) must be equal to zero. Since δi is a derivation, it is trivial on F̄ (ap ).Every element in F̄ (a) can be written as a polynomial c1 + c2 a + . . . cp ap−1 , whereci ∈ F̄ (ap ). Therefore, we can write δi = δi (a)∂/∂(a).

So, δip (a) = δi (a)∂/∂(a)(δip−1 (a)).Hence ∂/∂(a)(δip−1 (a)) = 0 and δip−1 (a) ∈ F̄ (ap ).If δip−1 (a) = 0, then let j be the maximal natural such that δij (a) = 0, δij (a) ∈F̄ (ap ). Now put a1 = δij−1 (a)(δij (a))−1 . Note that a1 generates Ā over F̄ Since δij (a) =∂/∂(a)(δij−1 (a))δi (a), we have δi (a1 ) = 1.So we can put a := a1 and assume δi = ∂/∂(a). Now the proof is by induction onk in the formula 2. For k = ip + 1 we haveδi1 . . .

δip (a) = 0(i1 ,...,ip )60By lemma 0.44, δi+1 = δil + cδi , so δi1 . . . δip (a) = 0 if ip = i. Therefore, we haveδi . . . δi δi+1 (a) = 0Therefore, there exists an element b ∈ Ā such that δi (b) = δi+1 (a) and by lemma 0.24there exists an element a2 = a + b2 z such thatza2 z −1 = a2 + z i + δi+2z i+2 + . . .Note that here the coeffitients on the right hand side belong to another embeddingof Ā given by element a2 . Since Ā is a C1 -field, Ā is generated by a¯2 over F̄ . So, thep-basis of Ā consists of 1 element. So, by classical Cohen’s theorem, any lifting of thiselement gives an embedding of Ā. Now using induction and completeness of A we getthat there exists an element a3 such thatza3 z −1 = a3 + z iand a¯3 generates Ā over F̄ .

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