Multidimensional local skew-fields, страница 3

. . ,where y0 ∈ K̄. Then we havez n az −n = zy0 z −1 + zy1 z −1 z + . . . = [y0α + y0δ1 z + . . .] + [y1α + y1δ1 z + . . .]z + . . .Putnz az−n=∞k=09wk z k .Then we havewd =dδj (yd−j ) + α(yd )j=1Since the sum of coefficients of yj is equal to sjn−1 , we getsdn=dsjn−1j=0Now let us show that ski+1−k = Cik if k < i + 1. The proof is by induction on i. Fori = 0 we have s01 = 1 = Ci0 . For arbitrary i we haveski+1−k =kk−10sli−k = Cik + Ci−1+ . . . + Ci−k=l=00(. . . (((Ci−k123+ Ci−k+1) + Ci−k+2) + Ci−k+3) + . . .

+ Cik ) =123k(. . . (((Ci−k+2) + Ci−k+2) + Ci−k+3) + . . . + Cik ) = Ci+1This completes the proof.2Corollary 1 Suppose α = Id. Then the following formula holdsδi (ab) = δi (a)b +iδi−k (a)lCi−k+1δj1 . . . δjl (b)(j1 ,...,jl )k=1is taken over all the vectors (j1 , . . .

, jl ) such thatwhere δ0 = α and the second sum jm = k.0 < l ≤ min{i − k + 1, k}, jm ≥ 1,In the sequel we will need the following definition.Definition 0.8 Let (α, β) be endomorphisms of a skew field L. A map δ : L → L ,where L ⊂ L is a subalgebra, is called a (α, β)-derivation if it is linear and satisfy thefollowing identityδ(ab) = δ(a)bα + aβ δ(b)where a, b ∈ L.We will say that (α, 1)-derivation is an α-derivation.For example δ1 is an (α2 , α)-derivation.If α = Id, then δ1 is an usual derivation; δ2 = δ12 + δ, where δ is a derivation.10Corollary 2 If δ1 = .

. . = δk−1 = 0, then δk is an (αk+1 , α)-derivation.The following corollary will be used in §3 of this chapter.Corollary 3 Let K̄ be a field, K̄ = k((u)), k ⊂ Z(K) and the maps δi , i ≥ 1 becontinuous if chark = 0. Thenδi (∞j=Njxj u ) =∞xj δi (uj ),xj ∈ kj=NSo, for every i the map δi is completely defined by elements δi (u) and δj (u) for j < i.Proof.

If chark = p = 0 and α = id the maps δi , i ≥ 1 are continuous, sinceiδi (ap ) = 0 for any a ∈ K̄. Since a topology on a 1-dimensional local field is uniquelydefined by its local structure, the continuity does not depend on the choice of local parameters (for more information about a relation between a topology and a parametrisation see [35]). If α = id one can use lemma 1.29 to reduce this case to the previousone.Let us show that α is a continuous map. In our case it suffice to show that αpreserves the valuation.

Our proof will not depend on a characteristic.It suffice to show that ν̄(α(u )) = 1 for any u , ν̄(u ) = 1. Consider the automorphismα :α (a) := z −1 azwhere a ∈ K̄ (we use the notation from proposition 1.7). It’s clear that α = α−1 .Let u be an arbitrary parameter. Put κ = ν̄(α(u )). We claim that |κ| ≤ 1 or|κ| = pq , q ∈ N. Assume the converse. Then κ = mpq , (m, p) = 1, |m| = 1 and thereexist c ∈ k, a ∈ K̄ such that α(u ) = cam . Therefore, we getu = α−1 (α(u )) = c(α−1 (a))m ,i.e.ν̄(u ) = 1 = ν̄(c(α−1 (a))m ) = mν̄(α−1 (a)),a contradiction.Let us show that κ ≥ 0. Assume the converse.

Consider the element u + u2 (u + u3if chark = 2). Then ν̄(α(u + u2 )) = 2κ < −1. If chark = 2 we get a contradictionwith the assertion |ν̄(α(u ))| = pq or |ν̄(α(u ))| ≤ 1 for any parameter u . If chark = 2one can apply the same arguments to the element u + u3 .Similarly, for κ := ν̄(α−1 (u )) the property 0 ≤ κ ≤ 1 or κ = pl holds.Let us show that κ = pq .

Assume the converse. Consider the following two cases:q1) Suppose κ ≤ 1. There exist r ∈ k, a1 ∈ k((u)) such that α(u ) = c2 u2 ap1 −2 .Therefore,1 = 2ν̄(α−1 (u )) + (pq − 2)ν̄(α−1 (a1 )),11i.e. (pq − 2)|1. It is possible only if p = 3, q = 1. In this case one can use the same−1arguments with α(u ) = c3 u5 a−21 . Then we get ν̄(α (a1 )) = 2, a contradiction (since0 ≤ κ ≤ 1 or κ = pl ).q2) Suppose κ = pl . Let α(u ) = cu ap −1 for some c ∈ k, a ∈ k((u)), ν̄(a) = 1. Thenwe haveν̄(u ) = 1 = ν̄(α−1 (u )) + (pq − 1)ν̄.But this contradicts with ν̄(α−1 (a)) ≥ 0.So, κ = 0 or κ = 1, i.e. for any parameter u we have ν̄(α(u )) = 0 or ν̄(α(u )) = 1.Suppose κ = 0.

Consider the element x = u + c1 u2 + c2 (u3 + c1 u4 )), where c1 = −w0−1if α(u ) = w0 + . . . and c2 is an element such that ν̄(α(x)) > 1 (it always exists sinceν̄(α(u + c1 u2 )) > 0). But this contradicts with ν̄(x) = 1. Therefore, κ = 1 and α is acontinuous map.jTo complete the proof it suffice to show that the series ∞j=N xj δi (u ) converges,because the topology on k((u)) is complete and separate. The proof is by inductionon i.

For i = 0 we have ν̄(α(uj )) = j and the series converges. For i = 1 we haveν̄(δ1 (uj )) = (j − 1)ν̄(δ1 (u)) and again the series converges.j−1)yi−k ,At last, by proposition 0.7 for j > 1 we have δi (uj ) = δi (uj−1 )y0 + i−1k=0 δk (uwhere ν̄(yk ) does not depend on j.By induction we have min{ν̄(δ0 (uj−1 )yi ), . . . , ν̄(δi−1 (uj−1 )y1 )} >min{ν̄(δ0 (uj−2 )yi ), . . .

, ν̄(δi−1 (uj−2 )y1 )} and ν̄(y0 ) = 1. So,min{ν̄(δi (uj−1 )y0 ), ν̄(δ0 (uj−1 )yi ), . . . , ν̄(δi−1 (uj−1 )y1 )} >min{ν̄(δi (uj−2 )y0 ), ν̄(δ0 (uj−2 )yi ), . . . , ν̄(δi−1 (uj−2 )y1 )}.Therefore, the series converges.20.2Splittable skew fields.In this section we will assume that K̄ is a field.For such a skew field one can define a notion of a canonical automorphism α.By definition there exist the following exact sequences:1 → O∗ → K ∗ −→ Z → 1νwhere O is a valuation ring;1 → 1 + ℘ → O∗ → K̄ ∗ → 1where ℘ is a maximal ideal.Consider the mapφ : K ∗ → Int(K),φ(x) = ad(x),12ad(x)(y) = x−1 yxwhere Int(K) is the group of inner automorphisms of the skew field K. Since innerautomorphisms preserve the valuation, this group acts on the ring O.

Moreover, itpreserve the ideal ℘. Therefore, there exists a map φ : K ∗ → Aut(O/℘) = Aut(K̄). Letus show that the action of φ(O∗ ) is trivial on K̄. To show it we use the second exactsequence. Since (1 + ℘)−1 x(1 + ℘) = x mod ℘ for any x ∈ O, the action of φ(1 + ℘)on K̄ is trivial. Therefore, there exists an action of K̄ on K̄. Namely, an element ā ∈ K̄acts on x̄ ∈ K̄ as a−1 xa mod ℘, where a, x are any lifts of ā, x̄ in O. Since K̄ is acommutative field, this action is trivial.Definition 0.9 An automorphism α of the field K̄ defined by the formulaα = φ(z)where z ∈ K ∗ and ν(z) = 1, is called a canonical automorphism.It does not depend on the choice of z.We want to classify all splittable two-dimensional local skew fields which have isomorphic last residue fields up to isomorphism. Let K and K be two splittable skewfields, K ∼= K̄ ((z )).

If K ∼= K , then one can represent an isomor= K̄((z)), K ∼phism ϕ : K → K as a compositum of an isomorphism φ : K −→ K such thatφ(u) = u , φ(z) = z , and of an automorphism ψ of the skew field K. Since every isomorphism in our paper preserve the local structure, every automorphism of a splittabletwo-dimensional local skew field is defined by change of parameters(z)u → u = c0 + c1 z + c2 z 2 + . . . ,z → z = a0 z + a1 z 2 + . .

. ,ν̄(c0 ) = 1a0 = 0where ai , ci ∈ K̄.It is easy to see that every change of parameters looks like above and can bedecomposed into a sequence of changes u → u , z → z; u → u , z → z = a0 z+a1 z 2 +. . .(or in a backward order). Also u → u can be decomposed into a sequence of changesu → u1 = c0 , u1 → u2 = u1 + c1 z, . . .

, ui → ui+1 = ui + ci z i , . . . and z → z can bedecomposed into a sequence of changes z → z1 = a0 z, z1 → z2 = z1 + a1 z 2 , . . . , zi →= zi + ai z i+1 , . . ..zi+1Remark. We must note that any change of parameters (z) defines a map f : K →K which is not always an automorphism. Indeed, assume the converse. Consider a mapwhich is given by f (z) = z , f (u) = u, where z is another parameter.

Then we musthavef (zu) = f (z)f (u) = z u = uα z + uδ1 z 2 + . . .f (zu) = f (uα z + uδ1 z 2 + . . .) = uα z + uδ1 z 2 + . . .13Hence, α = α ; δ1 = δ1 and so on, i.e. δi = δi ∀i.Consider the skew field ”((u))((z)) with the relation zu = (u + u2 )z and consider achange of parameters z → z = z + z 2 . Thenz u = (z + z 2 )u = (u + u2 )z + z(u + u2 )z = (u + u2 )z + [(u + u2 )z + (u + u2 )2 z]z =(u + u2 )z + [u + 2u2 + 2u3 + u4 − u − u2 ]z 2 = (u + u2 )z + [u2 + 2u3 + u4 ]z 2 + . . .So, δ1 = δ1 , a contradiction.Proposition 0.10 Let K be a splittable two-dimensional local skew field.

Suppose thecanonical automorphism α has infinite order.Then there exists a parameter z such that z a = aα z for any a ∈ K̄.Proof. We will show that there exists a sequence of parameters {zk } such that theequality zk azk−1 = aα mod ℘k holds and the sequence {zk } converges in K.We need some additional lemmas.Lemma 0.11 Suppose the following relation holds:=zaz −1 = aα + aδj z j + aδj+1 z j+1 + . .