Multidimensional local skew-fields, страница 5

If à commutes with A, thenf (A + Ã) = f (A) + f (A)à + P Ã2where P ∈ F [A, Ã]. We use Teilor’s formula here. Put à = −(f (A))−1 f (A). It’s clearthat à ∈ I and à commutes with A. Moreover, à commutes with every element in F ./ I, where X ∈ F [A, Ã].Thus, f (A + Ã) = P Ã2 ∈ I 2 and f (A + Ã) = f (A) + X à ∈−1Similarly we can find the element Ã2 = −(f (A + Ã)) f (A + Ã) ∈ I 2 , which commuteswith A, à and with every element in F and such thatf (A + à + Ã2 ) ∈ I 4Continuing this line of reason we can find the element  = A + à + Ã2 + . . ..

The sumis converge because of completeness of O.2Remark. If αn = Id, then the theorem is not true (see an example in §3).1the idea of the proof of this lemma was offered by N.I.Dubrovin19Corollary 5 Proposition 0.10 is true for any two-dimensional local skew field withαn = id for all n ∈ N.Theorem 0.16 Let K, K be two-dimensional local skew fields such that αn = Id,αn = Id for all n ∈ N, K̄, K̄ are commutative fields.

Then(i) K is isomorphic to a two-dimensional local skew field K̄((z)) where za = aα z,a ∈ K̄.(ii) K is isomorphic to K iff k ∼= k and there is an isomorphism f : K̄ → K̄ such−1 that α = f α f .Proof. The proof follows from corollary 5 and from the known classification of onedimensional local fields (see for example [30]).2Definition 0.17 Let K̄ be a one-dimensional local field with residue field k, charK̄ =−1mod ℘ ∈ k.chark, let α be an automorphism of the field K̄. Put a1 = α(u)uDefine iα ∈ N ∞ as follows:iα = 1 if a1 is not a root of unity in k elseiα = ν̄((αn − Id)(u)), where n ≥ 1: an1 = 1, am1 = 1 ∀m < n.Lemma 0.18 Let k be a field of characteristic 0.

Any k-automorphism α of a fieldk((u)) with α(u) = ξu + a2 u2 + . . ., where ξ n = 1, n ≥ 1, ξ m = 1 if m < n, is conjugatewith an automorphism β: β(u) = ξu + xuiα + yu2iα −1 , where x ∈ k ∗ , y ∈ k, x and ydepend on α.Moreover, iα = iβ .Proof. First we prove that α = f β f −1 whereβ (u) = ξu + xuin+1 + yu2in+1for some natural i. Then we prove that iα = iβ .Consider a set {αi : i ∈ N} where αi = fi αi−1 fi−1 , fi (u) = u + xi ui for some xi ∈ k,α1 = α. Writeαi (u) = ξu + a2,i u2 + a3,i u3 + . .

.One can check that a2,2 = x2 (ξ 2 − ξ) + a2,1 and hence there exists an element x2 ∈ ksuch that a2,2 = 0. Since aj,j+1 = aj,i , we have a2,j = 0 for all j ≥ 2. Further, a3,3 =x3 (ξ 3 − ξ) + a3,2 and hence there exists an element x3 ∈ k such that a3,3 = 0. Thena3,j = 0 for all j ≥ 3. Thus, any element ak,k can be made equal to zero if n |(k − 1)and so α = f α̃f −1 whereα̃(u) = ξu + ãin+1 uin+1 + ãin+n+1 uin+n+1 + .

. .20for some i, ãj ∈ k. Notice that ãin+1 does not depend on xi . Put x = x(α) = ãin+1 .Now we replace α by α̃. One can check that if n|(k − 1) thenaj,k = aj,k−1for2 ≤ j < k + inandak+in,k = xk x(k − in − 1) + ak+in + some polynomial which does not depend on xkFrom this fact it immediately follows that a2in+1,in+1 does not depend on xi and for allk = in + 1 ak+in,k can be made equal to zero. Then y = y(α) = a2in+1,in+1 .Now we prove that iα = iβ . Using the formulaβ (u) = u + nx(α)ξ −1 uin+1 + .

. .nwe get iβ = in + 1. Since f −1 αf = β , f −1 (αn − Id)f = β n − Id. Therefore, ν̄(f −1 (αn −Id)f (u)) = ν̄((β n − Id)(u)) = iβ . Suppose f (u) = u = f1 u + f2 u2 + . . ., f1 = 0. Letus show that ν̄f −1 (αn − Id)(u ) = iα . It suffice to check that ν̄(αn − Id)(u ) = iα .

Wehave(αn − Id)(u ) = [f1 (u + a¯iα uiα + . . .) + f2 (u + a¯iα uiα + . . .)2 + . . .] − [f1 u + f2 u2 + . . .] =[(f1 u + f1 a¯iα uiα + . u>iα ) + (f2 u2 + . u>iα ) + (f3 u3 + . u>iα ) + . . .]−[f1 u + f2 u2 + . . .] = f1 a¯iα uiα + . u>iαThe lemma is proved.2Proposition 0.19 Let barK be a one-dimensional local field with the residue field kand charK̄ = chark. Suppose k is algebraically closed and chark = 0. Let α, β beautomorphisms of the field K̄.Then K̄ = k((u)) and α = f −1 βf (where f is an automorphism of K̄) iff(a1 , iα , y(α)) = (b1 , iβ , y(β)).Proof. The ”only if” part is clear. We prove the ”if” part.It is easy to see that a1 = b1 if α = f −1 βf .If ξ is not a root of unity, then by lemma 0.18 α is conjugate with β: β(u) = ξu.Therefore, the ”if” part is proved for the case iα = iβ = 1.Suppose now iα = iβ = 1 and a1 = b1 are roots of unity.Lemma 0.20 Let β, β be k-automorphisms of the field k((u)): β(u) = ξu + xuin+1 +yu2in+1 , β (u) = ξu + x̄uin+1 + ȳu2in+1 , where x̄/x ∈ (k ∗ )in , ȳ = (x̄/x)2 y.Then β and β are conjugate.21−1Proof.

Put x0 = (x̄/x)(in) . Let f be an automorphism such that f (u) = x0 u.Then we havef β(u) = ξx0 u + x(x0 u)in+1 + y(x0 u)2in+1 = x0 ξu + x0 x̄uin+1 + x0 ȳu2in+1 = β f (u)2From this and previous lemmas we get the proof of the proposition.2Corollary 6 In the conditions of the proposition suppose k is not algebraically closedfield. Suppose αn = Id. Then there exists a parameter u in k((u)) such that α(u ) =a1 u .Proof. The proof follows from lemma 0.18.From the proposition we get also the following result:Theorem 0.21 Let K, K be two-dimensional local skew fields with the last residuefields k and k and with canonical automorphisms α, α . Suppose charK = chark,charK = chark , αn = Id, αn = Id for all n ∈ N, the fields k, k are algebraicallyclosed of characteristic 0.K is isomorphic to K iff k ∼= k and (a1 , iα , y(α)) = (a1 , iα , y(α )).Now let us study skew fields with canonical automorphisms of finite order.0.3Classification of two-dimensional local splittable skew fields of characteristic 0.In this part we assume thata two-dimensional local skew field K splits,k ⊂ K, k ⊂ K̄, k ⊂ Z(K),char(K) = char(k) = 0,αn = id for some n ≥ 1,for any convergent sequence (aj ) in K̄ the sequence (zaj z −1 ) converges in K (i.e.

themaps δi , i ≥ 1 are continuous, see corollary 3).We note that the continuity of the maps δi , i ≥ 1 does not depend on the choice ofparameters, as it follows from lemma 0.11 and corollary 3.220.3.1The case α = Id.Definition 0.22 Definei = ν((φz − 1)(u))r = ν̄[((φz − 1)(u))z −imod∈N℘]∞modi∈ Z/iZwhere u, z are arbitrary local parameters of K, φz : K → K, φz (a) = ad(z)(a).Proposition 0.23 i and r do not depend on the choice of parameters u and z.Proof. We fix some parameters u, z: K ∼= k((u))((z)).

Let u , z be other parameters.Thenu = (x0 u + x1 u2 + . . .) + c1 z + c2 z 2 + . . .z = a0 z + a1 z 2 + . . . ,wherexi ∈ k,ai ∈ k((u)),ci ∈ k((u)),, x0 = 0;a0 = 0Put z = a−10 z . It’s clear that ν((φz − 1)(u)) = ν((φz − 1)(u)). From the other handby corollary 4, ν((φz − 1)(u)) = ν((φz − 1)(u)). So, i does not depend on the choice ofparameter z.Now we prove that ν((φz − 1)(u )) = ν((φz − 1)(u)). One can obtain this propertyfrom the following lemma.Lemma 0.24 Suppose the following relation in K holds:zuz −1 = uα + uδj z j + .

. . ,where δ1 = . . . = δj−1 = 0, δj = 0. Then(i) for u = u + bz q we havezu z −1 = uα + uδ1 z + . . . uδq−1 z q−1 + uδq z q + . . . ,where uδq = uδq + bα − ∂/∂u(uα )b.(ii) Suppose α(u) = ξu, ξ ∈ k, ξ n = 1 for some natural n. Then for u = u + bz q ,n|q we havezu z −1 = ξu + . . . + (uδq + bα − ξb)z q + . . . + uδq+j−1 z q+j−1 + uδq+j z q+j + . . . ,where uδq+j = uδq+j + bδj − ∂/∂u(uδj )b(iii) If α = id, then for u = x0 u + x1 u2 + . . ., where xq ∈ k, x0 = 0, we havezu z −1 = u + (uδj23∂ ju )z + . . .∂uProof.

(i) We havezu z −1 = z(u + bz q )z −1 = uα + uδ1 z + . . . + (bα + bδ1 z + . . .)z q =uα + uδ1 z + . . . + (uδq + bα )z q + . . . = uα + uδ1 + . . . + (uδq + bα − ∂/∂u(uα )b)z q + . . . ,because uδ1 = (u + bz q )δ1 = x0 (u + bz q ) + x1 (u + bz q )2 + . . . = uδ1 + ∂/∂u(uδ1 )bz q + . . .if uδ1 = x0 u + x1 u2 + . . ..(ii) We havezu z −1 = z(u + bz q )z −1 = ξu + uδj z j + .

. . + (bα + bδj z j + . . .)z q =ξu + uδj z j + . . . + (uδq + bα )z q + uδq+1 z q+1 + . . . + uδq+j−1 z q+j−1 + (uδq+j + bδj )z q+j + . . . =∂ξu +. . .+(uδq +bα −ξb)z q +uδq+1 z q+1 +. . .+uδq+j−1 z q+j−1 +(uδq+j +bδj − (uδj )b)z q+j∂u(iii) We havezu z −1 = x0 (u + uδj z j + . . .) + x1 (u + uδj z j + . . .)2 + .

. . = u + (uδj∂ ju )z + . . .∂u2Remark. Note that this lemma works also in characteristic p > 0.So, i does not depend on the choice of parameters u and z.Now we prove it for r. Recall that in our proposition α = id (because i and r weredefined only for α = id). By lemma 0.24 for any parameter u we havezu z −1 = u + (uδi∂ iu )z + . . .∂uTherefore, ν̄[((φz − 1)(u ))z −i ] = ν̄(uδi ) = ν̄[((φz − 1)(u ))z −i ]If we change z by z we getz uz −1 = zuz −1mod ℘iHenceν̄[((φz − 1)(u))z −iν̄[((φz − 1)(u))z −imod ℘] = ν̄[((φz − 1)(u))z −i−imod ℘] + ν̄(a−i0 ) = ν̄[((φz − 1)(u))z2Definition 0.25 Definea = resui+1 2uδ2i − 2 δidu(uδi )224∈kmod ℘] =mod ℘] mod iProposition 0.26 auδi+1 , . .

. , uδ2i−1 .=a(uδi+1 , . . . , uδ2i−1 ), i.e. a depends only on the mapsProof. We comment on the statement first. The maps δj are uniquely defined byparameters u, z and they depend on the choice of these parameters. So it suffice to showthat a does not depend on the on the choice of parameters which preserve the mapsδi+1 , . . . , δ2i−1 .

We can assume that δi+1 = 0, . . . , δ2i−1 = 0, because we can change theparameters to make this maps to be equal to zero (see lemma 0.11).First we show that any change of the type u → u = u + c1 z + . . . + ci z i is equivalentto a change of parameters as follows: z → z = z + a1 z 2 + . . ., u → u = u + ci z i + . . .,i.e. we get the same maps δj in both cases. The proof is by induction.One can decompose the change u → u = u + c1 z + . .

. + ci z i in a finite number ofchanges u → u1 = u + ci z i , u1 → u2 = u1 + ci−1 z i−1 , . . . , ui−1 → ui = ui−1 + c1 z. So itsuffice to prove our assertion for any change of the type uj → uj+1 = uj + ci−j z i−j .For j = 1 the assertion is trivial. Consider an arbitrary case. By lemma 0.24, δ2i−j isthe first map which is not invariant under this change.