Multidimensional local skew-fields (792481), страница 19

Просмтор этого файла доступен только зарегистрированным пользователям. Но у нас супер быстрая регистрация: достаточно только электронной почты!

Текст из файла (страница 19)

Let’s use the calculationsbefore the formulas (1.9) and (1.10). The equation (1.9) is solvable with cm−1 = 0,(m−1)/iν̄(cm−1 ỹ1ᾱ(ỹ2−1 )) > i(ᾱ) − 1, m − 1 < i(α) − 1, and to complete the proof wecq , aq88only have to check that the properties of the coefficients cq , aq remain true by the−1α fm−1 . But this follows from the same arguments as in the case,conjugation fm−1(m−1)/i −1ỹ2 ) = 0, because we usedwhen fm−1 (u) = u + xm−1 z m−1 , fm−1 (z) = z, ν̄(xm−1 ỹ1(m−1)/i −1ỹ2 ) ≥ 0, which is true also in our case, becauseonly the inequation ν̄(xm−1 ỹ1xm−1 is a solution of an equation of the type (1.9). In the same way one can prove this(m−1)/i) ≥ 0.

From thefact in the case fm−1 (u) = u, fm−1 (z) = z + ym−1 z m , ν̄(ym−1 ỹ1other hand side, one can see from the equation (1.9) that the conjugation fm−1 (u) = u,fm−1 (z) = z+ym−1 z m does not change the coefficient cm−1 , so any conjugation fm−1 can, fm−1such that fm−1(z) = z,be decomposed into composition of two conjugations fm−1fm−1 (u) = u.Thus, we have proved that α is conjugate to α , where j > j, iα = i(α). Since ourarguments do not depend on j, we get the required assertion by induction.In the same way with Proposition 1.2 it is proved now that α = f −1 βf , where(i(α)−1)/i)=β(u) = ᾱ(u), β(z) = ã0 z + ai(α)−1 z i(α) + a2(i(α)−1) z 2i(α)−1 , where ν̄(ai(α)−1 ỹ12(i(α)−1)/ii(ᾱ) − 1, ν̄(a2(i(α)−1) ỹ1) = i(ᾱ) − 1 (i.e it is the case b) of the theorem).Case 2).

This case is divided into two ones:a) j = i(α) − 1,b) j < i(α) − 1.Let us look first at the case a). We show that α = f βf −1 , where β is defined in thecase a) of the theorem.−1αfmi , fmi (u) = u +To do that we make sequential substitutions α → α = fmi−mmimi+1, where xmi = ỹ1 ỹ2 (xmi )0 , ymi = ỹ1−m (ymi)0 . It isxmi z , fmi (z) = z + ymi zenough to show that for every m a corresponding automorphism α has coefficientscq , aq , which satisfy the property:q/iq/iν̄(ỹ1 ᾱ(ỹ2−1 )cq ) > iᾱ − 1, ν̄(ỹ1 aq ) > i(ᾱ) − 1, i|q, im ≤ q ≤ mi + j, q = j, 2j,cq = aq = 0, q < im,because then α can be reduced to the case, when the appropriate coefficients cq , aqare equal to zero. That is done using the same substitutions, as by deriving equations(1.9), (1.10), and with the help of result from the case 1). Since for every m the numberof necessary conjugations is finite, the desired automorphism f : α = f βf −1 exists.Let us write down the calculations for an arbitrary m:αf (z) = α(z)+α(ymi )α(z mi+1 ) = ã0 z+B1 ui(ᾱ)−1 z i(α) +B2 ui(ᾱ)−1 z 2i(α)−1 +ami+i(α)−1 z mi+i(α) +ᾱ(ymi )ãmi+1z mi+1 + ᾱ(0∂ymi )Aui(ᾱ)−1 z im+1+j +∂ui(ᾱ)−1ᾱ(ymi )z mi+i(α)(mi + 1)ãmi0 B1 umod ℘mi+i(α)+1f α (z) = f (ã0 )f (z) + f (aim )f (z im+1 ) + .

. . + f (aim+i(α)−1 f (z im+i(α) ) =ã0 z+∂(ã0 )xmi z mi+1 +ã0 ymi z mi+1 +aim z im+1 +. . .+aim+i(α)−1 z im+i(α) +i(α)ymi ai(α)−1 z im+i(α) +∂u89∂ )xmi z im+i(α) mod ℘mi+i(α)+1(1.16)(a∂u i(α)−1Because of the special form of yim , ᾱ(ymi )ãim+1= ã0 ymi . The coefficients aq , q <0im + i(α) − 1 can be chosen so that they have the pointed properties, in the same way,)0 :as in the case 1). For q = im + i(α) − 1 it is necessary to show, that there exists (ymim+(i(α)−1)/iỹ1(ami+i(α)−1 +ᾱ(∂iᾱ −1ᾱ(ymi )−i(α)ymi ai(α)−1 −(ymi ))Aui(ᾱ)−1 +(mi+1)ãmi0 B1 u∂u∂ (ai(α)−1 )xmi ) = 0 mod ℘¯i(ᾱ)∂u(1.17)Sincem+(i(α)−1)/im+(i(α)−1)/i∂ν̄(ỹ1Aᾱ( ∂uymi )ui(ᾱ)−1 ) > i(ᾱ) − 1, ν̄(ỹ1ami+i(α)−1 ) ≥ i(ᾱ) − 1,m+(i(α)−1)/im+(i(α)−1)/i∂i(ᾱ)−1ν̄(ỹ1B1 ᾱ(ymi )u) = i(ᾱ) − 1, ν̄( ∂u (ai(α)−1 )xmi ỹ1) ≥ i(ᾱ) − 1 −1 + i(ᾱ) > i(ᾱ) − 1,the element (ymi)0 exists and is defined uniquely if (im + 1) = i(α), i.e.

q = 2j. Further,miαf (u) = α(u) + α(xmi )α(z mi ) = ᾱ(u) + Aui(ᾱ)−1 z j + cmi+i(α)−1 z mi+j + ᾱ(xmi )ãmi0 z +ᾱ(∂B1 uiᾱ −1 ᾱ(xmi )z mi+i(α)−1xmi )Aui(ᾱ)−1 z mi+j + miãmi−10∂uf α (u) = f (ᾱ(u))+f (cim )f (z im )+. . .+f (cim+j )f (z im+j ) = ᾱ(u)+mod ℘mi+i(α)∂(ᾱ(u))xim z im +cim z im +. . .∂u∂ (cj )xmi z im+j + jymi cj z mi+j ,∂uwhence we get similarly that we must solve an equation over (xmi )0 :+ cim+j z im+j +m+j/i −1ỹ2 (cmi+jỹ1(1.18)+ miã0mi−1 B1 uiᾱ −1 ᾱ(xmi )−∂∂ (1.19)(cj )xmi − jymi cj + ᾱ( xmi )Aui(ᾱ)−1 ) = 0 mod ℘¯i(ᾱ)∂u∂uSince (ymi)0 was already defined (if mi = j, we can take (ymi)0 equal to a constant),m+j/i −1∂i(ᾱ)−1ν̄(ỹ1ỹ2 ᾱ( ∂u xmi )Au> i(ᾱ) − 1,m+j/i −1 ∂m+j/iν̄(ỹ1ỹ2 ∂u (cj )xmi ) > i(ᾱ) − 1, ν̄(ỹ1ỹ2 B1 ui(ᾱ)−1 ᾱ(xmi )) = i(ᾱ) − 1,so an element (xmi )0 does exist.Let us now examine the case b).

Now by the similar arguments as in a), we get thatα is conjugate to β,β(u) = ᾱ(u) + A1 z j + A2 z j+1 + . . . + Ai(α)−1−j z i(α)−1 ,q/iq/iβ(z) = ã0 z + B1 z i(α) + b2 z 2i(α)−1 , ν̄(Aq ỹ1 ᾱ(ỹ2−1 )) = i(ᾱ) − 1 or Aq = 0, ν̄(Bq ã−10 ỹ1 ) =i(ᾱ) − 1 if i(α) is finite, and90β(u) = ᾱ(u) + Az j ,β(z) = ã0 z if i(α) = ∞ (see cases c) and d) correspondingly).In fact, let us use formulas 1.16 and (1.18). Sincem+j/i+q−1∂ymi )ỹ1) > i(ᾱ) − 1, the arguments from deriving the formula (1.17)ν̄(Aq ᾱ( ∂uremain true for all coefficients aq , q ≥ i + i(α) − 1, q = 2i(α) − 1, and the property fromm+j/i+q−1 −1∂xmi )Aq ỹ1ỹ2 ) > i(ᾱ) − 1, we canthe case a) is realized.

Similarly, since ν̄(ᾱ( ∂uapply formula (1.19) for the coefficients cq , q ≥ i + i(α) − 1 and get the desired result.Remark. In the case b) of theorem, if ã0 = 1 or ã0 = 1 but y = 0, where yis a second parameter of the canonical representation of ᾱ, one can show by directcalculations that α is conjugate with β: β(u) = ᾱ(u) + Az j , β(z) = ã0 z + Bz i(α) ,where A satisfies (1.8) and B does not.

But, if ã0 = 1 and y = 0, then for any k ≥ 1∂(B)ỹ2 ∈ Im(ᾱ − Id), where B = cỹ2−1 u1+k(1−i(ᾱ)) , whence, by formulas (1.17) and∂u(1.19), one can derive that β does not exist and the number of parameters can not bedecreased.Remark. 1. In the case of characteristic p > 0 we have in general dim(kerT ) =dim(cokerT ), as it was shown in lemma 1.3. From this follows that automorphismscan not be parameterised by finite number of parameters in more cases than in thecase of chark = 0. For example, α can not be always redused to β, where β(u) =ᾱ(u) + A1 z j + . .

. + Ak uj+k : k may be equal to the infinity.2. The classification can be easily generalised to the case of n-dimensional localfield, because we used only the property dim(kerT ) = dim(cokerT ) and argumentswith valuations. In the case of multidimensional equal characteristics local fields ofcharacteristic 0 all our arguments can be carried over to the case of higher dimensionif we assume that the value group of ν̄ is Z ⊕ . .

. ⊕ Z.Now we only have to prove that the automorphisms β, β are conjugate if and onlyif β = β , where β, β are automorphisms from the formulation of theorem. It’s clearthat if β is conjugate with β , then ã0 = ã 0 and β̄(u) = β̄ (u) = ᾱ(u) is a nesessarycondition, whence β is defined up to the change u → x0 : ᾱ(x0 ) has the canonical viewand z → cz, c ∈ k ∗ .Then, β and β must have the same numbers j, j and i(α), iα . Indeed, if β andβ are conjugate, then β = f −1 β f , and f can be decomposed in a compositionof automorphisms f = f1 f2 . .

. fm , where fq (u) = u + xq z q , fq1 (z) = z + yq1 z q+1 .Then from (1.9), (1.10) follows that for q < min{j, j } we have xq ∈ kerTq,1 , forq1 < min{i(α) − 1, iα − 1} we have yq1 ∈ kerTq,2 . From the proof of the case a) followsthat the conjugations fq with this numbers preserve properties (1.8) of the coefficientscq , aq1 for q ≤ min{j, j }, q1 ≤ min{i(α) − 1, iα − 1}. Therefore, if j = j or i(α) = iα ,then the first nonzero coefficient of β(u) or β (u) or β(z) or β (z) must lie in the kernelof the map Tj(j ),1 (Ti(α)(iα ),2 ), but this contradicts to the choice of these coefficients.Therefore, j = j and i(α) = iα .

So, β and β are in the same class defined by the91pair (j, i(α)). In this case the equality follows from the special choice of coefficients ofz j , z i(α) and the proves of the corresponding cases.1.3Proof of the theorem IIILet ᾱn = 1. Then, by Proposition 1.2, there exists x0 such that ᾱ(x0 ) = ξx0 , where ξis a primitive root of 1. As in the theorem II, we consider that α(u) = ξu + cj z j + . .

.,α(z) = a0 z + ai(α)−1 z i(α) + . . .., then, as it was shown in theAt first we note that i = 1 or ∞. Indeed, if ai0 = ᾱ(y)yqntheorem II, we may suppose a0 = 1 + cu + . . .. But in this case ai0 = 1 + icuqn + . . ..Further,nᾱ2 (y). . . ᾱᾱn−1(y)= 1. So we have that:ai0 ᾱ(ai0 ) . . . ᾱn−1 (ai0 ) = ᾱ(y)y ᾱ(y)(y)ai0 ᾱ(ai0 ) .

. . ᾱn−1 (ai0 ) = (1 + icuqn + . . .) . . . (1 + icuqn + . . .) = 1 + nciuqn + . . . = 1, wherefrom a0 = 1 or i = ∞. Thus, α(u) = ξu + cj uj + . . ., α(z) = z + ai(α)−1 z i(α) + . . . (inthis case dim(kerTk,1 ) = ∞ = dim(kerTk,2 ) = dim(cokerTk,1,2 )).Further let us consider that cq ∈ uk((un )), aq ∈ k((un )), because by going over toconjugations as in (1.9) and (1.10), we can solve all the equationsᾱ(y) − ξy = cq mod uk((un )), ᾱ(y) − y = aq mod k((un ))(in theorem 2 we have reduced general case to a case aq , cq ∈ cokerTq,1 , Tq,2 in the sameway).As in Proposition 1, it is proved that if i = ∞, then takes place the case O) of thetheorem.Let i = 1. The case j ≥ i(α) coincide with the case j ≥ i(α) of the theorem 2:by writing over the formula (1.11), we get that holds (1.12) and there from holdscq = 0, q < j, and the equation (1.13) always has a solution xm ∈ uk((un )) , whenaq ∈ k((un )), cq , cq ∈ uk((un )).

Характеристики

Список файлов диссертации