Multidimensional local skew-fields (792481), страница 18
Текст из файла (страница 18)
To prove the theorem we must prove theexistence of an automorphism f such that α = f βf −1 and β is an automorphism, asdefined in the formulation of the theorem. Thereto it would be also proved, that the83automorphism β can be uniquely reconstructed by the automorphism α and any β-likeautomorphism gives its own conjugacy class.Assumef (u) = x0 + x1 z + x2 z 2 + . . .f (z) = y0 z + y1 z 2 + . . .We choose the parameter x0 ∈ K̄ in such a way that ᾱ(x0 ) has a canonical form, thati(ᾱ)2i(ᾱ)−1.
Recall that x is a representative of a class k ∗ /k ∗(i(ᾱ)−1) .is ᾱ(x0 ) = ξx0 +xx0 +yx01.2Proof of the theorems I and IILet a0 fulfil the assumptions of the theorem. We prove, that there exists an automorphism f such that αf (u) = f β(u); αf (z) = f β(z), where β is an automorphism, as defined in the theorem.
To do that, we prove by induction that αf (u) =f β(u) mod ℘m and αf (z) = f β(z) mod ℘m+1 for all m ∈ N.From (1.2), (1.1), (1.3) (which remain true also in the case of finite order automorphism ᾱ ) follows, that the set of representatives of classes of the elementsf¯−1 (a0 a¯0 −1 ᾱ(y0 )/y0 ) can be described as the set of the elements {uν̄(a0 ) (1 + an un +a2n u2n + . . .
ai(ᾱ)−1 ui(ᾱ)−1 ), anq ∈ k, ai(ᾱ)−1 = nlξ −1 x, where ξ is a primitive root from1 of a degree n, l ∈ Z\{0}}. From the definition of the element a0 follows that theelements anq are uniquely defined by automorphism α, that is, they don’t depend onthe choice of parameter z, and a¯0 is defined up to multiplication by an element ξ m ,m ∈ Z.Assume a˜˜0 = a¯0 uν̄(a0 ) (1+an un +a2n u2n +.
. . ai(ᾱ)−1 ui(ᾱ)−1 ). Then we have for m = 1thati(ᾱ)αf (u) = α(x0 ) = α(x0 ) = ξx0 + xx0 + . . . = f β(u); αf (z) = α(y0 )α(z) =˜0 )y0 z = f (ã˜0 )f (z) = f β(z) mod ℘2 .ᾱ(y0 )a0 z = f (ã−1For an arbitrary m we replace α by fm−2αfm−2 for a suitable automorphism fm−2(that is, for any automorphism with suitable coefficients x0 , . . .
, xm−2 , y0 , . . . , ym−2 ),˜0 ,and now can consider that c0 = ξu + xui(ᾱ) + . . ., c1 = . . . = cm−2 = 0, a0 = ãa1 = . . . = am−2 = 0, x0 = u, x1 = . . . = xm−2 = 0, y0 = 1, y1 = . . . = ym−2 = 0. Thenαf (u) = α(u)+α(xm−1 )α(z m−1 ) = ξu+xui(ᾱ) +. . .+cm−1 z m−1 +ᾱ(xm−1 )ãm−1z m−10modf β(u) = ξ(u + xm−1 z m−1 ) + x(u + xm−1 z m−1 )i(ᾱ) + . . . = ξu + xi(ᾱ) u + . . . + ξxm−1 z m−1 +i(ᾱ)xxm−1 ui(ᾱ)−1 z m−1 + . . . =ξu + xi(ᾱ) u + . . .
+ xm−1 (∂(ᾱ(u)))z m−1∂(u)84mod℘m℘mHence,+ cm−1 = xm−1 (ᾱ(xm−1 )ãm−10∂(ᾱ(u))).∂(u)(1.6)And in the same way,αf (z) = α(z)+α(ym−1 )α(z m ) = ã0 z+am−1 z m +(ᾱ(ym−1 )+. . .)(ã0 z+. . .)mf β(z) = f (ã0 )f (z) = (ã0 +ã0 z +mod℘m+1∂(ã0 )xm−1 z m−1 )(z + ym−1 z m ) =∂(u)∂(ã0 )xm−1 z m + ã0 ym−1 z m∂(u)mod℘m+1Hence,ᾱ(ym−1 )ãm0 + am−1 = ã0 ym−1 +∂(ã0 )xm−1 .∂(u)(1.7)By Corollary 12, if the conditions of the theorem are fulfilled, the equations (1.6),(1.7) have the unique solution with any cm−1 , am−1 and with any m, whence followsthe proof of the case 1). By Proposition 1.2 the proof of the case 2) is evident.Proof of the Theorem IIIf i = ∞, we can apply entirely the same arguments as in the Theorem I, andget that α is conjugate to the automorphism β, where β has the same form as in theTheorem I (i.e.
this case corresponds to the case d), when j = ∞). In order that theseiarguments remain true, we must only show that the element a := ξ + i(α)xx0ᾱ−1 + . . . in(1.6) can be represented in the form ᾱ(y)/y. But it follows directly from the relations(1.2), (1.3), (1.1).Let i < ∞. We prove that there exists such an automorphism f that αf (u) = f β(u),αf (z) = f β(z), where automorphism β is as defined in the theorem. The proof is thesame as in the Theorem I.The case m = 1 coincides with the case m = 1 from the Theorem I. Applying thesame arguments as there, we get equations of the form (1.6) and (1.7). By Corollary2, these equations are solvable if i |(m − 1).
They may be unsolvable if i|(m − 1). Sincechark = 0, the kernel and the cokernel of the maps− (ξ + xui(ᾱ)−1 + . . .)xm−1 ,Tm−1,1 = ᾱ(xm−1 )ãm−10Tm−1,2 = ᾱ(ym−1 )ãm0 − ã0 ym−1are one-dimensional if i|(m − 1).k/ik/iWe put xk = ỹ1 ỹ2−1 xk , yk = ỹ1 yk for k = iq, q ∈ N . Thenᾱ(xk )ãk0 −(ξ+xui(ᾱ)−1 +.
. .)xk=k/i−k/i ᾱ(ỹ1 ) ᾱ(ỹ2 ) −k/iᾱ(ỹ1 ỹ2 xk ) k/i −x ỹỹ2 k 1ỹ185−k/i= ᾱ(ỹ2 )ỹ1(ᾱ(xk )−xk ),ᾱ(yk )ãk+10− ã0 yk =k/i−k/i ᾱ(ỹ1 )ᾱ(ỹ1 yk ) k/i ã0ỹ1−k/i− ã0 yk ỹ1−k/i= ã0 ỹ1(ᾱ(yk ) − yk )Now we can write down the kernel and cokernel of these maps in the explicit form.−k/i−k/iFor Tk,1 the kernel is ỹ1 ỹ2 (xk )0 , where (xk )0 ∈ k, cokernel — cui(ᾱ)−1 ỹ1 ᾱ(ỹ2 ),−k/ic ∈ k; in the same way, for Tk,2 the kernel is ỹ1 (yk )0 , where (yk )0 ∈ k, cokernel —−k/ic1 ui(ᾱ)−1 ỹ1 ã0 ᾱ.Step 1 We show that α is conjugate to an automorphism α , which has all thecoefficients cq and aq , q ≥ 1 , satisfying the property:if i |q, then aq = cq = 0; if i|q, then ν̄(cq ỹ1 ᾱ(ỹ2−1 )) ≥ i(ᾱ) − 1, ν̄(aq ỹ1 ã−10 ) ≥ i(ᾱ) − 1(1.8)We even show that in (1.8) we have either equalities or cq = 0 (aq = 0).In fact, let α be such an automorphism that c0 = c0 = ξu + xui(ᾱ) + .
. ., a0 = ã0 .Let us find the rest coefficients satisfying these properties. Applying induction on m,we have for arbitrary m thatq/iq/iαf (u) = α(u) + α(xm−1 )α(z m−1 ) =ᾱ(u) + c1 z + . . . cm−1 z m−2 + cm−1 z m−1 + (ᾱ(xm−1 ) + . . .)(ã0 z + . . .)m−1mod ℘mf α (u) = f (ᾱ(u)) + f (c1 )f (z) + . . . + f (cm−2 )f (z m−2 ) + f (cm−1 )f (z m−1 ) =ᾱ(u) +∂(ᾱ(u))xm−1 z m−1 + c1 z + . .
. + cm−1 z m−1∂umod ℘mHence,∂(ᾱ(u))) + cm−1(1.9)∂uIf i |(m − 1), then cm−1 = 0 and by Corollary 2 the solution of this equation existsand is unique. If i|(m − 1), then for solvability of this equation it is enough to select−(m−1)/i(m−1)/iᾱ(ỹ2 ), i.e. ν̄(cm−1 ỹ1ᾱ(ỹ2−1 )) ≥ i(ᾱ) − 1.cm−1 in the form cui(ᾱ)−1 ỹ1Further,αf (z) = α(z) + α(ym−1 )α(z m ) =+ cm−1 = xm−1 (ᾱ(xm−1 )ãm−10mã0 z + a1 z 2 + . . . + am−2 z m−1 + am−1 z m + (ᾱ(ym−1 ) + . .
.)(ãm0 z + . . .) mod℘m+1f α (z) = f (ã0 )f (z) + f (a1 )f (z 2 ) + . . . + f (am−1 )f (z m ) =∂∂ (ã0 )xm−1 z m + ã0 ym−1 z m + (a1 +(a )xm−1 z m−1 )(z + ym−1 z m )2 + . . .∂u∂u 1∂∂+(am−1 + (am−1 )xm−1 z m−1 )(z+ym−1 z m )m = ã0 z+ (ã0 )xm−1 z m +ã0 ym−1 z m +a1 z 2 +. . .∂u∂umm+1+am−1 zmod ℘ã0 z +86Hence∂(ã0 )xm−1 + am−1 ,∂uand in the same to the previous case way we get the desired result.ᾱ(ym−1 )ãm0 + am−1 = ã0 ym−1 +(1.10)Step 2 Here two cases are possible:1) j ≥ i(α);2) j < i(α).Case 1). We show that α = f −1 β f , where β (u) = ᾱ(u).−1αfmi , where fmi (u) =To do that we find the sequential conjugations α → α = fmiu + xmi z mi , fmi (z) = z, m ≥ 1, xmi = ỹ1−m ỹ2 (xmi )0 . If m = (j − i(α) + 1)/i, we havefor the coefficients cq that:αf (u) = α(u) + α(xim )α(z im ) = ᾱ(u) + cj z j + .
. . +∂xim )z j + . . .)(ã0 z + ai(α)−1 z i(α) + . . .)im =∂ujimᾱ(u) + cj z + ᾱ(xim )ãim+ ᾱ(xim )ã0im−1 ai(α)−1 z j mod ℘j+10 z(ᾱ(xim ) + ᾱ(∂(ᾱ(u))z im +. . .+f (c(m+1)i z (m+1)i )+. . .∂u℘j+1(1.11)f α (u) = f (ᾱ(u))+f (c(m+1)i z (m+1)i )+. . . = ᾱ(u)+xim+ f (cj z j ) modSince xmi = ỹ1−m ỹ2 (xmi )0 , the equation at z mi has the formᾱ(xmi )ãim0 − xmi (∂(ᾱ(u))) = 0∂u(1.12)We show that all the coefficients cq in (1.11) can be chosen so that ν̄(ỹ1 ᾱ(ỹ2−1 )cq ) >i(ᾱ) − 1.In order to do that if q < j, we prove, applying induction on q/i, that all thecoefficients at z in degrees higher than im in f (ᾱ(u)), f (cli z li ) satisfy this property,supposing that cli satisfies this property at l < q/i.For f (ul ), l > 1 we have by Newton’s binomial formula thatq/illf (u ) = u +lul−k xkim z imk Clk ,k=1whenceν̄(ul−k xkim ỹ1mk ᾱ(ỹ2−1 )) = l−k+(k+1)ν̄(ỹ2 ) = l−k+(k−1)i(ᾱ) = l−k+(k−1)(i(ᾱ)−1) >(i(ᾱ) − 1) for k > 1,what proves our assertion for f (ᾱ(u)).
For f (cli z li ) we have f (cli z li ) = f (cli )z li , and,using Newton’s binomial formula again, we get87l > i(ᾱ)−1− ν̄(ỹ1l )+ ν̄(ỹ2 ) = i(ᾱ)−1− ν̄(ỹ1l )+i(ᾱ), where from l−1+(k−1)(i(ᾱ)−1) >i(ᾱ) − 1 − ν̄(ỹ1l ) for all k, what proves our assertion in this case also.At z j we have the equationai(α)−1 + cj = cj ,ᾱ(xim )ãim−10(1.13)and we must only solve the equationỹ1 ᾱ(ỹ2−1 cj + ỹ1 ᾱ(ỹ2−1 )ai(α)−1 ã0im−1 ᾱ(xim ) = 0 modj/ij/i℘¯i(ᾱ)(1.14)in order to finish the induction step for the coefficients cq .
We have:ỹ1 ᾱ(ỹ2−1 )cj + ỹ1 ᾱ(ỹ2−1 )ai(α)−1 ã0im−1 ᾱ(xim ) =j/ij/iỹ1 ᾱ(ỹ2−1 cj + ỹ1 ᾱ(ỹ2−1 )ᾱ(ỹ1−m )ᾱ(ỹ2 )(xmi )0 ai(α)−1 =j/ij/i(i(α)−1)/iỹ1 ᾱ(ỹ2−1 )cj + ỹ1j/i(i(α)−1)/iai(α)−1 (xmi )0mod℘¯i(ᾱ)Since ν̄(ỹ1ai(α)−1 ) = i(ᾱ) − 1 = ν̄(ỹ1 ᾱ(ỹ2−1 cj ), there exists a unique constant(xmi )0 , with which the equation (1.14) is solvable.j/iLet us show that the coefficients cq , q > j, aq , q ≥ 1 satisfy the properties (1.8).As for coefficients cq , it is remained to prove, that the coefficients at z d inα(xim )α(z im ) for d > j satisfy (1.8) .
It’s clear that (1.8) remains true if i |q. But ifi|q, then α(z im ) = z im D, where D is a series with coefficients of the same behaviouras aq . It follows from the Newton’s binomial formula. Applying the same argumentsas for f (cli )z li , we get that (1.8) holds for α(xim )z im , where from (1.8) also holdsimimqfor theim )z D, because (1.8) holds for each series α(xim )z dq z , whereproduct α(xD = q≥0,i|q dq z q .For the coefficients aq we haveαf (z) = α(z)f α (z) = f (ã0 )z + f (ai(α)−1 )z i(α) + . . .(1.15)where from, using calculations for f (ul ), we get that (1.8) holds for aq .
Therefore,q/isince ã0 = 1 mod ℘¯i(ᾱ)−1 , we have ν̄(aq ỹ1 ã−10 ) > i(ᾱ) − 1 for all q < i(α) − 1.To complete the induction, let’s show that α = f −1 α f , where the coefficientsof the automorphism α satisfy (1.8) and cq = 0, 1 ≤ q ≤ j, aq = 0, 1 ≤q < i(α) − 1. The proof is again by induction on m (℘m ).
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
