Multidimensional local skew-fields (792481), страница 17

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If ν̄(T (a0 − 1)) = (i(ᾱ) − 1), then may takeplace either the first or the second case.Indeed, we have seen thatᾱ(culn )= 1 + nlξ −1 xui(ᾱ)−1 + . . .culn(1.2)ᾱ(1 + cul )= 1 + c(ᾱ(ul ) − ul )(1 + cul )−1l1 + cu(1.3)Hence, if ν̄(T (a0 −1)) < (i(ᾱ)−1) or ν̄(T (a0 −1)) = (i(ᾱ)−1), but a0 = 1+wui(ᾱ)−1 +. . .and w = nlξ −1 x for all l, then 1) holds. If the rest inequalities hold, then one can seethat 2) may take place.Let the case 1) holds. Let’s show that T is surjective.Indeed, this case is equivalent to the property ν̄(ᾱ(y) − a0 y) ≤ i + ν̄(y), y ∈ K̄. Butthis means that< T (ul ), N ≤ l ≤ N1 >⊂< ul , N + min{i(ᾱ) − 1, i} ≤ l ≤ N1 + max{i(ᾱ) − 1, i} > forall integers N, N1 , N < N1 .From this follows that the cokernel of the map T cannot have infinite dimension.Suppose it has finite dimension, i.e.

it is not equal to zero. Choose an element of theminimal value κ of the valuation in the cokernel and choose a number N1 : N1 + i < κ.Complete a basis of the vector space < T (ul ), N ≤ l ≤ N1 > with respect to the basisof the vector space < ul , N + min{i(ᾱ) − 1, i} ≤ l ≤ N1 + max{i(ᾱ) − 1, i} >. Denote76these elements by ej , j ∈ {1, . . . , |i − i(ᾱ) + 1|}. Then for any integer Ñ1 > N1 we willhave< T (ul ), ej , N ≤ l ≤ Ñ1 , j ∈ {1, . . .

, |i − i(ᾱ) + 1|} >=< ul , N + min{i(ᾱ) − 1, i} ≤l ≤ Ñ1 + max{i(ᾱ) − 1, i} >, but this contradicts to the existence of elements of thecokernel.Let the case 2) holds. This case is a negation of the case 1); so, for any natural ithere exists N1 ∈ N such that < T (ul ), N ≤ l ≤ N1 >⊂< ul , N + min{i(ᾱ) − 1, i} ≤l ≤ N1 + max{i(ᾱ) − 1, i} >. Repeating the arguments of the case a0 = 1 we get thatthere exists a converge sequence in K̄ such that the maximal values of the valuation onpreimages of elements of this sequence tend to −∞. From this we get that the cokernelhas infinite dimension.The lemma is proved.2Corollary 13 In the notation of lemma let chark = 0.

Then d = 1 if and only if ᾱhas infinite order and a0 = ᾱ(x)/x for some x ∈ K̄; d = ∞ iff ᾱ has finite order anda0 = ξ j , j ∈ Z; d = 0 in the rest cases.Let α ∈ Autk (K). Then the automorphism ᾱ ∈ Autk (K̄) and its invariants ξ(ᾱ) ∈k , i(ᾱ), x(ᾱ) ∈ k ∗ /(k ∗ )i(ᾱ)−1 , y(ᾱ) ∈ k are defined (see def. and prop. 1.2 ). Puta0 = α(z)z −1 ∈ k((u)). Note that the number ν̄(a0 ) does not depend on the choice ofthe parameter z.∗/Theorem 1.4 (Theorem I) Let chark = 0. Let ν̄(a0 ) = 0 or ν̄(a0 ) = 0, but ā0 ∈{ξ(ᾱ)m , m ∈ Z}. Then1) The automorphism α is conjugate with an automorphism β given by the formulaβ(u) = ξu + xui(ᾱ) + x2 yu2i(ᾱ)−1β(z) = uν̄(a0 ) ā0 (1 + an un + a2n u2n + . .

. ai(ᾱ)−1 ui(ᾱ)−1 )z/where ξ = ξ(ᾱ), x = x(ᾱ), y = y(ᾱ), anq ∈ k, q ∈ {1, . . . , (i(ᾱ) − 1)/n}, ai(ᾱ)−1 ∈−1nξ(ᾱ) x(ᾱ)Z , Z = Z\{0}, n = ord(ξ(ᾱ)).(i(ᾱ)−1)2) ν̄(a0 ), ā0 , aj/x(ᾱ)j , ξ(ᾱ), x(ᾱ), y(ᾱ), i(ᾱ) is the complete system of invariantswith respect to the conjugation.Assumeα(u) = c0 + c1 z + c2 z 2 + . . . ,α(z) = a0 z + a1 z 2 + . . . ,ci ∈ k((u))ai ∈ k((u))Let us denote the additional notation:i ∈ N∪{∞} — such a minimal positive integer that ai0 = ᾱ(Y )/Y for some Y ∈ k((u)),j = minq {iq : ciq = 0}, q ≥ 0,77i(α) = minq {iq + 1 : aiq = 0},ã0 = 1 + an un + a2n u2n + . . .

+ ai(ᾱ)−1 ui(ᾱ)−1 ,f¯ ∈ Autk (K̄) — such an automorphism that f¯−1 ᾱf¯(u) = ξ(ᾱ)u + xui(ᾱ) + yu2i(ᾱ)−1 ,i(ᾱ)−12i(ᾱ)−2ỹ2 — such a solution of the equation ᾱ(Y )/Y = ξ(ᾱ)+i(ᾱ)xx0+(2i(ᾱ)−1)yx0,−ν̄(ỹ)¯2) = 1,x0 = f (u) that (ỹ2 uỹ1 — such a solution of the equation ᾱ(Y )/Y = ãi0 that (ỹ1 u−ν̄(ỹ1 ) ) = 1,−(i(α)−1)/iB1 = ỹ1ã0 ,−2(i(α)−1)/iã0 ,B2 = ỹ1−(j+q−1)/i ¯−1f (ᾱ(ỹ2 )).Aq = ỹ1Note that B1 , B2 , Aq are defined uniquely.Theorem 1.5 (Theorem II) Assume chark = 0 and let ν̄(a0 ) = 0 and ā0 ∈{ξ(ᾱ)m , m ∈ Z} and ᾱ be of infinite order.Then α is conjugate to β, that is defined according to the next four possible cases:a) i(α) − 1 = j, i(α) < ∞, soβ(u) = ξ(ᾱ)u + xui(ᾱ) + x2 yu2i(ᾱ)−1 + r1 A1 ui(ᾱ)−1 z jβ(z) = ã0 z + s1 B1 ui(ᾱ)−1 z i(α) + s2 B2 ui(ᾱ)−1 z 2i(α)−1where r1 ∈ k ∗ /(k ∗ )j , s2 ∈ k, s1b) i(α) − 1 < j, i(α) < ∞j(i(ᾱ)−1) (i(ᾱ)−2)(i(α)−1)x(i( barα)−1)(i(α)−1)/r1∈ k.β(u) = ξ(ᾱ)u + xui(ᾱ) + x2 yu2i(ᾱ)−1β(z) = ã0 z + s1 B1 ui(ᾱ)−1 z i(α) + s2 B2 ui(ᾱ)−1 z 2i(α)−1where s1 ∈ k ∗ /k ∗(i(α)−1,i(ᾱ)−1) , s2 ∈ k.c) i(α) − 1 > j, i(α) < ∞β(u) = ξ(ᾱ)u+xui(ᾱ) +x2 yu2i(ᾱ)−1 +r1 A1 ui(ᾱ)−1 z j +.

. .+ri(α)−1−j Ai(α)−1−j ui(ᾱ)−1 z i(α)−1β(z) = ã0 z + s1 B1 ui(ᾱ)−1 z i(α) + s2 B2 ui(ᾱ)−1 z 2i(α)−1where r1 ∈ k ∗ /k ∗j , s1d) i(α) = ∞ (j ≤ ∞ )j(i(ᾱ)−1) (i(ᾱ)−2)(i(α)−1)x(i(ᾱ)−1)(i(α)−1)/r1∈ k, rq , s2 ∈ k, q = 1.β(u) = ξ(ᾱ)u + xui(ᾱ) + x2 yu2i(ᾱ)−1 + r1 A1 ui(ᾱ)−1 z jβ(z) = ã0 zwhere r1 ∈ k ∗ /k ∗j .78We denote j(α) := j in the cases a), c), d), and j(α) := ∞ in the case b). Then(i(ᾱ)−1)/jν̄(a0 ), aj/x(ᾱ), ξ(ᾱ), x(ᾱ), y(ᾱ), i(ᾱ), i(α), j(α), i, and the elements rq , s1 , s2with the relations defined in the items a)-d) are complete system of invariants withrespect to the conjugation.Moreover, i|j(α), i|(i(α) − 1) (we accept that in the case i = ∞ i|j means, thatj = ∞, i(α) = ∞, i.e. there are no elements with z j , z i(α) ).

i = ∞ if and only ifã0 = 1 + ai(ᾱ)−1 ui(ᾱ)−1 , where ai(ᾱ)−1 = qξ −1 x, q ∈ ”.Let us introduce the additional notation:qa := −ν̄(cj ) mod j, i.e. 0 ≥ qa > −j;qb := qa + min{ν̄(aj ) − qa ; −1};in the case of qb − qa = −1 we denotecb /ca := {resu (cj /aj ) if aj = 0,cb /ca := 1 otherwise };in the case of cb /ca ∈ ” we denoteq1 := 1 if cb /ca ∈ Z,q1 is a denominator of the fraction cb /ca = p1 /q1 , where (p1 , q1 ) = 1, q1 > 0 otherwise;in the last case let us denote by p1 ∈ Z the numerator of this fraction;in the case q1 < j(α), q1 |j(α) we denote byn1 ∈ N a number that satisfies the properties n1 < q1 , q1 |(j(α) − n1 );in the case, when the equation (x + 1)/j = p1 /q1 is solvable, we denote byib ∈ N such a number that ib − qa + 1 is a solution of this equation.Consider the equations0 = −(1 + w)n21 p1 + n1 jw − (q − 1)(2 + w)q1 n1 + q1 (−2 + (1 + w)qa )+q1 [p1 (j(q −1)(w+2)+j −(q −1)2 q1 +(q −1)q1 +2q1 )+2j(qa −1)+(q −1)q1 ((1+w)qa −2)](1.4)(p21 (−1+q−q1 (−1+q−2q 2 +qq1 ))−qq12 (qa −1)+p1 q1 (1−3q−(q−1)qq1 +qqa )) = 0 (1.5)Theorem 1.6 (Theorem III) Let chark = 0 and let ν̄(a0 ) = 0 and ā0 ∈ {ξ(ᾱ)m , m ∈Z} and ᾱn = Id.Then α is conjugate to β defined in one of the following ways, depending on thepossible cases:O) i = ∞.

Thenβ(u) = ξuβ(z) = B0 z79where B0 ∈ k((un )), that is, β has a form of a canonical automorphism of onedimensional local field F ((z)), where F = k((un )) in the appropriate case.O’) i < ∞. Then i = 1 and this case is divided into two ones:I) j > i(α) − 1, i.e. j(α) = ∞β(u) = ξuβ(z) = z + B1 z i(α) + B12 B2 z 2i(α)−1where B1 , B2 ∈ k((un )), B ∈ k((un ))∗ /k((un ))∗(i(α)−1) , i.e. β has the form of acanonical automorphism of one-dimensional local field F ((z)), where F = k((un )).II) j ≤ i(α) − 1 (and in this case j = i(α) − 1). This case has two subsections:A) qb − qa < −1. Thenβ(u) = ξu + ruqa z jβ(z) = z + s1 uqb z j+1 + s2 B2 z 2j+1where B2 = uqb −1 if qa = 0, and 0 otherwise,j(q −1)the numbers r = s1 (cb /ca )−1 ∈ k ∗ /(k ∗ )(j,qa ) , s2 , s1 a /r(i(α)−1)qb ∈ k.B) qb − qa = −1.

This case has two possibilities:1) cb /ca ∈/ ”. Thenβ(u) = ξu + ruqa z jβ(z) = z + s1 uqa −1 z j+1j(qa −1)where numbers r ∈ k ∗ /(k ∗ )(j,qa ) , s1/r(i(α)−1)qb ∈ k.2) cb /ca ∈ ”. This case has three possibilities:a) q1 = j. Thenβ(u) = ξu + ruqa z jβ(z) = z + s1 B1 z j+1 + s2 u−p1 −1+qa z 2j+1 + s3 u−p1 +2qa −2 z 3j+1where B1 = uqa −1 if (x + 1 − qa )/j = p1 /q1 for all x ∈ N,and B1 = uqa −1 + rib uib , ib > qa − 1 otherwise,j(q −1)s2 , s3 , rib ∈ k, r ∈ k ∗ /(k ∗ )(j,qa ) , s1 a /r(i(α)−1)qb ∈ k.80b) q1 < j. Here we have two sub-cases:i) q1 |j.

Thenβ(u) = ξu + ruqa z j−1β(z) = z+s1 B1 z j+1 +(s2,1 u−p1 q1−1sj/q1 u−p1 q1(q1 −j)+2qa −2+s2,2 u−p1 −1+qa )z j+1+q1 +s3 u−2p1 −1+qa z j+1+2q1 +((j/q1 +1)q1 −j)+2qa −2 j+1+(1+j/q1 )q1z+ sq2 Bq2 z j+1+q1 (1+q2 )where B1 = uqa −1 if (x + 1 − qa )/j = p1 /q1 for all x ∈ N andB1 = uqa −1 + rib uib , ib > qa − 1 otherwise,Bq2 = u−p1 (1+q2 )−1+qa if −qa + 1 − (q2 − 1)p1 = 0, and 0 otherwise,j(q −1)s2 , s3 , sj/q1 , sq2 ∈ k, r ∈ k ∗ /(k ∗ )(j,qa ) , s1 a /r(i(α)−1)qb ∈ k.ii) q1 |j. Then we again have two cases:−1i’) n1 |q1 and −q1 n−11 (p1 q1 (n1 − j) + qa ) = −p1 + 1.

Thenβ(u) = ξu + ruqa z j−1β(z) = z + s1 uqa −1 z j+1 + s2 u−p1 q1(n1 −j)+2qa −2 j+1+n1z+ sq Bq z j+1+lq1 ,where Bq = u−p1 q−1+qa if (1.4) is fulfilled and 0 otherwise,l is a solution of an equation (1.4);j(q −1)s2 , sq ∈ k, r ∈ k ∗ /(k ∗ )(j,qa ) , s1 a /r(i(α)−1)qb ∈ k.−1ii’) n1 |q1 or −q1 n−11 (p1 q1 (n1 − j) + qa ) = −p1 + 1. Thenβ(u) = ξu + ruqa z j−1β(z) = z + s1 uqa −1 z j+1 + s2 u−p1 q1(n1 −j)+2qa −2 j+1+n1z+ s3 u−p1 −1+qa z j+1+q1 +s4 u−p1 −1+qa z j+1+2q1 + sq Bq z j+1+lq1where Bq = u−p1 q−1+qa if −qa + 1 − (q − 1)p1 = 0 and 0 otherwise,l is a solution of the equation −qa + 1 − (q − 1)p1 = 0,j(q −1)s2 , sq , s3 , s4 ∈ k, r ∈ k ∗ /(k ∗ )(j,qa ) , s1 a /r(i(α)−1)qb ∈ k.c) q1 > j.

There are two possibilities:i) j |q1 . Thenβ(u) = ξu + ruqa z jβ(z) = z + s1 uqa −1 z j+1 + s2 u−p1 −1+qa z j+1+q1 + s3 u−p1 −2+2qa z 2j+1+q1 +s4 u−2p1 −1+qa z j+1+2q1 + sq Bq z j+1+lq181where Bq = u−p1 q−1+qa if −qa + 1 − (q − 1)p1 = 0, and 0 otherwise,l is a solution of the equation −qa + 1 − (q − 1)p1 = 0,j(q −1)s2 , sq , s3 , s4 ∈ k, r ∈ k ∗ /(k ∗ )(j,qa ) , s1 a /r(i(α)−1)qb ∈ k.ii) j|q1 . It also has two possible cases:i’) −p1 + qa − 2 = q1 (qa − 1)/j.

Thenβ(u) = ξu + ruqa z jβ(z) = z+s1 uqa −1 z j+1 +s2 u−p1 −1+qa z j+1+q1 +s3 u−p1 −2+2qa z 2j+1+q1 +s4 u−2p1 −1+qa z j+1+2q1 +sq Bq z j+1+lq1where Bq = u−p1 q−1+qa if −qa + 1 − (q − 1)p1 = 0, and 0 otherwise,l is a solution of an equation −qa + 1 − (q − 1)p1 = 0,j(q −1)s2 , sq , s3 , s4 ∈ k, r ∈ k ∗ /(k ∗ )(j,qa ) , s1 a /r(i(α)−1)qb ∈ k.ii’) −p1 + qa − 2 = q1 (qa − 1)/j. Thenβ(u) = ξu + ruqa z jβ(z) = z + s1 uqa −1 z j+1 + s2 u−p1 −1+qa z j+1+q1 + sqn ,1 Bqn ,1 z 2j+1+l1 q1 + sqn ,2 Bqn ,2 z 2j+1+l2 q1 +sqm ,1 Bqm ,1 z j+1+l1 q1 + . . .

+ sqm ,w Bqm ,w z j+1+lw q1where Bqn ,i = u−p1 li −2+2qa if (1.5) is satisfied, and 0 otherwise,Bqm ,j = u−p1 lj −1+qa if lq are defined, and 0 otherwise,l1 , l2 are the solutions of equation (1.5), l1 , . . . , lw are solutions of some equation ofdegree w = q1 /j,j(q −1)s2 , sqn ,i , sqm ,j ∈ k, r ∈ k ∗ /(k ∗ )(j,qa ) , s1 a /r(i(α)−1)qb ∈ k.ν̄(a0 ), ξ(ᾱ), i(ᾱ), i(α), j(α), i, qa , qb , q1 , n1 , ib , and those elements with relationsthat were defined in all items are the complete system of invariants with respect to theconjugation.82 Corollary 14 For given ν̄(a0 ), n, i(ᾱ), i(α), j(α), i, qa , qb , q1 , n1 , ib , the set of conjugacy classes of the automorphism α is parametrised by only finite number of parameters,except the cases Th.III O), O’)(I), Th.I i(ᾱ) = ∞.Proof of theorems (and of corollary) Recall that ᾱ is an automorphism onthe field K̄, ᾱ = α mod ℘.It is clear that if two automorphisms α, β are conjugate, then the automorphismsᾱ, β̄ are conjugate in the group Autk (K̄).

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