Multidimensional local skew-fields (792481), страница 20
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It holds from (1.15) that by conjugation the coefficientsaq = 0, q < i(α) − 1. All other arguments from the theorem 2 should be applied herealso, and in the same way as with the case O), we get the case O’) I).Let now j + 1 = i(α) and let f (z) = y0 z, y0 ∈ k((un )). Then we have:αf (z) = α(y0 )α(z) = (y0 +∂(y0 )cj z j + . . .)(z + ai(α)−1 z i(α) + . . .)∂uf α (z) = f (z) + f (a1 z 2 ) + .
. . = y0 z + f (a1 )y02 z 2 + . . .(1.20)j+1∂(y0 )cj + y0 ai(α)−1 . If theSo we get from here that aq = 0, q < j, f¯(aj )y0 = ∂u∂equation ∂u (y0 )cj + y0 ai(α)−1 = 0 isn’t solvable, then dlog(y0 ) = −ai(α)−1 /cj , where/ Z. If it isfrom ν̄(ai(α)−1 /cj ) < −1 or ν̄(ai(α)−1 /cj ) = −1, but res(−ai(α)−1 /cj ) ∈92solvable, then we can consider that ν̄(ai(α)−1 /cj ) = −1, res(ai(α)−1 /cj ) ∈ ”, settingy0 = u.The case j + 1 < i(α) is reduced to a case j + 1 = i(α): indeed, by setting y0 = u,we get that aj = 0 , and it’s the first non equal to zero coefficient in the decompositionα (z), while ν̄(aj /cj ) = −1, res(aj /cj ) ∈ ”.The next part of the proof is following the proof of the theorem 2: in order toprove the rest items of the theorem, we shall go over to conjugations, and as a resultget formulas of the kind (1.17) and (1.19).
However, we cannot completely repeat thearguments from the previous case, because the kernel and cokernel of correspondingmaps Tk,1 and Tk,2 are infinite-dimensional in our case. The formulas (1.17) and (1.19)are now written down as∂∂(ym )A + (m + 1 − i(α))Bym −(B)xm = a∂u∂u(1.21)∂∂(1.22)(A)xm − jym A +(xm )A = b∂u∂uwhere A = cj , B = ai(α)−1 , a and b — arbitrary elements from uk((un )) and k((un ))correspondingly, xm ∈ uk((un )), ym ∈ k((un )). It turns out that the solutionsof this system strongly depend on the properties of the numbers, defined beforethe formulation of the theorem 3.
From now on in the proof we are going to investigate the solvability of this system in dependence from the behaviour of these numbers.mBxm −First of all we note that we can put A = c1 uk , k ≤ 0, |k| < j, c1 ∈ k. Indeed, wewrite down the conjugation f −1 αf , f (z) = y0 z, y0 = uq , where q ≥ 0 is a minimalpositive integer such that qi ≥ ν̄(A), f (u) = x0 , ν̄(x0 ) = 1. Thenαf (u) = α(x0 ) = ξx0 +∂(x0 )cj z j + . . .∂uf α (u) = f (ξu) + f (cj z j ) + . . . = ξx0 + f¯(cj )y0j z j + .
. . ,f¯(cj )∂= ∂u(x0 ). We consider cj = A = c1 uk , kcj∂(x−k+1). We can choose c1 ∈ k so, that the0∂u∂also aj = f¯−1 ( ∂u(y0 )y0−j−1 cj + y0−j aj ).where from we get y0j= −ν̄(y0j /cj ). Then(−k + 1)c1 y0j /cj =solvable. And hereequation would beWe show that in all the cases (except the cases 2) a), 2) b) i) of the theorem) sucha conjugation could be found, that it holds A = c1 uk , B = c2 uk1 . By that it appears,that the coefficients A1 , B1 in all the cases of the theorem have the form, as mentionedabove.Let ν̄(B/A) = −1, res(B/A) = p1 /q1 ∈ ”, (p1 , q1 ) = 1.93We note that q1 doesn’t depend on conjugation. Indeed, A and B change only byconjugation f −1 αf , f (u) = x0 , ν̄(x0 ) = 1, f (z) = y0 z.
But then from the (1.20) follows,that res(B /A ) − res(B/A) ∈ Z, hence we get that q1 doesn’t depend on conjugations.Let us now show that there is such a conjugation, that B/A = res(B/A)u−1 , ifq1 |j or res(B/A) < 0. Therefore we look for a conjugation f , f (u) = x0 , f (z) = y0 z,so that the automorphism α = f −1 αf would have A = c1 uk , k < 0, B = c2 uk−1 . Forthat, considering (1.20), we must solve a system∂f¯(A )y0j =(x0 )cj ,∂u∂f¯(B )y0j+1 =(y0 )cj + y0 aj∂uDividing the first equation by the second, we get: f¯(B /A )y0 =fromc2 −1 ∂x(x0 ) =c1 0 ∂u∂(y )∂u 0y0+ B/A,c1 y0j A−1 =∂(y )+y0 B/A∂u 0∂(x0 )∂u, where∂−1 ∂) (1.23)(x0 )x−k(x1−k0 = (1 − k)∂u∂u 0We look for x0 , y0 in a form of x0 = u(1 + ε1 u + ε2 u2 + .
. .), y0 = uλ (1 + ω1 u + . . .). Letc = c2 /c1 , B/A = cba u−1 + γ0 + γ1 u + . . .. Then from the first equation (1.23) we get:cu−1 + c(ε1 + 2ε2 u + 3ε3 u2 + . . .)(1 + ε1 u + ε2 u2 + . . .)−1 =λu−1 + (ω1 + 2ω2 u + . . .)(1 + ω1 u + . . .)−1 + (cba u−1 + γ0 + γ1 u + . . .)Suppose c = λ + cba = 0 (we can always find such λ ≥ 0). By comparing the coefficientsin the left and right sides, we get linear equations of the formcεi = ωi + i−1 γi−1 + i−1 ψi ,where ψi — certain polynomial from εq , ωq , q < i (they are determined from theprevious equations).
From the second equation we get:−ka= u−k + (2 − k)ε1 u1−k + (3 − k)(ε2 + . . .)u2−k + . . .c1 uλj (1 + ω1 u + . . .)j c−1a u(where A = ca uka ). Suppose c1 = ca , −k = λj − ka . Then k ≤ 0. Because of cba = 0, wecan put λ = 0. Hence k = ka > −j. Comparing the coefficients, we get linear equationsof the form(1.24)jωi = (i + 1 − k)εi + ψ̃i = (i + 1 + λj − ka )εi + ψ̃iFor every i the system has a solution, if (i + 1 − ka )/j = cba = p1 /q1 , what holds truealways under the condition that q1 |j or cba = res(B/A) < 0. If these conditions arenot fulfilled, then B can have the form B = c2 uk−1 + cib uib , what is evident from thearguments, mentioned above.If res(B/A) ∈/ ”, then applying the same thoughts, we also get the same result.94Let ν̄(B/A) < −1.Then we look for B in the form B = c2 uk+ν̄(B/A) .
System (1.23) will now have theform∂(y0 )∂ν̄(B/A) ∂+ B/A, c1 y0j A−1 = (1 − k)−1 (x1−k)cx0(x0 ) = ∂u∂uy0∂u 0Hencecuν̄(B/A) + cuν̄(B/A)+1 (ε1 + 2ε2 u + . . .)(1 + ε1 u + . . .)−1 =λu−1 + (ω1 + 2ω2 u + . . .)(1 + ω1 u + . . .)−1 + (cba uν̄(B/A) + γν̄(B/A)+1 uν̄(B/A)+1 + . .
.)whence c = cba and equation i looks like following:ciεi = ωi+ν̄(B/A)+1 (i + ν̄(B/A) + 1) + ψiwhere ωi+ν̄(B/A)+1 = 0 if (i + ν̄(B/A) + 1) ≤ 0. Equations (1.24) are written overwithout changes, where from we get that every system i is solvable, and our propositionis proved.Let us now go back to a system (1.21), (1.22). We show, that system of the equations(1.21), (1.22) is solvable, if m = j. It holds:ym = (mB/(jA) −∂∂(A)/(jA))xm +(xm )/j − b/(jA)∂u∂u(1.25)Hence∂∂∂∂2(xm ) +(xm )((2m − j)B/A −(A)/A) + xm ((m − j) (B)B/(BA)−2∂u∂u∂u∂u∂∂2∂∂2(2m − j) (A)B/A − ( 2 (A) (A))/( (A)A)+∂u∂u∂u∂u∂∂( (A))2 /A2 + (m − j)mB 2 /A2 ) −(b/A) − (m − j)Bb/A2 − ja/A = 0∂u∂uWe set q = ν̄(B/A).
From this: q ≤ −1, if q = −1, then res(B/A) ∈/ Z.Let us show that the equation(1.26)∂2∂∂∂(xm )((2m − j)B/A −(A)/A) + xm ((m − j) (B)B/(BA)−(xm ) +2∂u∂u∂u∂u(2m − j)(∂∂2∂∂(A)B/A2 − ( 2 (A) (A))/( (A)A)+∂u∂u∂u∂u∂(A))2 /A2 + (m − j)mB 2 /A2 ) = cuk∂u95mod ℘¯k+1(1.27)is solvable, if q < −1 or q = −1, but res(B/A) ∈/ ”, for all k ∈ Z and every constantc ∈ k. From here we immediately get the solvability of the equation (1.26) for all b anda, and also of a system (1.21), (1.22). From here will follow the proof of the items A),B) 1).If q < −1, then∂(A)/A) = ν̄(B/A) (if 2m = j) and ≥ −1 (if 2m = j),ν̄((2m − j)B/A − ∂u∂∂∂2∂∂ν̄((m − j) ∂u (B)B/(BA) − (2m − j) ∂u(A)B/A2 − ( ∂u2 (A) ∂u (A))/( ∂u (A)A) +∂222222( ∂u (A)) /A + (m − j)mB /A ) = ν̄(B /A ), because m = j.
Thus,ν̄(∂2∂∂∂(xm )((2m − j)B/A −(A)/A) + xm ((m − j) (B)B/(BA)−(xm ) +2∂u∂u∂u∂u∂∂2∂∂∂(A)B/A2 −( 2 (A) (A))/( (A)A)+( (A))2 /A2 +(m−j)mB 2 /A2 ) =∂u∂u∂u∂u∂u2∂∂ν̄( 2 (xm ) +(xm )(2m − j)c1 uq + xm (m − j)mc2 u2q ) = ν̄(xm (m − j)mc2 u2q )∂u∂uwhere from immediately follows solvability of the equation (1.27).If q = −1,then we put qa = ν̄(A), qb = ν̄(B), k = ν̄(xm ), x = res(B/A). And now for thesolvability of the equation (1.27) is necessary to show that the equation(2m−j)k(k − 1) + k(2m − j)x − kqa + ((m − j)qb − (2m − j)qa )x + qa + (m − j)mx2 = 0 (1.28)doesn’t have a solution.a, − k−1(and if m = j, then oneThis quadratic equation has the critical points − k−qm−jmk−1of the points is − m ), so if res(B/A) ∈/ ”, then our assertion is proved.
Moreover, in thecase when q = −1, res(B/A) ∈/ ” we have proved the solvability of the equation (1.26),and through that also of a system (1.21), (1.22) for all m, by this proving the case B) 1).If m = j, q < −1, qa = 0, then the equation (1.27) has the form∂2∂∂∂(xm ) +(xm )(mB/A −(A)/A) + xm (−mB (A)/A2 −2∂u∂u∂u∂u∂∂∂∂2(A) (A))/( (A)A) + ( (A))2 /A2 = cuk mod ℘¯k+1 ,2∂u∂u∂u∂u∂that is always solvable, because ν̄(xm B ∂u (A)/A2 ) < ν̄(xm B/A) < ν̄(xm ).If qa = 0, then this equation isn’t solvable with k = qa − 1 + q. Thus,if q < −1, α is the conjugation to automorphism β: β(u) = ξu + Auj ,β(z) = z + Bz i(α) + cuν̄(A)−1+q z 2i(α)−1 (see case A)).(96The case res(B/A) ∈ ” should be studied precisely.
Recall that in this caseν̄(B/A) = −1.Let res(B/A) = p1 /q1 (= cb /ca ), (p1 , q1 ) = 1. The following proof of the theoremwould be divided into three cases (which do not coincide with the corresponding casesfrom the formulation of the theorem), in order to make the proof easier:a) q1 |j, q1 = jb) q1 |jc) q1 = j.a) (see the case B) 2) b) i)).a2 −1= − km= pq11 . Then there exist c1 , c2 ∈ k such that the equationHere − km11−q−j1(1.26) + c1 uk1 −2 + c2 uk2 −2 = 0 has solutions with m = m1 , what follows from the solvability of the equation (1.27) for all k, except of k = k1 − 2, k = k2 − 2, and m1 — is thefirst index, when the system (1.21), (1.22) isn’t solveable in a general case.
Also in thiscase the space of solutions of the homogeneous equation (1.26) is generated by x1 andx2 , ν̄(x1 ) = k1 , ν̄(x2 ) = k2 . Thus, automorphism α is conjugate to the automorphism α ,α (u) = ξu + Az j + cj+1+2m1 z j+1+2m1 + . . .,α (z) = z + Bz j+1 + B2 z j+1+m1 + . . ., where B2 = c1 uk1 −2+qa + c2 uk2 −2+qa .Now let us investigate behaviour of the values k1,mq , k2,mq for different mq , for whichthe equation (1.26) has no solutions, where k1,mq , k2,mq are solutions of the equation(1.28).Obviously, mq = qq1 , q ∈ N. Note that (k1,mq − k2,mq ) doesn’t depend on mqkkq −qaq −1(q = j/q1 ). Indeed, 1,m= 2,m= − pq11 . Hence k2,mq = −p1 q + 1, k1,mq =mq −jmq−p1 (q − j/q1 ) + qa , and (k1,mq − k2,mq ) = p1 j/q1 + qa − 1.
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