Multidimensional local skew-fields (792481), страница 21
Текст из файла (страница 21)
We observe, thatk2,mq = k2,mq−1 + k2,m1 − 1, k1,mq = k1,mq−1 + k2,m1 − 1 = k2,mq−1 + k1,m1 − 1.We write down the formula (13) for the case, whenym1 = ω1 ym1 ,1 + ω2 ym1 ,2 , xm1 = ω1 xm1 ,1 + ω2 xm1 ,2 ,where ω1 , ω2 ∈ k, xm1 , ym1 are solutions of the homogeneous system (1.21), (1.22)for m = m1 .
Because of ν̄(xm1 ,1 ) = k1,m1 , ν̄(xm1 ,2 ) = k2,m1 , we have ν̄(ym1 ,1 (xm1 ,1 )) =k1,m1 −1, ν̄(ym1 ,2 (xm1 ,2 )) = k2,m1 −1. Indeed, from the formula (1.25), ν̄(ym ) = ν̄(xm )−1,if mp1 /q1 − qa + k = 0, where k = ν̄(xm ). Let be mp1 /q1 − qa + k = 0. Then p1 /q1 =k−q− 1,mm11 a , whence j = 0. It’s a contradiction. Analogously for k2,m1 qa = 1, but qa ≤ 0,also a contradiction. So we have:α f (z) = z + Bz i(α) + B2 z i(α)+m1 + a2m1 +i(α)−1 z 2m1 +i(α) + .
. . + ym1 z m1 +1 +∂(ym1 )Az m1 +i(α) + (m1 + 1)Bym1 z m1 +i(α) + (m1 + 1)B2 ym1 z 2m1 +i(α) + . . .∂u∂f α (z) = z + ym1 z m1 +1 + Bz i(α) + B2 z i(α)+m1 + B3 z m2 +i(α) +(B)xm1 z m1 +i(α) +∂u97∂∂222(B)x2m1 z 2m1 +i(α) +Ci(α)ymBz 2m1 +i(α) + (B2 )xm1 z i(α)+2m1 +12∂u∂u∂(i(α) + m1 )ym1 B2 z 2m1 +i(α) +(B)i(α)xm1 ym1 z 2m1 +i(α) + a2m1 +i(α)−1 z 2m1 +i(α) + . . .∂u∂(xm1 )Az m1 +j +α f (u) = ξu + Az j + c2m1 +j z 2m1 +j + .
. . + ξxm1 z m1 +∂um1 Bxm1 z m1 +j + m1 B2 xm1 z 2m1 +j + . . .i(α)ym1 Bz m1 +i(α) +2−1∂∂2(A)xm1 z m1 +j +jym1 Az m1 +j +2−1 2 (A)x2m1 z 2m1 +j +∂u∂u∂2Cj2 ymAz 2m1 +j +(A)jxm1 ym1 z 2m1 +j + c2m1 +j z 2m1 +j + . . .1∂uAs m1 < j, in the expression for α f (z), α f (u) there is the only term at z 2m1 +i(α) .For ymq = ω1 ymq ,1 + ω2 ymq ,2 , xmq = ω1 xmq ,1 + ω2 xmq ,2 the formula 1.16 should havethe formf α (u) = ξu+ξxm1 z m1 +Az j +α f (z) = z+Bz i(α) +B2 z i(α)+m1 +B3 z i(α)+m2 +amq +m1 +i(α)−1 z mq +m1 +i(α) +.
. .+ymQ z mq +1 +∂(ymq )Az mq +i(α) + (mq + 1)Bymq z mq +i(α) + (mq + 1)B2 ymq z mq +m1 +i(α) +∂u(mq + 1)B3 ymq z mq +m2 +i(α) + . . .f α (z) = z + ymq z mq +1 + Bz i(α) + B2 z i(α)+m1 + B3 z i(α)+2m1 +∂(B)xmq z mq +i(α) + i(α)ymq Bz mq +i(α) +∂u∂(B2 )xmq z i(α)+m1 +mq + (i(α) + m1 )ymq B2 z m1 +mq +i(α) + amq +m1 +i(α)−1 z m1 +mq +i(α) +∂u∂(1.29)(B3 )xmq z i(α)+m2 +mq + (i(α) + m2 )ymq B3 z m2 +mq +i(α) + . . .∂u∂α f (u) = ξu + Az j + cmq +j+m1 z mq +j+m1 + .
. . + ξxmq z mq +(xmq )Az mq +j +∂umq +jmq +m1 +jmq Bxmq z+ mq B2 xmq z+ ...f α (u) = ξu+ξxmq z mq +Az j +∂(A)xmq z mq +j +jymq Az mq +j +cmq +m1 +j z mq +m1 +j +. . .∂uWhence follows:2−1 (2−1 (∂2(B)x2m1 ) = Cq2a −1 (cb uqa −3 + . . .)(ω12 x21 + 2ω1 ω2 x1 x2 + ω22 x22 )∂u2∂2(A)x2m1 ) = Cq2a (ca uqa −2 + . . .)(ω12 x21 + 2ω1 ω2 x1 x2 + ω22 x22 )∂u298(1.30)2ymB = (cb uqa −1 + . .
.)((ω12 y12 + 2ω1 ω2 y1 y2 + ω22 y22 )12A = (ca uqa + . . .)((ω12 y12 + 2ω1 ω2 y1 y2 + ω22 y22 )ym1∂(B)i(α)xm1 ym1 = (qa − 1)i(α)(cb uqa −2 + . . .)(ω12 x1 y1 + ω22 x2 y2 + ω1 ω2 (x1 y2 + x2 y1 ))∂u∂(A)jxm1 ym1 = qa j(ca uqa −1 + . . .)(ω12 x1 y1 + ω22 x2 y2 + ω1 ω2 (x1 y2 + x2 y1 )) (1.31)∂uymq B2 = ω1 c1 ymq ,1 uk1,mq −2+qa +ω1 c2 ymq ,1 uk2,mq −2+qa +c1 ω2 ymq ,2 uk1,mq −2+qa +c2 ω2 ymq ,2 uk2,mq −2+qaxmq B2 = ω1 c1 xmq ,1 uk1,mq −2+qa +ω1 c2 xmq ,1 uk2,mq −2+qa +c1 ω2 xmq ,2 uk1,mq −2+qa +c2 ω2 xmq ,2 uk2,mq −2+qa(1.32)∂( (B2 ))xmq = (k1,mq −2+qa )ω1 c1 xmq ,1 uk1,mq −3+qa +(k2,mq −2+qa )ω1 c2 xmq ,1 uk2,mq −3+qa +∂u(k1,mq − 2 + qa )c1 ω2 xmq ,2 uk1,mq −3+qa + (k2,mq − 2 + qa )c2 ω2 xmq ,2 uk2,mq −3+qa(1.33)Let (k1,mq − k2,mq ) < 0.
We shall show, that in formulas (1.30)- (1.33) monomialswith valuation (k1,mq + k1,m1 + qa − 3), belong to the image of the map (1.27), it meansthat the equation (1.27) with the right side in a form of these monomials is solvable.Indeed, in a case of (k1,mq −k2,mq ) < 0 we have p1 /q1 < (−qa +1)/j < (−qa +1+i)/j,if i ≥ 1. But then A = c1 uqa , B = c2 uqb = c2 uqa −1 , and ymq = ω1 uk1,mq −1 + ω2 uk2,mq −1 ,xmq = ω1 uk1,mq +ω2 uk2,mq , because all the coefficients of the homogeneous system (1.21),(1.22) have the monomial form.
Since (k1,mq + k1,m1 − 1) < k1,mq + k2,mq − 1 = k1,mq+1 <k2,mq+1 , the equation (1.27) has monomial solutions of a form, mentioned above.If (k1,mq − k2,mq ) > 0, so (k1,mq + k1,m1 − 1) > k1,mq + k2,m1 − 1 = k1,mq+1 > k2,mq+1 ,where from follows the same result.If (k1,mq − k2,mq ) = 0, then ymq and xmq consist of the only monomial, i.e. ω1 = 0and expressions in (1.30)-(1.33) are simplified to the one monomial, which is in thegeneral case not in the image of the map (1.27).Now we show that for all q except q = 1, q = j/q1 , q = (1 − qa )/p1 + 1 there existthe coefficients ω1 , ω2 (coefficient ω2 , if k1,mq = k2,mq ) are such that∂∂(c)/j − cm1 +mq +j ∂u(A)/(jA) + (mq+1 −amq +m1 +i(α)−1 +∂u mq +m1 +j∂−1 j)p1 u cmq +m1 +j /(jq1 ) + (1 − i(α))B2 ymq − ∂u (B2 )xmqbelongs to the image of the map (1.27), i.e.
the equation (1.27) with the right side inthe form of these expressions is solvable.According to (1.32), (1.33), we need to show that(1−i(α)+mq−1 )c2 ω2 ymq ,2 uk2 −2+qa −(k2,m1 +qa −2)c2 ω2 xmq ,2 uk2 −3+qa +∂∂(b)/j+b (A)/(jA)∂u∂u+(mq+1 − j)p1 u−1 b/(jq1 ) = 0 mod ℘¯k2,mq+1 +qa −199where b = mq c2 ω2 xmq ,2 uk2,m1 −2+qa , only if (q − 1)q1 (1 − qa − (q − 1)p1 ) = 0 and(1−i(α)+mq−1 )ω1 c2 ymq ,1 uk2 −2+qa −(k2,m1 −2+qa )c2 ω1 xmq ,1 uk2 −3+qa +∂∂(b)/j+b (A)/(jA)∂u∂u+(mq+1 − j)p1 u−1 b/(jq1 ) = 0 mod ℘¯k1,mq+1 +qa −1where b = mq c2 ω1 xmq ,1 uk2,m1 −2+qa only if −(q − 1)2 q1 p1 + (q − 1)(qa − 1)q1 + (1 −j/q1 )(jp1 + q1 (qa − 1)) = 0 (we remark that ω2 does not depend on ω1 ).p −q +kSince ymq ,1,2 = 1 a j 1,2,mq xmq ,1,2 u−1 + .
. ., it is necessary to show that−(1 − i(α) + mq−1 )mq (k2,mq+1 − 1 + qa )p1 − qa + k2,mq− (k2,m1 − 2 + qa ) +−jjmq qa (mq+1 − j)p1 q+= 0jjif (q − 1)q1 (1 − qa − (q − 1)p1 ) = 0,−(1 − i(α) + mq−1 )mq (k1,mq+1 − 1 + qa )p1 − qa + k1,mq− (k2,m1 − 2 + qa ) +−jjmq qa (mq+1 − j)p1 q+= 0jjif −(q − 1)2 q1 p1 + (q − 1)(qa − 1)q1 + (1 − j/q1 )(jp1 + q1 (qa − 1)) = 0. But−(1−i(α)+mq−1 )mq (k2,mq+1 − 1 + qa ) mq qa (mq+1 − j)p1 qp1 − qa + k2,mq−(k2,m1 −2+qa )+−+jjjj=−(1−i(α)+mq−1 )=(q − 1)q1 (1 − qa − (q − 1)p1 ),jmq (k1,mq+1 − 1 + qa ) mq qa (mq+1 − j)p1 qp1 − qa + k1,mq−(k2,m1 −1+qa )+−+jjjj−(q − 1)2 q1 p1 + (q − 1)(qa − 1)q1 + (1 − j/q1 )(jp1 + q1 (qa − 1))jWe observe here, that −(q − 1)2 q1 p1 + (q − 1)(qa − 1)q1 + (1 − j/q1 )(jp1 + q1 (qa − 1)) = 0.In fact, if −(q − 1)2 q1 p1 + (q − 1)(qa − 1)q1 + (1 − j/q1 )(jp1 + q1 (qa − 1)) has solutionsin integers, then its discriminant must be equal to q12 l2 , where l ∈ Z. But D = (qa −1)2 q12 + 4(qa − 1)q1 p1 (q1 − j) + 4p21 j(q1 − j), whence follows, that j(q1 − j) = (q1 − j)2 ,what is wrong.So, we have shown that α is conjugated to α :α (u) = ξu + Az j ,100α (z) = z + Bz i(α) + B2 z i(α)+q1 + B3 z i(α)+2q1 + Bj/q1 z i(α)+(1+j/q1 )q1 + Bq2 z i(α)+q1 (1+q2 ) ,k−2+qawhere Bj/q1 = cu 1,mj/q1 +1, Bq2 = c̃uk2,m1+q2 −2+qa if q1 (1 − qa − (q − 1)p1 ) = 0 andBq2 = 0 otherwise.Let’s show now, that B3 can be taken as cb3 uk2,m2 −2+qa .
In order to do that, weexhibit that in formulas (1.30)-(1.33) monomials with ω22 belong to the image of themap (1.27). Then the case q = 1 is equivalent to a general case, and since q = 1 is oneof the solutions of (q − 1)q1 (1 − qa − (q − 1)p1 ) = 0, B3 is defined in the same way asBq . For that, according to (1.30)-(1.33), we must show that∂∂(b)/j − b (A)/(jA) + (mq+1 − j)p1 u−1 b/(jq1 ) + a = 0∂u∂u2where b = Cj2 ω22 y2,mc q /p1 u2k2,m1 −2+qa + Cq2a cb q1 /p1 ω22 u2k2,m1 −2+qa + qa jcb q1 /p1 ω22 (p1 −1 b 12(p1 − qa + k2,m1 )2 /j 2 + (qa − 1)i(α)(p1 − qa + k2,m1 )/j.qa + k2,m1 )/j, a = Cq2a −1 + Ci(α)In fact,Cj2qa jqa j(p1 − qa + k2,m1 )2 q12 q1 (2k2,m1 − 2 + qa )(2k−2+q)+C+2,ma1qaj 3 p1p1 jq1 (p1 − qa + k2,m1 )qa 2 (p1 − qa + k2,m1 )2 q1q1(2k−2+q)−+ Cq2a +(Cj2,m1a22p1 jjj p1p1q1 (p1 − qa + k2,m1 ) m2 − j 2 (p1 − qa + k2,m1 )2p1 − qa + k2,m1+Cq2a +qa j)+(Cj)+Cq2a −1 +2p1 jjjj2Ci(α)(p1 − qa + k2,m1 )2 /j 2 + (qa − 1)i(α)(p1 − qa + k2,m1 )/j = 0,and it proves our assumption.The case k1,mq = k2,mq is more simple, and all the arguments remain true.b)1 −qaIn this case the system − km−j= pq11 , − k2m−1 = pq11 is incompatible, that is why forall m the ”cokernel” of the map (1.27) is one-dimensional, A = c1 uqa , B = c2 uqa −1 .a= pq11 , and as k2,mq theLet denote as k1,nq the solution of the equation − kn1q−q−j2 −1solution of the equation − km= pq11 .
It is clear that mq = qm1 = qq1 , as in the case a),qand nq+1 = nq +m1 only if nq +m1 = j. But in this case the next value of nq+1 is nq +2m1 ,so we consider this recurrence relation to be true always. Further, (k1,nq −k2,mq ) doesn’tdepend on q, as in a), and k2,mq+1 = k2,mq + k2,m1 − 1, k1,nq+1 = k1,nq + k2,m1 − 1.The proof, that follows, would be divided into three cases:1) q1 < j (see case B) 2) b) ii)),2) q1 > j, j |q1 (see case B) 2) c) i)),3) j|q1 (see case B) 2) c) ii)).We put B2 = c1 uk1,n1 −2+qa , B3 = c2 uk2,m1 −2+qa , if n1 < m1 , and B2 = c1 uk2,m1 −2+qa ,B3 = c2 uk1,n1 −2+qa otherwise.101In the case 1)n1 , m1 < j, n1 < m1 .
The idea of the following proof is the following: we look forthe sequentional conjugations fnq and fmq , where fnq (u) = u + xnq z nq , fnq (z) = z +ynq z nq , and xnq , ynq are solutions of the homogeneous system (1.21), (1.22) with m =nq ; fmq (u) = u + xmq z mq , fmq (z) = z + ymq z mq , where xmq , ymq are solutions of thehomogeneous system (1.21), (1.22) with m = mq . We choose xnq , ynq , xmq , ymq so thatwith these conjugations the equation (1.26) become solvable with m = nq+1 , m = mq+1 .For the automorphism fmq we can use the results from the case a), because xmqand ymq here have the form ω2 xmq ,2 , ω2 ymq ,2 , ν̄(ymq ,2 ) = k2,mq − 1, ν̄(xmq ,2 ) = k2,mq ,and m1 = q1 < j, j = mq .
We rewrite for fnq the formula (1.29):if q > 1, thenα fnq (u) = ξu + Az j + cnq +j+m1 z nq +j+m1 + . . . + ξxnq z nq +∂(xnq )Az nq +j +∂unq Bxnq z nq +j + nq B2 xnq z nq +n1 +j + nq B3 xnq z nq +m1 +j + . . .∂(A)xnq z nq +j + jynq Az nq +j +∂u+ cnq +m1 +j z nq +m1 +j + .
. .fnq α (u) = ξu + ξxnq z nq + Az j +cnq +n1 +j z nq +n1 +j(since nq > m1 , there are no more terms with z nq +m1 +j ),α fnq (z) = z+Bz i(α) +B2 z i(α)+n1 +B3 z i(α)+m1 +anq +m1 +i(α)−1 z nq +m1 +i(α) +. . .+ynq z nq +1 +∂(yn )Az nq +i(α) +(nq +1)Bynq z nq +i(α) +(nq +1)B2 ynq z nq +n1 +i(α) +(nq +1)B3 ynq z nq +m1 +i(α)∂u q∂fnq α (z) = z+ynq z nq +1 +Bz i(α) +B2 z i(α)+m1 +B3 z i(α)+n1 + (B)xnq z nq +i(α) +i(α)ynq Bz nq +i(α) +∂u∂∂(B2 )xnq z i(α)+n1 +nq + (i(α) + n1 )ynq B2 z n1 +nq +i(α) +(B3 )xnq z i(α)+m1 +nq +∂u∂u(i(α) + m1 )ynq B3 z m1 +nq +i(α) + anq +n1 +i(α)−1 z nq +n1 +i(α) + anq +m1 +i(α)−1 z nq +m1 +i(α) + . . .The formula remains true for q = 1 also, as it is seen from the calculations, similar toq = 1 in case a).If n1 |m1 and k1,nq + (m1 /n1 − 1)k1,n1 = k2,mq , i.e. m1 /n1 k1,n1 = k2,m1 , then thecoefficient anq +m1 +i(α)−1 (cnq +m1 +i(α)−1 ) depends on anq +n1 +i(α)−1 (cnq +n1 +i(α)−1 ) in ageneral case.
In this situation for almost all q the conjugation fnq can be chosen so thatthe equation (1.26) is solvable for m = mq , and fmq so that equation (1.26) is solvablefor m = nq+1 .Thus, the arguments, similar to the case a), tell us that α is conjugated to β:β(u) = ξu + Az j ,β(z) = z + Bz i(α) + B2 z i(α)+n1 + Bq z i(α)+mq + Bq2 z i(α)+nq2 ,102where Bq = cuk2,mq −2+qa or equals to zero. It depends on that, if at least one expressionfrom∂∂(b)/j − b (A)/(jA) + (mq+1 − j)p1 u−1 b/(jq1 ) + a∂u∂uequals to zero or not, with b = uk1,nq +wk1,n1 −2+qa nq , a = (k1,n1 −2+qa )uk1,nq +wk1,n1 −3+qa +(1−i(α)+mq−1 )uk1,nq +wk1,n1 −3+qa (p1 −qa +k1,nq )/j, w has the values from 1 to q1 /n1 −1,in other words, if equals to zero at least one of expressions−(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)](1.34)k1,nq2 −2+qaor zero, in accordance with equality to zero of the expresFurther, Bq2 = cusion∂∂(b)/j − b (A)/(jA) + (nq2 − j)p1 u−1 b/(jq1 ) + a∂u∂uwith b = nq2 −1 uk2,mq2 −1 +k1,n1 −2+qa , a = (k1,n1 − 2 + qa )uk2,mq2 −1 +k1,n1 −3+qa + (1 − i(α) +k+k−3+qa(p1 − qa + k2,mq2 −1 )/j, i.e.mq2 −2 )u 2,mq2 −1 1,n1− (q2 − 2)2 q12 p1 + (q2 − 2)q1 jp1 + p1 j(j − n1 ) + (3j + n1 )q1 (qa − 1)(1.35)We note, that this equation doesn’t have solutions in integers, i.e.
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
