Multidimensional local skew-fields (792481), страница 14

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Therefore, ap3 commutes with z, from here follows that a3is a purely inseparable element and F (a3 ) is an ”unramified” extension.The corollary is proved.2This corollary concludes the proof of theorem 0.36.0.5Classes of conjugate elementsLet K be a splittable local skew field of characteristic 0 whose first residue skew fieldis commutative and whose last residue skew field k is contained in its centre. We haveclassified these skew fields in the preceding section.

In this section we give necessaryand sufficient conditions for two elements of K to be conjugate.We fix a representation of K in the form k((u))((z)).Definition 0.56 Let α = Id. A residue resi,r on K is defined to be a map resi,r :k((u))((z)) → kxiresi,r (X) = res δi duuwhere X = l xl z l .Proposition 0.57 Let α = Id. Let L, M ∈ K, ν(L) = ν(M ) = −1,M = b−1 z −1 + b0 + b1 z + . .

.,L = a−1 z −1 + a0 + a1 z + . . ..The following assumptions are equivalent:61(i) there is an S ∈ K, ν(S) = 0, such that M = S −1 LS(ii) a−1 = b−1 , a0 = b0 , . . . , ai−2 = bi−2 ;resai−1 − bi−1du ∈ Z anduδi a−1uai−1 − bi−1∈ k[[u]]uδi a−1−1resi,r (M j ) = resi,r (Ljj ) for all j ≥ 1, where Lj = S˜j Lj−1 S˜j , L0 := L, S˜j =S˜j (M, Lj−1 ).Proof K has the form k((u))((z)) with the relation zuz −1 = u + uδi z i + .

. .. Thuswe have:SM = s0 b−1 z−1+ (s0 b0 + s1 b−1 ) + . . . + (i−2bj si−2−j )zi−2+(j=−1−1LS = s0 a−1 z +(s0 a0 +s1 a−1 )+. . .+(i−2i−1bj si−1−j )z i−1 + . . .j=−1aj si−2−j )zi−2j=−1i−1δi+(−a−1 s0 +aj si−1−j )z i−1 +. . .j=−1It follows that the condition a−1 = b−1 , a0 = b0 , . . . , ai−2 = bi−2 is necessary for Mand L to be conjugate.

Another necessary condition is given by the following equationfor s0 :sδ0iai−1 − bi−1=s0a−1Since δi is a differentiation, we have∂s∂u 0s0=ai−1 − bi−1uδi a−1Thus we obtain the second necessary condition:resai−1 − bi−1du ∈ Z anduδi a−1uai−1 − bi−1∈ k[[u]]uδi a−1Conversely, if these two conditions hold, then there is an s0 ∈ k((u)) such that thefirst i + 1 summands in L1 = s−10 Ls0 are the same as those in M . It is clear that Land M are conjugate if and only if L1 and M are conjugate. The conjugating elementS̃ has the form 1 + · · · (S̃ can be written as (1 + s1 z)(1 + s2 z 2 ) . .

..) Note that for everyx−1 z −1 + x0 + x1 z + . . . ∈ K holds:(1 + sj z j )−1 (x−1 z −1 + x0 + x1 z + . . .)(1 + sj z j ) = x−1 z −1 + x0 + x1 z + . . . + xi+j−2 z i+j−2 +isj + x−1 sδji )z i+j−1 + . . .(xi+j−1 + jxδ−162since the proof of lemma 0.11, (ii) implies that(1 + sj z j )−1 (x−1 + x0 z + x1 z 2 + . . .)(1 + sj z j ) = x−1 + x0 z + . . . + xi+j−2 z i+j−1 +isj )z i+j + . .

.),(xi+j−1 + jxδ−1and(1+sj z j )−1 z −1 (1+sj z j ) = (1+sj z j )−1 (z −1 +sj z j−1 −sδji z i+j−1 +. . .) = z −1 −sδji z i+j−1 +. . .It follows that(s1 a−1 )δi = bi − ai(j = 1),if M = S̃ −1 L1 S̃, where ai is the coefficient of L1 . This equation is soluble if and only ifresb i − aidu = 0,uδithat is, resi,r (M ) = resi,r (L1 ).Conversely, if the residues are equal then there is an s1 ∈ k((u)) such that the firsti + 2 summands in L2 = (1 + s1 z)−1 L1 (1 + s1 z) are the same as those in M .Proceeding by induction, we obtain at the kth step that if M = S̄ −1 Lk S̄, theni+ a−1 sδki = bi+k−1 − ai+k−1 .ksk aδ−1To solve this equation, we substitute sk = a−k−1 s into it and obtain the equations = ak−1−1bi+k−1 − ai+k−1,uδiak−1 awhich is solvable if and only if res −1 uδi+k−1= resiikcoefficient of z in M has the formu−r ak−1−1 bi+k−1.u δiOn the other hand, thekak−1−1 bi+k−1 + fMwhere fM is a polynomial in bi+k−2 , .

. . , b−1 and the values of δj at these points. Thecorresponding coefficient in Lkk has the formkak−1−1 ai+k−1 + fLkand fLk = fM , since aj = bj for j ≤ i + k − 2. It follows that resi,r Lkk = resi,r M k if andonly if res2ak−1−1 ai+k−1u δi= resak−1−1 bi+k−1.u δiwhich completes the proof of the proposition.63Definition 0.58 Let α = Id. We say that the residue resα ofX =zero, ifx0 ∈ im(α − Id)lxl z l is equal toWe say that two elements have the same residue if the residue of their difference isequal to zero.−1We define ϕ : k((u)) → k((u)), ϕ(x) = xα /x.Proposition 0.59 Let α = Id.

Let L, M ∈ K, ν(L) = ν(M ) = −1,M = b−1 z −1 + b0 + b1 z + . . .,L = a−1 z −1 + a0 + a1 z + . . ..The following conditions are equivalent:(i) there exists an S ∈ K, ν(S) = 0, such that M = S −1 LS(ii) b−1 /a−1 ∈ imϕ;−1resα (M j ) = resα (Ljj ) for all j ≥ 1, where Lj = S˜j Lj−1 S˜j , L0 := L, S˜j =S˜j (M, Lj−1 ).Proof is similar to that of the preceding proposition. We haveSM = s0 b−1 z −1 + (s0 b0 + s1 bα−1 ) + . .

.−1−1LS = a−1 sα0 z −1 + (a0 s0 + a−1 sα1 ) + . . .−1Therefore, s0 b−1 = a−1 sα0 , that is b−1 /a−1 ∈ imϕ. If this condition holds, then we putL1 = s−10 Ls0 . The first coefficients in L1 and M are equal.Now we observe that(1 + sj )−1 (x−1 z −1 + x0 + x1 z + . . .)(1 + sj z j ) = x−1 z −1 + . . .j−1+xj−2 z j−2 + (xj−1 + sj xα−1 − x−1 sαj )z j−1 + . . .for any x−1 z −1 + x0 + x1 z + .

. . ∈ K, which follows from the calculation in the proof ofLemma 0.11, (i).The arguments used in the proof of the preceding proposition yield at the first stepthe following condition that is necessary for conjugacy:s1 aα−1 − a−1 sα1−1−1−1= α(sα1 a−1 ) − (sα1 a−1 ) = b0 − a0This equation is soluble if and only if (b0 − a0 ) ∈ im(α − Id).

which is equivalent tothe equality resα M = resα L1 .64At the jth step we have the conditionjsj aα−1 − a−1 sαj−1= aj−1 − bj−1Hence,2j−12jj−1(aα−1 aα−1 . . . aα−1 )(aj−1 − bj−1 ) = (aα−1 aα−1 . . . aα−1 )sj − (a−1 . . . aα−1 )sαjj−1−1j−1α((a−1 . . . aα−1 )sαj ) − (a−1 . . . aα−1 )sαj−1=−1j−1This equation is soluble if and only if (a−1 . . .

aα−1 )(aj−1 − bj−1 ) ∈ im(α − Id), whichis equivalent to the equality resα (M j ) = resα (Ljj ), since the first (j − 1) coefficientsin Lj are equal to the corresponding coefficients in M , and the coefficient of the 0thpower of z in M j is−j+2a−1 . . . aα−1−j+1bαj−1j−1+ bj−1 aα−1 . . . aα−1 +a sum of monomials with indices < j − 1The corresponding coefficient in Ljj is−j+2a−1 .

. . aα−1−j+1aαj−1j−1+ aj−1 aα−1 . . . aα−1 +a sum of monomials with indices < j − 1Hence,−j+2(a−1 . . . aα−1−j+1bαj−1−j+2− a−1 . . . aα−1−j+2([a−1 . . . aα−1−j+1aαj−1−j+1bαj−1j−1j−1+ bj−1 aα−1 . . . aα−1 − aj−1 aα−1 . . . aα−1 ) =−j+2− a−1 + . . . aα−1−j+1aαj−1 ]−α[. . .] + α[. . .] − α2 [. . .] + α2 [. . .] . . . + αj−1 [. . .]+j−1j−1bj−1 aα−1 .

. . aα−1 − aj−1 aα−1 . . . aα−1 ) =j−1(2[aα−1 . . . aα−1 (aj−1 − bj−1 )])2Remark It was shown in [18], that for the residue res1,0 in the skew field ofpseudodifferential operators holds res1,0 [X, Y ] = 0, where [X, Y ] is the commutator oftwo pseudodifferential operators. The next statements provide other examples of skewfields with this property.Lemma 0.60 Let K be a skew field such that αn = Id or αn = Id, in = ∞. LetX, Y ∈ K. Then resα [X, Y ] = 0.65Proof It is sufficient to prove the assertion for X = ul z k , Y = um z q .If k + q = 0, then resα (XY ) = resα (Y X) = 0.

In the case k + q = 0 we have:kXY − Y X = ul (um )α − um (ul )α−k−k= αk (um (ul )α ) − um (ul )α−k∈ im(α − Id)2In this case our propositions can be stated as follows:Corollary 10 Let K be a skew field such that α = Id, i = 1, r = 0, a = 0 ((Inthis case K is the ring k((u))((∂ −1 )) of pseudodifferential operators.) Let L, M ∈ K,ν(L) = ν(M ) = −1,M = b−1 z −1 + b0 + b1 z + . . .,L = a−1 z −1 + a0 + a1 z + . . ..The following conditions are equivalent:(i) there is an S ∈ K, ν(S) = 0, such that M = S −1 LS(ii) a−1 = b−1 ;resa0 − b0du ∈ Z anda−1u(a0 − b0 )∈ k[[u]]a−1res1,0 (M j ) = res1,0 (Lj ) for all j ≥ 1.Corollary 11 Assume that αn = Id for all n ∈ N. Let L, M ∈ K, ν(L) = ν(M ) = −1,M = b−1 z −1 + b0 + b1 z + .

. .,L = a−1 z −1 + a0 + a1 z + . . ..The following conditions are equivalent:(i) there is an S ∈ K, ν(S) = 0, such that M = S −1 LS(ii) b−1 /a−1 ∈ imϕ;resα (M j ) = resα (Lj ) for all j ≥ 1.The following examples show that the identity res... ([X, Y ]) = 0 does not hold inother cases.Example (i) Let K be a skew field with α = 1, a(0, . . .

, 0) = 0, r = 1. We assumethat K has the form specified in Theorem 0.35. Let M = z −1 , L = z −1 + z i ∈ k((z)) ⊂K. If resi,r ([X, Y ]) = 0 holds, then M and L are conjugate by Proposition 0.57. LetS = 1 + s1 z + . . .. We haveSM = z −1 + s1 + s2 z + . . . = LS = (z −1 + z i )(1 + s1 z + . . .) =66δ 2 −δ2i 2i(z −1 +s1 +s2 z+. .

.)+(z i −sδ1i z i )+(s1 z i+1 −sδ2i z i+1 )+. . .+(si z 2i +s1iiz −sδi+1z 2i )+. . .Hence, 1 − sδ1i = 0. Since r = 1, this equation is soluble, and s1 = (1 − r)−1 c−1 u1−r .Solving the next equations, we obtain s2 , s3 , . . .. Each of these elements consists of asingle monomial whose valuation is different from r − 1.δ 2 −δiz 2i = 0. By Theorem 0.35, if a(0, . . . , 0) = 0,Further, we have si z 2i + s1i 2i z 2i − sδi+1δ 2 −δthen s1i 2i contains a monomial whose valuation is equal to r − 1.

Therefore, the equation is insoluble with respect to si+1 , and M is not conjugate to L. This contradictioncompletes the proof of the assertion.(ii) Let K be a skew field with α = 1, a(0, . . . , 0) = 0. In this case i > 1, since r = 0for i = 1, and we obtain the ring of pseudodifferential operators. We assume that Khas the form specified in Theorem 0.35.

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