Multidimensional local skew-fields (792481), страница 11
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. .+cr−3 δ r−3 for w −j(n+1)−s <(r − 2)j(n + 1). Since j(n + 1) − i − mp = 0 mod p, j(n+1)−i−mp δj(n+1) (t) = 0 andj(n+1)−i−mp δj(n+1)+i+mp (t) = 0 (here we use also i’) and ii’)). Therefore,r−2and lemma 0.44 completes the proof.j(n+1)−i−mp δw−j(n+1) |F (t) = c1 δ + . . . + cr−2 δ48The same arguments show that i+mp δ̃rj(n+1) |F (t) = c1 δ + . .
. + cr−1 δ r−1 . Therefore,p−1only if s = j(n + 1) or s > j(n + 1)i+mp δζ−s · i+mp δ̃s |F (t) = c1 δ + . . . + cp−1 δand s = rj(n + 1) + i + mp. Note that cp−1 = 1r i δ̃j(n+1)+i+mp (t)(i δj(n+1) (t))p−2 if s =rj(n + 1) + i + mp.Using the same arguments for the maps i+mp δζ−rj(n+1) , we get i+mp δζ |F (t) = c1 δ +. . . + cp−1 δ p−1 , wherecp−1 = −(1 +(1 + .
. . +11+ ... +)i δ̃j(n+1)+i+mp (t)(i δj(n+1) (t))p−2 −2p−11)i δ̃j(n+1)+i+mp (t)(i δj(n+1) (t))p−2 − . . . − i δ̃j(n+1)+i+mp (t)(i δj(n+1) (t))p−2 =p−2−i δ̃j(n+1)+i+mp (t)(i δj(n+1) (t))p−2 = 0.This completes the proof of ii) and iii).The lemma is proved.2Case 2. δi (δ̃2i+mp (u)) = 0.Lemma 0.48 Suppose i + mp ≥ i. Put t = up . Then there exists a parameter z suchthat the following properties hold:i)j(1) δ̃j(1)+i+mp+(p−1)j(1)is the first map such thatj(1) δ̃j(1)+i+mp+(p−1)j(1) |F (tp )ii) j(1) δ̃r |F (tp ) = 0 for j(2) < r < j(2) + i andj(2) = j(1) + i + mp + (p − 1)j(1).iii)pj(1) δ̃j(2)+i (t )∈ Z(D̄),pj(1) δ̃j(2) (t )pj(1) δ̃j(2)+i (t )= 0.= 0, where∈ Z(D̄).Proof.
To prove i) one can repeat the proof of i) in lemma 0.47. Note thatpp−2∈ Z(D̄).j(1) δ̃j(2) (t ) = −j(1) δ̃j(1)+i (i δj(1) (t))ii) As in the case 1 we can find a parameter z̄ such that δ¯q |F (t) = δq |F (t) for q ≤j(1) + i + mp, ˜δ̄ q (tp ) = 0 for q = 2i mod p, q > j(2).For r = 2i mod p, by proposition 0.41 we havepj(1) δ̃r (t) = j(1) δ̃j(1)+i+mp (t)p−2p−1−q q)t +i+mp δr−j(1)−i−mp (tq=0r−1k=j(1)+i+mp+1j(1) δ̃k (t)p−2q=049p−1−q q)tk−j(1) δr−k (tBy lemma 0.44, k−j(1) δr−k |F (t) = c1 δ + . .
. + cp−2 δ p−2 if r − k < (p − 1)j(1).p−1−q qNote that j(1) δ̃r−(p−1)j(1) (t) p−2)t = 0.q=0 r−pj(1) δ(p−1)j(1) (tIndeed, r −pj(1) = 2i mod p. Therefore by lemma 0.44 (ii), r−pj(1) δ(p−1)j(1) |F (t) =c1 δ + . . . + cp−2 δ p−2 .The same arguments as in lemma 0.47 (ii) show that m δs |F (t) = c1 δ + . . . + cp−2 δ p−2for (p − 1)j(1) < s < (p − 1)j(1) + i.Let us prove that i+mp δ(p−1)j(1)+i |F (t) = c1 δ + . .
. + cp−1 δ p−1 with cp−1 = 0.First let us show that i+mp δ̃ζ |F (t) = c1 δ + . . . + cr δ r for ζ = (r + 1)j(1) + i, r ≤ p − 2.For r = 0, i+mp δ̃ζ |F (t) is a derivation. Since j(1) δ̃ζ (t) = 0 for ζ = j(1) + i and j(1) =i + mp mod p and p > 2, we have i+mp δ̃ζ (t) = 0 and i+mp δ̃ζ |F (t) = 0.For arbitrary r we haveqi+mp δ̃ζ (tq−1) = q i+mp δ̃ζ (t)t+ i+mp δ̃j(1) (t)q−2q−1−r r)t +j(1)−i−mp δζ−j(1) (tr=0ζ−1i+mp δ̃k (t)k−i−mp δζ−k |F (t)j(1)−i−mp δζ−j(1)q−1−r r)tk−i−mp δζ−k (tr=0k=j(1)+i+1By lemma 0.44,j(1) + i + 1.By definition,q−2= c1 δ + . .
. + cr−1 δ r−1 for ζ − k < rj(1), i.e. k ≥=lCj(1)−i−mpδj1 . . . δjl ,(j1 ,...,jl ),l≥pbecause j(1) − i − mp = 0 mod p. Since l ≥ p, there exist jk , jk1 such that jk < j(1)and jk1 < j(1). Thus, j1 + . . . + jˆk + . . . + jˆk1 + . . . + jl < rj(1) and δj1 . . . δjl |F (t) =c1 δ + . . . + cr−1 δ r−1 .
Hence by lemma 0.44, i+mp δ̃ζ |F (t) = c1 δ + . . . + cr δ r .Now we have(p−1)j(1)i+mp δ(p−1)j(1)+i+ i+mp δ̃(p−1)j(1)+i +i+mp δk· i+mp δ̃(p−1)j(1)+i−k = 0k=iwith i+mp δk · i+mp δ̃(p−1)j(1)+i−k |F (t) = c1 δ + . . . + cp−2 δ p−2 for k = (p − 2)j(1) + i.Therefore,i+mp δ(p−1)j(1)+i |F (t)= c1 δ + .
. . + cp−2 δ p−2 − i+mp δ(p−2)j(1)+i · i+mp δ̃j(1) |F (t)So by induction,i+mp δ(p−1)j(1)+i |F (t)= c̃1 δ+. . .+c̃p−2 δ p−2 +i+mp δi (i+mp δj(1) )p−1 |F (t) = c1 δ+. . .+cp−2 δ p−2 +50p−1 p−1)δi+mp δi ((i δj(1) (t))Since i δj(1) (t) = −j(1) δ̃j(1) (t) = i δ̃2i+mp (u)(i δi (u))p−1 , we have i+mp δi (i δj(1) (t)) = 0,which completes the proof of ii).Finally, j(1) δ̃j(2)+i (tp ) = −j(1) δ̃j(1)+i+mp (t)i+mp δi (i δj(1) (t))(i δj(1) (t))p−2 ∈ Z(D̄),because i+mp δi is a derivation and i δj(1) (t) ∈ Z(D̄) .
This proves iii).2The following lemma completes the proof of Case 2 and of Theorem.Lemma 0.49 Suppose the following conditions hold:j(1) δ̃j(n)+i |F (upn )i’)ii’)j(1) δ̃j(n) (upn= 0 andj(1) δ̃r |F (upn )= 0 for j(n) < r < j(n) + i, n ≥ 1;n), j(1) δ̃j(n)+i (up ) ∈ Z(D̄).Then there exists a parameter z such thati)j(1) δ̃j(n)+i+(p−1)j(n)is the first map such thatj(1) δ̃j(n)+i+(p−1)j(n) |F (upn+1 )= 0;n+1ii) j(1) δ̃j(n+1)+i (up ) = 0 and j(1) δ̃r |F (upn+1 ) = 0 for j(n + 1) < r < j(n + 1) + i,where j(n + 1) = j(n) + i + (p − 1)j(n);iii)j(1) δ̃j(n+1) (upn+1), j(1) δ̃j(n+1)+i (upn+1) ∈ Z(D̄).nProof.
By induction, j(n) = i mod p. Put a = j(1) δ̃j(n) (up ). We have akj(n)+1 =lap ∈ Z(D) for some k ∈ Z. Put z = (a−kj(1) z j(1) )1/j(1) . We claim that i’), ii’) hold for˜˜˜j(1) δ j(n)+i , i.e. j(1) δ j(n)+i |F (upn ) = 0, and j(1) δ r |F (upn ) = 0 for j(n) < r < j(n) + i, andpnpnpn˜˜˜j(1) δ j(n)+i (u ), j(1) δ j(n) (u ) ∈ Z(D̄). Moreover, j(1) δ j(n) (u ) ∈ Z(D).Note that α = id, because a ∈ Z(D̄). We havez−j(1) pn j(1)u znnn= z −j(1) up z j(1) = up + az j(n) + j(1) δ̃j(n)+i (up )z j(n)+i + .
. .Let us show thatzWe havej(n)= a−kj(n) z j(n)zj(n)= zmod ℘j(n)+i+1j(n)−j(1) −kj(1) j(1)az.It is easy to see that z = a−k z + xz i+1 + . . ., x ∈ D̄. Since j(n) − j(1) = 0 mod p,z j(n)−j(1) = (a−k z)j(n)−j(1) mod ℘j(n)−j(1)+i+1 . Now we have(a−k z)j(n)−j(1) a−kj(1) = a−kj(n) z j(n)−j(1) + xz j(n)−j(1)+i + . . . ,51where x = [−kj(1) − k(j(1) + 1) − .
. . − k(j(n) − 1)]a−kj(n)−1 δi (a) = 0. Therefore,z−j(1) pn j(1)u znl= up + a p z j(n)n+ j(1) δ̃j(n)+i (up )ak(j(n)+i) z j(n)+i+ ...nnnland j(1) δ˜ j(n)+i (up ) = j(1) δ̃j(n)+i (up )ak(j(n)+i) ∈ Z(D̄), j(1) δ˜ j(n) (up ) = ap ∈ Z(D). Soi’), ii’) hold.Now to prove i) one can repeat the proof of i) in lemma 0.47. Note thatpp−2∈ Z(D̄).j(1) δ̃j(n+1) (t ) = −j(1) δ̃j(n)+i (i δj(n) (t))nii) We use the same arguments as in ii) of lemma 0.48. Put t = up .For r = 2i mod p, by proposition 0.41 we havepj(1) δ̃r (t) = j(1) δ̃j(n) (t)p−2p−1−qj(n)−j(1) δr−j(n) (tq)t +j(1) δ̃j(n)+i (t)q=0p−2p−1−q q)t +j(n)−j(1)+i δr−j(n)−i (tq=0r−1j(1) δ̃k (t)p−2p−1−q q)tk−j(1) δr−k (tq=0k=j(n)+i+1By lemma 0.44, k−j(1) δr−k |F (t) = c1 δ + .
. . + cp−2 δ p−2 if r − k < (p − 1)j(n).Note that j(n)−j(1) δr−j(n) |F (t) = c1 δ + . . . + cp−2 δ p−2 . Indeed, by proposition 0.41 wehaveqj(n)−j(1) δr−j(n) (tq−1) = q j(n)−j(1) δr−j(n) (t)t+j(n)−j(1) δj(n) (t)q−2q−1−s s)t +2j(n)−j(1) δr−2j(n) (ts=0j(n)−j(1) δj(n)+i (t)q−2q−1−s s)t2j(n)−j(1)+i δr−2j(n)−i (t+ ...+s=0j(n)−j(1) δj(n)+2i (t)q−2q−1−s s)t2j(n)−j(1)+2i δr−2j(n)−2i (t+ ...s=0Recall that r ≤ pj(n) + 2i. By lemma 0.44, m δs |F (t) = c1 δ + . . . + cp−3 δ p−3 ifs < (p − 2)j(n).
Since j(1) δ̃j(n) (t) ∈ Z(D), we have δj(n) (t) ∈ Z(D). Since j(n) − j(1) =0 mod p and δj(n) (t) ∈ Z(D) and charF > 2, we have j(n)−j(1) δj(n)+ei (t) = 0 fore ≤ p − 1, which completes theproof.p−1−q q)t = 0.Note that j(1) δ̃r−(p−1)j(n) (t) p−2q=0 r−(p−1)j(n)−j(1) δ(p−1)j(n) (tIndeed, r − (p − 1)j(n) − j(1) = 2i mod p. Therefore by lemma 0.44 (ii),p−2.r−(p−1)j(n)−j(1) δ(p−1)j(n) |F (t) = c1 δ + . . .
+ cp−2 δThe same arguments as in lemma 0.47 (ii) show that m δs |F (t) = c1 δ + . . . + cp−2 δ p−2for (p − 1)j(n) < s < (p − 1)j(n) + i.Let us prove that j(n)−j(1)+i δ(p−1)j(n)+i |F (t) = c1 δ + . . . + cp−1 δ p−1 with cp−1 = 0.Note that j(n)−j(1)+i δ̃rj(n)+i |F (t) = c1 δ + . . . + cr δ r with cr = 0 for any 1 ≤ r ≤ p − 1.52Indeed, by proposition 0.41 we haveqj(n)−j(1)+i δ̃rj(n)+i (tq−1) = q j(n)−j(1)+i δ̃rj(n)+i (t)t+ j(n)−j(1)+i δ̃j(n) (t)q−2s=0q−1−s s)t + j(n)−j(1)+i δ̃j(n)+i (t)j(1)−i δ(r−1)j(n)+i (tq−2q−1−s s)tj(1) δ(r−1)j(n) (t+ ...s=0By lemma 0.44, j(1) δ(r−1)j(n) |F (t) = c1 δ + .
. . + cr−1 δ r−1 with cr−1 = 0 and m δs |F (t) =c1 δ + . . . + cr−2 δ r−2 for s < (r − 1)j(n).Let us prove that j(1)−i δ(r−1)j(n)+i |F (t) = c1 δ + . . . + cr−2 δ r−2 . By proposition 0.41we haveqj(1)−i δ(r−1)j(n)+i (tq−1) = q j(1)−i δ(r−1)j(n)+i (t)t+j(1)−i δj(n) (t)q−2q−1−s s)t +j(n)+j(1)−i δ(r−2)j(n)+i (ts=0j(1)−i δj(n)+i (t)q−2q−1−s s)tj(n)+j(1) δ(r−2)j(n) (t+ ...s=0Since j(1) δ̃j(n) (t) ∈ Z(D), we have δj(n) (t) ∈ Z(D).
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