Multidimensional local skew-fields (792481), страница 10
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. .r=0m1 δs |F (u)= c1 δ + . . . + cp−3 δ p−3 for s ≤ ζ − 2i − mp.43To show that m δζ−i |F (u) = c1 δ +. . .+cp−3 δ p−3 we use induction on r, where i+mp ≤ζ − (p − 2 − r)i < 2i + mp + ri. For arbitrary r we can use the same calculations, sowe only have to prove that m δζ−(p−2−r)i |F (u) = 0 for some r ≥ 0.There exists r ≥ 0 such that i+mp ≤ ζ −(p−2−r)i < 2i+mp. If 2i+mp > 2i, thenkδm ζ−(p−2−r)i = cδi , k > p if i|ζ and m δζ−(p−2−r)i = 0 otherwise. So, m δζ−(p−2−r)i |F (u) = 0.If 2i+mp ≤ 2i, then (p−1)i < ζ < pi. So, i < ζ −(p−2)i < 2i+mp and m δζ−(p−2)i = 0.Let us prove that 2i+mp−κ δ(p−2)i+2i+mp |F (u) = c1 δ +. . .+cp−1 δ p−1 with cp−1 = 0.
Notethat 2i+mp−κ δ̃2i+mp+ri |F (u) = c1 δ + . . . + cr+1 δ r+1 with cr+1 = 0 for any 0 ≤ r ≤ p − 2.Indeed, by proposition 0.41 we have2i+mp−κ δ̃2i+mp+ri (uq) = q 2i+mp−κ δ̃2i+mp+ri (u)uq−1+ 2i+mp−κ δ̃i (u)q−2t=0lCκ−i−mpδj1. . . δjl (uq−1−tt)u + 2i+mp−κ δ̃2i+mp (u)(j1 ,...,jl ),l≥pq−2κ δri (uq−1−t)ut + .
. .t=0By lemma 0.44, κ δri |F (u) = c1 δ +. . .+cr δ r with cr = 0 and m δs |F (u) = c1 δ +. . .+cr−1 δ r−1for s < ri.If there exists jk ≥ 2i+mp, then j1 +. . .+ jˆk +. . .+jl ≤ ri; so there exists jt < i andδj1 . . . δjl = 0. If there are no jk ≥ 2i + mp, then δjk = cδik , c ∈ ” and δj1 . . . δjl |F (u) = 0,because l ≥ p.Hence by lemma 0.44, 2i+mp−κ δ̃2i+mp+ri |F (u) = c1 δ + . .
. + cr+1 δ r+1 with cr+1 =11rδ̃(u)(i δi (u))r = r+1i δ̃2i+mp (u)(i δi (u)) = 0.r+1 2i+mp−κ 2i+mpWe have2i+mp−κ δ(p−2)i+2i+mp+ 2i+mp−κ δ̃(p−2)i+2i+mp + 2i+mp−κ δi · 2i+mp−κ δ̃(p−3)i+2i+mp + . . . +2i+mp−κ δ(p−2)i· 2i+mp−κ δ̃2i+mp + 2i+mp−κ δ(p−3)i+2i+mp · 2i+mp−κ δ̃i = 0We have 2i+mp−κ δri · 2i+mp−κ δ̃(p−2−r)i+2i+mp |F (u) = c1 δ + . . . + cp−1 δ p−1 with cp−1 =1δ̃(u)(i δi (u))p−2 .p−1−r i 2i+mpSince (2i+mp−κ δ̃i )p |F (u) = 0, and using induction, we get 2i+mp−κ δ(p−2)i+2i+mp |F (u) =c1 δ + . .
. + cp−1 δ p−1 withcp−1 = −(1 + . . . +1)i δ̃2i+mp (u)(i δi (u))p−2 −p−11)i δ̃2i+mp (u)(i δi (u))p−2 −. . .−i δ̃2i+mp (u)(i δi (u))p−2 = −i δ̃2i+mp (u)(i δi (u))p−2 = 0p−2p−1−qNote that κ δ̃w−(p−1)i (u) p−2)uq = 0 only if w = i mod p.q=0 w−(p−1)i−κ δ(p−1)i (uIndeed, suppose w − (p − 1)i − κ = i mod p. Therefore by lemma 0.44, (ii),p−2.w−(p−1)i−κ δ(p−1)i |F (u) = c1 δ + . . . + cp−2 δ(1+. . .+44Let us prove thati−κ δw−i (uqp−2q=0 i−κ δw−i (up−1−q) = q i−κ δw−i (u)uq−1)uq = 0. By proposition 0.41 we have+ i−κ δi (u)q−22i−κ δw−2i (uq−1−r)ur +r=0w−i−1i−κ δs (u)q−2s+i−κ δw−i−s (uq−1−r)urr=0s=2i+mpSince i−κ = 0 mod p, i−κ δs (u) = 0 for s < 2i+mp+(p−1)i. For s ≥ 2i+mp+(p−1)iwe have w − i − s ≤ mp and s+i−κ δw−i−s = cδik , c ∈ ”.
But m ≤ 0 by our assumptionin the beginning of Step 2, so s+i−κ δw−i−s = 0.So, we have κ δ̃w (up ) = 0 only if w = i mod p or w = 2i+mp+(p−1)i+i+mp. Bylemma 0.11, (ii),(see the same arguments before this lemma, for example) there exists aparameter z such that the map κ δ̃w (up ) becomes equal to zero on up if w = i mod p.Since 2i + mp + (p − 1)i + i + mp − w ≤ i by our assumption, the change from lemma0.11 does not change the map κ δ̃2i+mp+(p−1i+i+mp .
So, we get the proof of (ii).Now we have κ δ̃2i+mp+(p−1i+i+mp (up ) = −κ δ̃2i+mp (u)(−i δ̃2i+mp (u)(i δi (u))p−2 ) ∈Z(D̄), which proves (iii).The lemma is proved.2Consider the following two cases.Case 1. δi (δ̃2i+mp (u)) = 0 or i + mp < i. In this case we have shown thatδi (i δ̃j(1) (up )) = 0 and δi (i δ̃j(1)+i+mp (up )) = 0.Lemma 0.47 Let δj(n+1) be the first map such that δj(n+1) |F (upn+1 ) = 0. Suppose thefollowing conditions hold:i’) j(n) δ̃j(n+1)+i+mp (upj(n + 1) + i + mp;n+1)|F (upn+1 ) = 0 andii’) δi (j(n) δ̃j(n+1)+i+mp (upn+1j(n) δ̃r |F (upn+1 ))) = 0 and δi (j(n) δ̃j(n+1) (upn+1= 0 for j(n + 1) < r <)) = 0.Then there exists a parameter z such that the following conditions hold:i)j(n+1) δ̃j(n+1)+i+mp+(p−1)j(n+1)j(n+1) δ̃j(n+1)+i+mp+(p−1)j(n+1) |F (uis the first map such that) = 0;pn+2n+2ii) j(n+1) δ̃j(n+1)+i+mp+(p−1)j(n+1)+i+mp (up ) = 0 and j(n+1) δ̃r |F (upn+2 ) = 0 forj(n + 1) + i + mp + (p − 1)j(n + 1) < r < j(n + 1) + i + mp + (p − 1)j(n + 1) + i + mp;45n+2iii) δi (j(n+1) δ̃j(n+2) (up )) = 0 and δi (j(n+1) δ̃j(n+2)+i+mp (up2) = j(n + 1) + i + mp + (p − 1)j(n + 1).n+2)) = 0, where j(n +Proof.
First we prove that there exists a parameter z̄ such that δ̄q |F (upn+1 ) =n+1δq |F (upn+1 ) for q ≤ j(n + 1) + i + mp and j(n+1) ˜δ̄ q (up ) = 0 only if q = 2i mod p forq > j(n + 1) + i + mp; here δ̄q are the maps given by the parameter z̄.n+1Suppose j(n+1) ˜δ̄ q (up ) = 0, q > j(n + 1) + i + mp and q = 2i mod p. Byn+1n+1n+1definition, j(n+1) δ̃q (up ) = −j(n + 1)δq (up ) + δk1 . .
. δkl (up ), where ki < q. Bylemma 0.11, (ii), for any a ∈ D̄ there exists a parameter z̄q such thatz̄q upn+1z̄q−1 = upn+1+ δj(n+1) (upn+1)z̄qj(n+1) + . . . + δq−1 (upn+1)z̄qq−1 + az̄qq + . . .n+1Therefore there exists an element a ∈ D̄ such that j(n+1) δ̃q (up ) = 0. It is easy to seethat the sequence {z̄q } converges in D. So, z̄ = limz̄q .n+1Now we prove that j(n+1) δ̃j(n+1)+i+mp (up ) = 0 and j(n+1) δ̃r |F (upn+1 ) = 0 for j(n +n+11) < r < j(n + 1) + i + mp and δi (j(n+1) δ̃j(n+1)+i+mp (up )) = 0.We have j(n + 1) = j(n) mod p. Therefore,z −j(n+1) upn+1z j(n+1) = z −pk (z −j(n) upn+1j(n) δ̃j(n+1)+i+mp (uz −pk upn+1upz pk +z −pk j(n) δ̃j(n+1) (upn+1+ j(n) δ̃j(n+1) (upn+1n+1z j(n) )z pk = z −pk (uppn+1n+1+j(n) δ̃j(n+1) (upn+1)z j(n+1) +)z j(n+1)+i+mp + .
. .)z pk =)z pk z j(n+1) +j(n) δ̃j(n+1)+i+mp (up)z j(n+1) + j(n) δ̃j(n+1)+i+mp (upn+1n+1)z j(n+1)+i+mp +. . . =)z j(n+1)+i+mp + . . . ,n+1because ii’) provide δi (δr (up )) = 0 for j(n + 1) < r < j(n + 1) + i + mp. So,n+1n+1j(n+1)+2i+mpz −pk up z pk = upmod MD.n+1i) Put w = j(n + 1) + i + mp + (p − 1)j(n + 1), t = up . By proposition 0.41 wehavep−2pp−1−q q)t +i+mp δ(p−1)j(n+1) (tj(n+1) δ̃w (t ) = j(n+1) δ̃j(n+1)+i+mp (t)q=0w−1k=j(n+1)+i+mp+1j(n+1) δ̃kp−2p−1−q q)tk−j(n+1) δw−k (tq=0By lemma 0.44, k−j(n+1) δw−k |F (t) = c1 δ + . . . + cp−2 δ p−2 for w − k < (p − 1)j(n + 1)and i+mp δ(p−1)j(n+1) |F (t) = c1 δ + . . . + cp−1 δ p−1 with cp−1 = 0.
Therefore, j(n+1) δ̃w (tp ) =−j(n+1) δ̃j(n+1)+i+mp (t)cp−1 = 0.The same arguments show that j(n+1) δ̃w is the first map such that j(n+1) δ̃w (tp ) = 0.n+1ii) Put t = up . Using the same arguments as above, we can find a parameter z̄such that δ¯q = δq for q ≤ j(n + 1) + i + mp + (p − 1)j(n + 1) and j(n+1) ˜δ̄ q (tp ) = 046for q > j(n + 1) + i + mp + (p − 1)j(n + 1), q = 2i mod p.
Since j(n+1) ˜δ̄ r |F (tp ) ,j(n + 1) + i + mp + (p − 1)j(n + 1) < r < j(n + 1) + i + mp + (p − 1)j(n + 1) + i + mp arederivations (see lemma 0.44), j(n+1) ˜δ̄ r |F (tp ) = 0 for j(n + 1) + i + mp + (p − 1)j(n + 1) <r < j(n + 1) + i + mp + (p − 1)j(n + 1) + i + mp.For j(n+1)+i+mp+(p−1)j(n+1) < w ≤ j(n+1)+i+mp+(p−1)j(n+1)+i+mp,w = 2i mod p, by proposition 0.41 we havepj(n+1) δ̃w (t) = j(n+1) δ̃j(n+1)+i+mp (t)p−2p−1−q q)ti+mp δ(p−1)j(n+1)+i+mp (t+ ...+q=0j(n+1) δ̃w−(p−1)j(n+1) (t)p−2p−1−q q)t +w−(p−1)j(n+1)−j(n+1) δ(p−1)j(n+1) (tq=0w−1j(n+1) δ̃k (t)p−2p−1−q q)tk−j(n+1) δw−k (tq=0k=w−(p−1)j(n+1)+1By lemma 0.44, k−j(n+1) δw−k |F (t) = c1 δ + .
. . + cp−2 δ p−2 if w − k < (p − 1)j(n + 1).p−1−q qTherefore, p−2)t = 0.q=0 k−j(n+1) δw−k (tp−2Note that j(n+1) δ̃w−(p−1)j(n+1) (t) q=0 w−(p−1)j(n+1)−j(n+1) δ(p−1)j(n+1) (tp−1−q )tq = 0.Indeed, w − (p − 1)j(n + 1) − j(n + 1) = 2i mod p. Therefore by lemma 0.44 (ii),p−2.w−(p−1)j(n+1)−j(n+1) δ(p−1)j(n+1) |F (t) = c1 δ + . . . + cp−2 δLet us prove that k−j(n+1) δζ |F (t) = c1 δ + .
. . + cr δ r for (r + 1)j(n + 1) < ζ <(r + 1)j(n + 1) + i + mp, r ≤ p − 2.The proof is by induction on r. By ii’) and i’), δs (t) = 0 for j(n + 1) < s <j(n + 1) + i + mp. Therefore for r = 0, k−j(n+1) δζ |F (t) = 0.For arbitrary r we haveqk−j(n+1) δζ (t ) = q k−j(n+1) δζ (t) + k−j(n+1) δj(n+1) (t)q−2q−1−r r)t +k δζ−j(n+1) (tr=0ζ−1k−j(n+1) δs (t)q−2q−1−r r)ts+(k−j(n+1)) δζ−s (tr=0s=j(n+1)+i+mpBy lemma 0.44, s+(k−j(n+1)) δζ−s |F (t) = c1 δ + . .
. + cr−1 δ r−1 for ζ − s < rj(n + 1).For any m m δζ−j(n+1) |F (t) = 0 if r = 1, and m δζ−j(n+1) |F (t) = c1 δ + . . . + cr−1 δ r−1 byinduction and lemma 0.44, becauseqm δζ−j(n+1) (tq−1) = q m δζ−j(n+1) (t)t+ m δj(n+1) (t)q−2l=047q−1−l l)t +m1 δζ−2j(n+1) (tζ−j(n+1)−1m δsq−2s=j(n+1)+i+mpq−1−l l)t ,ms δζ−j(n+1)−s (tl=0and by lemma 0.44, ms δζ−j(n+1)−s |F (t) = c1 δ + . .
. + cr−2 δ r−2 for s ≥ j(n + 1) + i + mp.The same arguments show that k−j(n+1) δ̃ζ |F (t) = c1 δ + . . . + cr δ r for (r + 1)j(n + 1) <ζ < (r + 1)j(n + 1) + i + mp.Let us show that i+mp δ(p−1)j(n+1)+i+mp |F (t) = c1 δ + . . . + cp−1 δ p−1 , cp−1 = 0. Putζ = (p − 1)j(n + 1) + i + mp. We havei+mp δζ + i+mp δ̃ζ +w−k−1i+mp δζ−s· i+mp δ̃s = 0s=1First we prove that i+mp δ̃rj(n+1)+i+mp |F (t) = c1 δ + .
. . + cr δ r withcr = 1r i+mp δ̃j(n+1)+i+mp (t)(j(n+1) δj(n+1) (t))r−1 = 0. We use the same arguments as above.The proof is by induction on r. For r = 0, since i + mp < j(n + 1), i+mp δ̃i+mp |F (t) = 0.Put w = rj(n + 1) + i + mp. For arbitrary r we haveqi+mp δ̃w (tq−1) = q i+mp δ̃w (t)t+ i+mp δ̃j(n+1) (t)q−2q−1−r r)t +j(n+1)−i−mp δw−j(n+1) (tr=0w−1q−2i+mp δ̃s (t)q−1−r r)ts−i−mp δw−s (tr=0s=j(n+1)+i+mpBy lemma 0.44, s−i−mp δw−s |F (t) = c1 δ + . . . + cr−2 δ r−2 for w − s < (r − 1)j(n + 1) andr−1with cr−1 = (j(n+1) δj(n+1) (t))r−1 = 0.j(n+1) δ(r−1)j(n+1) |F (t) = c1 δ + . . . + cr−1 δBy proposition 0.41 we haveqj(n+1)−i−mp δw−j(n+1) (t )= q j(n+1)−i−mp δw−j(n+1) (t)tq−1 +j(n+1)−i−mp δj(n+1) (t)q−2q−1−r r)t +m1 δw−2j(n+1) (tr=0w−j(n+1)−1s=j(n+1)+i+mpj(n+1)−i−mp δs (t)q−2q−1−r r)ts+j(n+1)−i−mp δw−j(n+1)−s (tr=0By lemma 0.44, s+j(n+1)−i−mp δw−j(n+1)−s |F (t) = c1 δ +.
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