Multidimensional local skew-fields (792481), страница 9
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. . δjl (b),σ(Si+m−1(j1 ,...,jl )where l, j1 , . . . , jl were defined in our proposition. This proves (ii).The proposition is proved.238Definition 0.42 Let D be a splittable division algebra. For any element a ∈ D̄ definethe numbersi(a) = max w(zaz −1 − α(a)) ∈ N ∪ ∞,j(a),zwhere j : D̄ → D, z — parameter in D;(z)dD (a) = max w(zaz −1 − α(a) − δij (a) z ij (a) ) ∈ N ∪ ∞,j(a),zwhere ij (a) = w(zaz −1 − α(a)) for a given embedding j.
It does not depend on thechoice of z as lemma 0.11 shows.The following theorem is a main technical result of this subsection.Theorem 0.43 Let D be a good splittable division algebra. Let u ∈ Z(D̄) be a purelyinseparable element over Z(D) and charF > 2. Then dD (u) = 2i(u) + np, where n > 0,and up ∈ F̄ only if dD (u) = ∞.Proof. By proposition 0.39 we can assume that u ∈ D̄ ⊂ D. Without loss ofgenerality we can assume that Z(D̄)/F̄ is a purely inseparable extension.
Moreover, itcan be assumed that Z(D̄) = F̄ (u). Then by Scolem-Noether theorem we can choosea parameter z such that α = id. Supposezuz −1 = u + δi (u)z i + . . . ,ki.e. δ1 |F (u) = . . . = δi−1 |F (u) = 0, δi (u) = 0. Suppose up ∈ F .Without loss of generality it can be assumed that the following property holds:ψ) let δj , j > i be the first map such that δj = 0 if j is not divisible by i andj/iδj = cj/i δi for some cj/i ∈ D̄ otherwise; then j = 2i mod p.Indeed, let δj be the first map such that δj = 0 and i |j, j = 2i mod p. Then bylemma 0.11, (ii) there exists a parameter z such that δj (u) = 0.
Therefore by corollary1, δj |F (u) = 0. By Scolem-Noether theorem, δj is an inner derivation. Therefore bylemma 0.11 (ii) there exists a parameter z such that δj = 0.j/iIf δj is the first map such that i|j, δj |F (u) = cj/i δi |F (u) and j = 2i mod p, then byj/ilemma 0.11, (ii) there exists a parameter z such that δj (u) = cj/i δi j/i (u) = cj/i δi (u),where(i + 1) . .
. (i(j/i − 1) + 1)(1)cj/i =(j/i)!One can easily show that δj |F (u) = cj/i δi j/i |F (u) . By Scolem-Noether theorem, (δj −j/icj/i δj ) is an inner derivation. Therefore by lemma 0.11 (ii) there exists a parameter zj/isuch that δj = cj/i δi .Let’s divide our proof in two steps.39Step 1. First we prove that (i, p) = 1.In this Step we don’t use the condition D is a good splittable algebra. We use onlya condition that D is splittable.Lemma 0.44 Let j be the minimal positive integer such that δj |F (upl ) = 0, l ≥ 0. Thenthe maps n δm , kj ≤ m < (k + 1)j, k ∈ {1, .
. . , p − 1} satisfy the following properties:i) there exist elements cm,k ∈ D̄ such that(n δm − cm,1 δ − . . . − cm,k δ k )|F (upl ) = 0,where δ : D̄ → D̄ is any F -linear map such that δ|F (upl ) is a derivation, δ(uj ) = 0 forllj∈/ pl N, δ(up ) = 1, ckj,k = c(δj (up ))k , c ∈ Fp .ii) ckj,k = 0 iff n, n + j, . .
. , n + (k − 1)j = 0 mod p.llProof. i) The proof is by induction on k. Let a, b ∈ F (up ). Put t = up . For k = 1,by proposition 0.41, (ii) we haven δm (ab)= n δm (a)b + an δm (b)lbecause all the maps δq , q < j are equal to zero on F (up ). Hence, n δm is a derivationlon F (up ) and cj,1 = n δj (t) = nδj (t).For arbitrary k, by proposition 0.41, (i) and by the induction hypothesis we have(∗∗)q−2q−2qq−1k−1q−1−l l+n δj (t)( (c1 δ+. . .+ck−1 δ )(t)t )+.
. .+n δs (t)( (g1 δ)(tq−1−l )tl ),n δm (t ) = q n δm (t)tl=0l=0where cj , gj ∈ D̄, s > m − 2j. Therefore, n δm (tp ) = 0, because k ≤ p − 1 andp−2j p−1−l l)t = 0 for j ≤ p − 2. Hence, n δm |F (t) = cm,1 δ + . . . + cm,p−1 δ p−1l=0 cj δ (tand we only have to show that cm,q = 0 for q > k.Using (**) we can calculate cm,j . We havecm,1 = n δm (t);112n δm (t ) − 2cm,1 t = n δj (t)(c1 δ(t)) + . . .
+ n δs (t)(g1 δ(t))2!2...q−2q−21= (n δj (t)(cq−1 δ q−1 (tq−1−l )tl ) + . . . + n δm−q+1 (t)(gq−1 δ q−1 (tq−1−l )tl ))q!l=0l=0cm,2 =cm,q1= (n δj (t)cq−1 + . . . + n δm−q+1 (t)gq−1 )q40Hence, cm,k+1 = . . . = cm,p−1 = 0 and ckj,k = cδj (t)ck−1 = c̃(δj (t))k , c, c̃ ∈ Fp . Note thatc(p−1)j,p = (j δj (t))p−1 .ii) Suppose n = 0 mod p. For k = 1 we have n δj |F (t) = nδj |F (t) = 0. For arbitraryk we haveqn δkj (tq−1) = q n δkj (t)t+n δj (t)q−2kj−1q−1−rn+j δ(k−1)j (tr)t +. .
.+r=0n δl (t)q−2q−1−r r)tn+l δkj−l (tr=0l=j+1Since n δj (t) = 0 and m1 δh |F (t) = c1 δ + . . . + ck−2 δ k−2 for h < (k − 1)j, the samearguments as in i) show that ckj,k = 0.Suppose n+rj = 0 mod p, r < k −1. The same arguments as above show that inthis case ckj,k (n δkj ) = 0 iff c(k−1)j,k−1 (n+j δ(k−1)j ) = 0. So, by induction, ckj,k (n δkj ) = 0iff c(k−r)j,k−r (n+rj δ(k−r)j ) = 0, which proves ii).The lemma is proved.2kLemma 0.45 If p|i, then there exists a map δj such that δj (up ) = 0.Proof. We claim that δpq i is the first map such that δpq i |F (upq ) = 0. The proof is byq−1induction on q.
For q = 0, there is nothing to prove. For arbitrary q, put t = up . Byproposition 0.41 we havepδpq i (t ) = δpq−1 i (t)p−2pq i−1p−1−r1+pq−1 i δpq−1 i(p−1) (tr)t +l=pq−1 i+1r=0δl (t)p−2p−1−r r)t1+l δpq i−l (tr=0By induction and lemma 0.44, 1+l δpq i−l |F (t) = c1 δ + . . . + cp−2 δ p−2 for l > pq−1 i.p−2p−1−r r)t = 0. By lemma 0.44, (ii), 1+pq−1 i δpq−1 i(p−1) |F (t) =Therefore,r=0 1+l δpq i−l (tc1 δ + . . . + cp−1 δ p−1 with cp−1 = 0. Hence, δpq i (tp ) = −cp−1 δpq−1 i (t) = 0.The same arguments show that δj (tp ) = 0 for j < pq i. So, δpq i is the first non-zeroqmap on F (up ).2Step 2.From now on (i, p) = 1. Note that δi (u) ∈ Z(D̄). Indeed, for any a ∈ D̄ we haveδi (au) = δi (a)u + aδi (u) = δi (ua) = uδi (a) + δi (u)aTherefore, aδi (u) = δi (u)a.
Since (i, p) = 1, there exists k1 ∈ N such that pk divides1 + k1 i. Therefore by lemma 0.11 (iii) there exists a parameter z such that δi (u) =(δi (u))1+k1 i ∈ Z(D̄).So we can assume that δi (u) ∈ Z(D̄) and ψ) holds.41Assume that dD (u) ≤ 2i(u).Then to prove our theorem it is sufficient to show that there exists a parameter zsuch that the maps δj satisfy the following property:(∗)If j is not divisible by i, then δj |F (u) = 0.
If j is divisible by i, then δj |F (u) =j/icj/i δi |F (u) with some cj/i ∈ D̄.To show it we prove that if property (∗) does not hold, then there exists a map δjksuch that δj (up ) = 0.Suppose (∗) does not hold and δ2i+mp is the first map which does not satisfy (∗). So,δ2i+mp (u) = 0. Note that δq (u) = 0 for i < q < 2i + mp.Note that δ2i+mp (u), δ̃2i+mp (u) ∈ Z(D̄). Indeed, by proposition 0.41, i δ̃2i+mp |F (u) is aderivation. Therefore, i δ̃2i+mp (u) ∈ Z(D̄). Since δi (u) ∈ Z(D) and δq (u) = 0 for i < q <2i + mp, i δ̃2i+mp (u) = iδ̃2i+mp (u) and δ̃2i+mp (u) ∈ Z(D̄).
Therefore, δ2i+mp (u) ∈ Z(D̄).First we prove that there exists a parameter z̄ such that δ¯q = δq for q ≤ 2i + mpand 2i+mp+(p−1)i δ˜¯q (u) = 0 for q = 2i mod p, q > 2i + mp; here δ¯q are maps given bythe parameter z̄. Put j(1) = 2i + mp + (p − 1)i.Suppose j(1) δ˜q (u) = 0, q > 2i + mp and q = 2i mod p. By definition,˜j(1) δq (u) = −j(1)δq (u) +δk1 .
. . δkl (u),where ki < q. By lemma 0.11, (ii) for any a ∈ D̄ there exists a parameter z̄q such thatz̄q uz̄q−1 = u + δi (u)z̄qi + . . . + δq−1 (u)z̄qq−1 + az̄qq + . . .Therefore, there exists an element a ∈ D̄ such thatthe sequence {z̄q } converges in D. So, z̄ = limz̄q .˜¯j(1) δq (u)= 0. It is easy to see thatLemma 0.46 Put κ = j(1) = 2i + mp + (p − 1)i. Then there exists a parameter zsuch that the following properties hold:(i) κ δ̃2i+mp+(p−1)i is the first map such that κ δ̃2i+mp+(p−1)i |F (up ) = 0.(ii) κ δ̃2i+mp+(p−1)i+i+mp (up ) = 0 and κ δ̃r |F (up ) = 0 for j(1) < r < j(1) + i + mp.(iii) κ δ̃2i+mp+(p−1)i (up ) ∈ Z(D̄), κ δ̃2i+mp+(p−1)i+i+mp (up ) ∈ Z(D̄).Proof.i) Put w := 2i + mp + (p − 1)i. By proposition 0.41 we haveκ δ̃w (up) = κ δ̃i (u)p−2i−κ δw−i (up−1−qq)u + κ δ̃2i+mp (u)q=0p−2q=0422i+mp−κ δ(p−1)i (up−1−q)uq +w−1κ δ̃k (u)p−2k−κ δw−k (up−1−q)uqq=0k=2i+mp+1By lemma 0.44, k−κ δw−k |F (up ) = c1 δ + .
. . + cp−2 δ p−2 for w − k < (p − 1)i andp−1with cp−1 = (i δi (u))p−1 = 0.2i+mp−κ δ(p−1)i |F (up ) = c1 δ + . . . + cp−1 δBy proposition 0.41 we havei−κ δw−i (uq) = q i−κ δw−i (u)uq−1+ i−κ δi (u)q−2m1 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+mpBy lemma 0.44, s+i−κ δw−i−s |F (u) = c1 δ+. .
.+cp−3 δ p−3 for w−i−s < (p−2)i. Since i−κ =0 mod p, i−κ δi (u) = 0 and i−κ δ2i+mp (u) = 0. So, i−κ δw−i |F (u) = c1 δ + . . . + cp−2 δ p−2 .Hence,pp−1= −i δ̃2i+mp (u)(i δi (u))p−1 = 0κ δ̃w (u ) = −κ δ̃2i+mp (u)(i δi (u))and κ δ̃w (up ) ∈ Z(D̄). 0.The same arguments show that κ δ̃w is the first map such that κ δ̃w |F (up ) =ii) For j(1) < w ≤ 2i + mp + (p − 1)i + i + mp, by proposition 0.41 we haveκ δ̃w (up) = κ δ̃i (u)p−2i−κ δw−i (up−1−qq)u +κ δ̃2i+mp (u)q=0κ δ̃w−(p−1)i (u)p−2p−22i+mp−κ δw−2i−mp (up−1−q)uq +.
. . +q=0p−1−q)uq +w−(p−1)i−κ δ(p−1)i (uq=0w−1κ δ̃k (u)p−2k−κ δw−k (up−1−q)uqq=0k=w−(p−1)i+1By lemma 0.44, k−κ δw−k |F (u) = c1 δ + . . . + cp−2 δ p−2 for w − k < (p − 1)i.Let us prove that 2i+mp−κ δζ |F (u) = c1 δ + . . . + cp−2 δ p−2 for (p − 1)i < ζ < (p − 1)i +i + mp.If (p − 1)i < ζ < 2i + mp, then it is clear that 2i+mp−κ δζ |F (u) = 0. By proposition0.41, for ζ ≥ 2i + mp and q < p we have2i+mp−κ δζ (u2i+mp−κ δi (u)q−2m δζ−i (uq−1−rq) = q 2i+mp−κ δζ (u)uq−1 +r)u + 2i+mp−κ δ2i+mp (u)r=0Since ζ − 2i − mp < (p − 2)i,q−2m1 δζ−2i−mp (uq−1−r)ur + .
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