Multidimensional local skew-fields (792481), страница 4

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. ,a ∈ K̄where δ1 = . . . = δj−1 = 0, δj = 0. Then(i) for z = z + bz q+1 we havez az −1 = aα + . . . + aδq−1 z q−1 + (aδq + baαq+1− aα b)z q + . . .q+1i.e. aδq = aδq + baα − aα b.(ii) Suppose αn = id, n ≥ 1. Then for z = z + bz q+1 , n|q we havez az −1 = aα + . . . + aδq+j−1 z q+j−1 +(aδq+jδj αq+ b(a )δj αj−a b+bqαk δj αq−k((a ) )−ak=1δjj−1kbα )z q+j + .

. .k=0(iii) for z = bz, b ∈ K̄, b = 0 we havejz az −1 = aα + aδj (b−1 )α . . . (b−1 )α z j + . . .Corollary 4 If α = Id, thenz az −1 = a + . . . + aδq+j−1 z q+j−1 + (aδq+j + (q − j)aδj b)z q+j + . . .14Proof of lemma.(i) We havez az −1 = (1 + bz q )zaz −1 (1 + bz q )−1 = (zaz −1 + bz q zaz −1 )(1 − bz q + bz q bz q − . . .) =(zaz −1 − zaz −1 bz q + . . .

+ bz q zaz −1 − . . .) =(zaz −1 − [aα + aδj z j + . . .]bz q + bz q [aα + aδj z j + . . .] + . . .) =j(zaz −1 − [aα b + aδj bα z j + . . .]z q + baα(zaz −1 + (−aα b + baαq+1q+1z q + . . .) =)z q + . . .) = aα + . . . + aδq−1 z q−1 + (aδq + baαq+1− aα b)z q + . . .(ii) We havez az −1 = (1 + bz q )zaz −1 (1 + bz q )−1 = (zaz −1 + bz q zaz −1 )(1 + bz q )−1 =(aα + aδj z j + . . .

+ aδq+j z q+j + . . . + bz q (aα + aδj z j + . . .))(1 + bz q )−1 =α(a +baαq+1 qδj jz +a z +. . .+aδq+j q+jzqkq−kq+. . .+b((aα )δj )α z q+j +b(aδj )α z q+j +. . .)(1+bz q )−1 =k=1αδj ja +(a z +. . .+aδq+j q+jzqkq−kq+. . .+b((aα )δj )α z q+j +b(aδj )α z q+j +. . .)(1−bz q +bz q bz q −. . .) =k=1αδj ja +a z +. . .+aδq+j q+jz+. . .+bqk((aα )δj )αq−kqjz q+j +b(aδj )α z q+j +.

. .−aδj bα z q+j +. . . =k=1qjaα + . . . + aδq+j−1 z q+j−1 + (aδq+j + b(aδj )α − aδj bα + bbecause z j = z j +(iii) We havej−1k=0qk((aα )δj )αq−k− aδjk=1j−1kbα )z q+j ,k=0αkb z q+j + . . ..jjz az −1 = bzaz −1 b−1 = aα + baδj (b−1 )α z j + . . . = aα + aδj (b−1 )α . . . (b−1 )α z j + . . .2Lemma 0.12 Let δ be an (α, β)-derivation of a field K̄ and α = β.Then δ is an inner derivation, i.e. there exists d ∈ K̄ such thatδ(a) = daα − aβ dfor all a ∈ K̄.15Proof.

Put d = δ(a)/(aα − aβ ), where a is any element such that aα = aβ . Putδin (x) = dxα − xβ d. We claim that δ = δin . Indeed, consider the map δ̄ = δ − δin . It isan (α, β)-derivation. Take arbitrary b ∈ K̄. Then δ̄(ab) = δ̄(ba). But we haveδ̄(ab) = δ̄(a)bα + aβ δ̄(b) = aβ δ̄(b),andδ̄(ba) = δ̄(b)aα + bβ δ̄(a) = aα δ̄(b)Therefore, δ̄(b) = 0 for any b.2Proof of proposition.Letzaz −1 = aα + aδ1 z + aδ2 z 2 + . . .By proposition 1.7 and corollary 1, δ1 is an (α2 , α)-derivation. Since α2 = α, by lemma20.12 it is an inner derivation, say δ1 (a) = d1 aα − aα d1 . By lemma 0.11, (i) for aparameter z2 = z − d1 z 2 we havez2 az2−1 = aα + aδ2 z22 + . . .

.Note that δ1 = 0. By corollary 1, δ2 is an (α3 , α)-derivation. Since α3 = α, by lemma0.12 it is an inner derivation. By lemma 0.11, (i) there exists a parameter z3 = z2 −d2 z23such that z3 az3−1 = aα mod ℘3 .By induction for arbitrary k ∈ N we havezk azk−1 = aα + aδk zkk + . . .and δj = 0 for j < k. By corollary 1, δk is an (αk+1 , α)-derivation. Since αk+1 = α, itis an inner derivation. By lemma 0.11, (i) there exists a parameter zk+1 = zk − dk zkk+1−1such that zk+1 azk+1= aα mod ℘k+1 .It is clear that the sequence {zn }: zn+1 = zn − dn znn+1 converges in K. Since K is acomplete and separate field, there exists a unique limit z. It is clear that zaz −1 = aα .The proposition is proved.2Theorem 0.13 Let K be a two-dimensional local skew field.

If αn = id for all n ∈ Nthen(i) charK = charK̄(ii) K splits.16Proof.If charK = charK̄ then charK̄ = p > 0. Hence ν(p) = r > 0. Then for any elementt ∈ K with ν(t) = 0 we have ptp−1 = αr (t̄) mod ℘ where t̄ is the image of t in K̄.But on the other hand, pt = tp, a contradiction.The proof of (ii) we will divide in three steps.Step 1.

Let π be the prime field in K. Since charK = charK̄ the field π is asubring of O.kLemma 0.14 There exists an element c ∈ K̄ such that cα = c for all k ∈ N.Proof. We claim that there exists a sequence {cji }, ji , i ∈ N, cji ∈ Ō such that(i) ν̄(cji ) > ν̄(cji−1 ) ∀i(ii) if k = 0 mod j2 . . . jl and kcj1 , . . . , αk (cjl−1 ) = cjl−1 , αk (cjl ) = cjl and= 0 mod j2 . . . jl+1 , then αk (cj1 ) =ν̄[(αk − Id)(cjl )] < ν̄(cjl+1 )Let us construct it.

Take an element cj1 such that α(cj1 ) = cj1 , and ν̄(cj1 ) ≥ 1. Suchan element always exists. Indeed, consider an element u with ν̄(u) = 1. If α(u) = u, thenone can put cj1 = u. If α(u) = u, then take any element c˜j1 such that α(c˜j1 ) = c˜j1 . Ifν̄(c˜j1 ) = 0, then put cj1 = c˜j1 u. Then we have ν̄(cj1 ) = 1 and α(c˜j1 u) = α(c˜j1 )u = c˜j1 u.Put j1 = 1.Let j2 be a minimal positive integer such that (αj1 )j2 (cj1 ) = cj1 , and let k1 =max{ν̄[(αj1 )m (cj1 ) − cj1 ], m ∈ {1, . . . , j2 − 1}}.Take any c˜j2 such that (αj1 )j2 (c˜j2 ) = c˜j2 . Put cj2 = c˜j2 cjk11 +1 .

Then (αj1 )j2 (cj2 ) = cj2 andν̄[(αj1 )m (cj1 ) − cj1 ] < ν̄(cj2 ) ∀m < j2 .By induction we geta sequence which satisfy (i) and (ii).∞kNow put c =i=1 cji . Then for all k we have α (c) = c. Indeed, let k =kand k = 0 mod j2 . . . jl+1 . By (ii), αk (c) − c = α(cjl ) − cjl +0 mod j2 .

. . jl ∞∞kkk( i=l+1 cjl ) − i=l+1 cjl . But ν̄(α (cjl ) − cjl ) < ν̄(cjl+1 ) ≤ ν̄(α ( ∞αi=l+1 cjl ) −∞kc).Therefore,α(c)−c=0.i=l+1 jl2Consider the field F̄ = π(c) ⊂ K. Let us show that this field can be embedded inO.Take any lift c ∈ O of the element c: c mod ℘ = c. It is clear that c commutewith any element from π. It is easy to see that c is a transcendental element over π.Indeed, assume the converse. Then its equation modulo ℘ must have infinite numberkof solutions, because cα = c ∀k, a contradiction. Therefore, π[c ] ℘ = 0. So, the fieldof fractions F̄ can be embedded in O.Let L̄ be a maximal field extension of F̄ which can be embedded in O. Denote byL its image in O.

Take ā ∈ K̄, ā ∈/ L̄. We claim that there exists a lifting a ∈ O of āsuch that a commutes with every element in L.17Step 2. Take any lifting a in O of ā. For every element x ∈ L we haveaxa−1 mod ℘ = x. If z is a parameter of K we can writeaxa−1 = x + xδ1 z,where xδ1 ∈ O. The map δ¯1 : x ∈ L → δ1 (x) ∈ K̄ is an α-derivation.

Indeed,a(x1 + x2 )a−1 = (x1 + x2 ) + (x1 + x2 )δ1 zδδδδa(x1 + x2 )a−1 = ax1 a−1 + ax2 a−1 = x1 + x11 z + x2 + x21 z = (x1 + x2 ) + (x11 + x21 )z¯δδTherefore, (x1 + x2 )δ1 = x11 + x21 . Then, we havea(x1 x2 )a−1 = (ax1 a−1 )(ax2 a−1 )Henceδδδδδδx1 x2 + (x1 x2 )δ1 z = (x1 + x11 z)(x2 + x21 z) = x1 x2 + x1 x21 z + x11 zx2 + x11 zx21 zδδ≡ x1 x2 + x1 x21 z + x11 xα2 zδδmod ℘2 = x1 x2 + (x11 xα2 + x1 x21 )zmod ℘2Therefore,δδδδ(x1 x2 )δ1 = x11 xα2 + x1 x21 = x11 xα2 + x1 x21By lemma 0.12, δ¯1 is an inner α-derivation, say δ¯1 (x) = d(xα − x).

Put a˜1 :=(1 + a1 z)a, where a1 mod ℘ = −d. Using the same calculations as in lemma 0.11 wehave(1 + a1 z)axa−1 (1 + a1 z)−1 = x + (xδ1 + a1 xα − xa1 )z mod ℘2Since xδ1 + a1 xα − xa1 = 0 mod ℘, we get a˜1 xa˜1 −1 = x + xδ2 z 2 . Using the samearguments as above one can check that δ¯2 : L → K̄ is an α2 -derivation. By inductionwe can find an an element ãi = (1 + ai z i ) . . . (1 + a1 z)a such thatãi xãi −1 = x + xδi+1 z i+1 ,¯¯and δi+1: L → K̄ is an αi+1 -derivation. By lemma 0.12, δi+1is an inner αi+1 -derivation.i+1So there exists an element ai+1˜ = (1 + ai+1 z )ãi such thatai+1˜ xai+1˜ −1 = x + xδi+2 z i+2for any x ∈ L.

It is clear that the sequence {ãi } converges in K. Since ãi mod ℘ = ā,the limit of this sequence is a needed lifting.Step 3. Now suppose ā is a transcendental over K̄. Then by step 2 there exists alifting a ∈ O such that a commutes with every element in L. Then L[a] ℘ = 0 andthe field of fractions L(a) can be embedded in O, which contradicts the maximality18of L. So we can assume that K̄ is algebraic over L. Suppose ā is an algebraic andseparable element over L̄. Using a generalisation of Hensel’s lemma (see below) we canfind a lifting a of a such that a commutes with elements of L and a is algebraic overL, which again leads to a contradiction.kFinally, let ā be purely inseparable over L̄, āp = x̄, x ∈ L.

Let a be its lifting whichkcommutes with every element of L. Then a p − x commutes with every element of L.kIf ν(a p − x) = r = ∞ then similarly to the beginning of this proof we deduce thatkkthe image of (a p − x)c(a p − x)−1 in K̄ is equal to αr (c), where c is an element fromklemma 0.14. Since αr (c) = c, we get a contradiction. Therefore, a p = x and the fieldL(a ) can be embedded in O, which contradicts the maximality of L. Thus, L̄ = K̄.The theorem is proved.2Proposition 0.15 (Hensel’slemma) 1 Let O be a complete valuation ring in K, I nbe the valuation ideal, I = 0, and let F be a subfield in O. Let A ∈ O be such that∀l ∈ F Al = lA. Let f (X) ∈ F [X], f (A) ∈/ I and f (A) ∈ I.Then there exists an element  ∈ O such thata)  commutes with A,b)  − A ∈ I,c) f (Â) = 0d) Âl = l ∀l ∈ FProof.

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