Multidimensional local skew-fields, страница 6

By lemma 0.11,(ii) there existsa parameter z = z + ai−j z i−j+1 such that the compositum of uj → uj+1 = uj + ci−j z i−jand z → z = z + ai−j z i−j+1 does not change this map. To use the induction hypothesisand complete the proof we have to show that there exists a parameter u = u + bz isuch that the compositum of uj → uj+1 = uj + ci−j z i−j , z → z = z + ai−j z i−j+1 andthe map which is givenu → u = u + bz i does not change the map δ2i . Denote by δ2iby the compositum uj → uj+1 = uj + ci−j z i−j , z → z = z + ai−j z i−j+1 . By lemma 0.24there exists such a parameter u iffresuj+1(δ2i− δ2i )(uj+1 )duj+1 = 0.i(uδj+1)2We have uj+1 = ci−j z i−j + .

. . + ci z i . One can decompose the change u → uj+1 intwo changes: u → u = u + ci−j z i−j and u → u = u + ci−j+1 z i−j+1 + . . . + ci z i . Thesecond change does not change the map δ2i−j , so by the induction hypothesis it sufficeto prove that the residue is equal to zero for the compositum of u → u = u + ci−j z i−jand z → z = z + ai−j z i−j+1 .∂((juδi )−1 ci−j )uδi . Note that ifUsing lemma 0.24, we can calculate ai−j : ai−j = ∂uν̄(ci−j ) = r is big enough then the residue is equal to zero. One can show it with helpof lemmas 0.11 and 0.24.

We denote by r the minimal positive integer which satisfythis property. hLet ci−j = rh=N xh uh + ∞h=r+1 xh u . Then we can decompose the change u →finite number of changes u → u1 = u + xN uN z i−j , . . . , ur−N −1 →u = u + ci−j z i−j inh i−j. It is clear that it suffice to prove our assertion forur−N = ur−N −1 + ∞h=r+1 xh u zeach change. So we have to prove it for an arbitrary change u → u + xuh z j , x ∈ k.We have to check that the compositum of u → u + xuh z j and z → z − (j − i)−1 (h −25r)xuh−1 z j+1 does not change the map δ2i , i.e.

the residue above is equal to zero. Put−(j − i)−1 (h − r)xuh−1 = b, xuh = b .Let us show that such a compositum change only the maps δi+qj , q ∈ N. Moreover,we claim that δi+qj (u) = const · ur+q(h−1) . Indeed, if u = u + b z j we havezu z −1 = u + uδi z i + uδ2i z 2i + . . . + (b + bδi + bδ2i z 2i + . . .)z j =u + uδi z i + bδi z i+j + uδ2i z 2i + . . . =∂∂ δi i+j1 ∂ 2 δi 2 i+2j1 ∂ 3 δi 3 i+3j(u)bz−(u )b z−...(u )b )z −u + uδi z i + ( (b )uδi −∂u∂u2! ∂u23! ∂u31 ∂ e δi e 2i(u )b )z+(uδ2i −e! ∂uewhere ej = i if j|i. If j | i, uδ2i does not change.Therefore,∂ δi∂ δi (b )u −(u )b ,uδi+j =∂u∂uand ν̄(uδi+j ) = r + (h − 1).Then∂1 ∂ 2 δi 2(u )buδi+2j = − (uδi+j )b −∂u2! ∂u2and ν̄(uδi+2j ) = r + 2(h − 1),uδi+qj = −∂ δi+(q−1)j 1 ∂ 2 δi+(q−2)j 21 ∂ q δi q)b −(u)b−...−(u )b(u∂u2! ∂u2q! ∂uqand ν̄(uδi+qj ) = r + q(h − 1).If z → z = z + bz j+1 we havez u = (z + bz j+1 )u = uz + uδi z i+1 + uδi+j z i+j+1 + .

. .+uδ2i z 2i+1 + . . . + buz j+1 + (j + 1)buδi z i+j+1 + . . . + (j + 1)buδ2i−j+1 + m. withz >2i+1 =uz + uδi z i+1 + uδi+j z i+j+1 + . . . + uδ2i z 2i+1 + . . . =u(z + bz j+1 ) + uδi (z + bz j+1 )i+1 + uδi+j (z + bz j+1 )i+j+1 + . . . + uδ2i (z + bz j+1 )2i+1 + . . .Hence,uδi+j = uδi+j + b(j − i)uδiand ν̄(uδi+j ) = r + (h − 1),21uδi+2j = uδi+2j − Ci+1b2 uδi − Ci+j+1buδi+j + (j + 1)buδi+j26and ν̄(uδi+2j ) = r + 2(h − 1),qq−11uδi+qj = uδi+qj −Ci+1bq uδi −Ci+j+1bq−1 uδi+j −. .

.−Ci+(q−1)j+1buδi+(q−1)j +(j +1)buδi+(q−1)jand ν̄(uδi+qj ) = r + q(h − 1).So if j | i, uδ2i does not change. If j|i but e(h − 1) − r = −1, then the residue isequal to zero ( note that e(h−1)−r = −1 if (r −1, i) = 1). At last, if e(h−1)−r = −1,then one can check the assertion by direct calculations.So we have shown that the change u → u = u + c1 z + . . . + ci z i is equivalent tothe change z → z = z + a1 z 2 + .

. ., u → u = u + ci z i + . . .. By lemma 0.24 thechange u → u = u + ci z i + . . . does not change a. By lemma 0.11 the change z → z =z + a1 z 2 + . . . does not change δi+1 , . . . , δ2i−1 only if a1 = a2 = . . . = ai−1 = 0. But thenit does not change also a. Therefore, any change of the type z → z = z + a1 z 2 + . . .,u → u = u + c1 z + . . ., which does not change δi+1 , . . . , δ2i−1 , does not change also a.To complete the proof we only have to show that changes of the type u → u =x0 u + x1 u2 + . . ., xj ∈ k and z → z = a0 z, a0 = 0 ∈ k((u)) don’t change a. It is clearfor the first change. For the second change we haveδ2iδi δi−i i−1 δii−1 δi+ ia0 (a−1uδ2i = a−2i0 [u0 ) u − a0 (a0 a0 + .

. . + a0 (a0 ) )] =δi δiδ2ia−2i+ i(i + 1)/2a−10 [u0 a0 u ] 22δi−2i−1 δi δiu(δi ) = a−2ia0 u0 u − ia0Therefore, 22uδ2i − (i + 1)/2u(δi )uδ2i − (i + 1)/2uδi==a(uδi )2(uδi )2The proposition is proved.2Remark. If a two-dimensional skew field K does not split, then the numbers i, r, acan not be defined as the following example shows (cf. also the remark after theorem0.13).Example.2 Let ”((u)) < x1 , x2 > be a free associative algebra over ”((u)) withgenerators x1 , x2 . Let I =< [[x1 , x2 ], x1 ], [[x1 , x2 ], x2 ] >.

It is easy to see that the quotientS = ”((u)) < x1 , x2 > /Iis a ”-algebra which has no non-trivial zero divisors, and in which z = [x1 , x2 ] + I is acentral element. The elements z, ui = xi + I (i = 1, 2) are algebraically independent.Any element of S can be uniquely represented in the formf0 + f1 z + f2 z 2 + . . . + fm z m2I thank N.I. Dubrovin for showing me this example27where f0 , . . .

fm are polynomials in the variables u1 , u2 :a + bu1 + cu2 + d1 u21 + d2 u1 u2 + d3 u22 + . . .S is an Ore domain (see [4]), and the quotient skew field K has a discrete valuation νsuch that ν(ui ) = 0, ν(”((u))) = 0, ν(z) = 1. The completion of K with respect to ν isa two-dimensional local skew field which does not split as the following lemma shows.Lemma 0.27 Suppose there exist elements u1 , u2 in the valuation ring O of a twodimensional local skew field K such that the element z = u1 u2 − u2 u1 is a parameterand for any m ∈ zO \ z 2 O the elements [ui , m] = ui m − mui (i = 1, 2) belong to z 2 O.Then K does not split.Proof. Assume the converse.

Let π : K̄ → K be an embedding. Consider elementsf ∈ π(u1 ), g ∈ π(u2 ). Then m1 = f − u1 , m2 = g − u2 ∈ zO and0 = [u1 + m1 , u2 + m2 ] = [u1 , u2 ] + [m1 , u2 ] + [u1 , m2 ] + [m1 , m2 ] =z + [m1 , u2 ] + [u1 , m2 ] + [m1 , m2 ]Note that the second and the third summands belong to z 2 O and [m1 , m2 ] ∈ z 2 O,because m1 m2 , m2 m1 ∈ z 2 O. But then∞ = ν(0) = ν(z + [m1 , u2 ] + [u1 , m2 ] + [m1 , m2 ]) = ν(z) = 1,a contradiction.2In this skew field we have i = ∞, and r, a are not defined.

From the other hand, ifwe consider the change z → u1 z, then i become equal to 1, r = 0, a = 0.So these numbers depend on the choice of parameters in this unsplittable skew field.Proposition 0.28 Let K be a skew field which satisfy the conditions in the beginningof this paragraph. Let char k = 0, α = 1 and i ≥ 1. Then K is isomorphic to a skewfield k((u))((z)) with zuz −1 = u + uδi z i + uδ2i z 2i , where δi (u) = cur , c ∈ k ∗ /(k ∗ )e , where(u) = (a(0, . .

. , 0) + r(i + 1)/2)u−1 (δi (u))2 and δj (u) = 0 for j = i, 2i.e = (r − 1, i); δ2iδiProof. Consider the change z → z = a0 z. By lemma 0.11, (iii) we have uδi = a−i0 u .δiSo, u can be made to be equal toδuδi = c0 uν̄(u i ) modi,where c0 ∈ k ∗ /(k ∗ )i . By lemmas 0.24 and 0.11, c0 depend only on changes of the typez → z = a0 z, u → u = x0 u, where a0 , x0 ∈ k.

So, c0 can be made to be equal to−r+1. Therefore, c ∈ k ∗ /(k ∗ )e , where e = (r − 1, i).c = c0 a−i0 x028Let us show that there exists a change z → z = z + a1 z 2 + . . . such that δj (u) = 0for 2i > j > i. Indeed, it can be done by a sequence of changes of the type z → z =z + bz j+1 . By corollary 4, for any such j there exists b such that δj (u) = 0.Let us show that there exists a change such that it changes the map δ2i as follows:(u ) = (a + r(i + 1)/2)u−1 (uδi )2 . To show it we use lemma 0.24, (ii). By this lemmaδ2iit suffice to show that there exists an element b such thatuδ2i − (a + r(i + 1)/2)u−1 (uδi )2 + bδi − (uδi ) b = 0, where the prime ’ denote the derivation by u.

By corollary 2, δi is a derivation.Therefore, we can rewrite the equation above as followsuδ2i − (a + r(i + 1)/2)u−1 (uδi )2 + b uδi − (uδi ) b = 0One can find a solution of this equation in the form b = uδj b̃. The equation has asolution if b̃ + uδ2i (uδi )−2 − (a + r(i + 1)/2)u−1 = 0. The last equation holds, becauseδ 2 (u)resu (δii(u))2 du = r.Using now the same arguments as in the previous paragraph, we can complete theproof.20.3.2The general case.Consider now the case αn = Id for some natural n > 1.Lemma 0.29 Suppose the canonical automorphism α of a local skew field K satisfy theproperty αn = 1 for some natural n > 1.

Then there exists a parameter z = z+a1 z 2 +. . .such thatz uz −1 = uα + uδn z n + uδ2n z 2n + . . .Here δj = 0 if n |j.Proof. Letzuz −1 = uα + uδ1 z + uδ2 z 2 + . . .By corollary 2, δ1 is a (α2 , α)-derivation.Since n > 1, α2 = α. Therefore, by lemma 0.12, δ1 is an inner derivation and2δ1 (u) = duα − uα d. Put z = z − dz 2 . By lemma 0.11, (i) we havez uz −1 = uα + uδ2 z 2 + . . .By corollary 2, δ2 is a (α3 , α)-derivation. If n = 2 then it is inner and we can applylemma 0.11. By induction we get that there exists a parameter z such thatz uz −1 = uα + uδn z n + uδn+1 z n+1 + . . .29where δn is a (αn+1 , α) = (α, α)-derivation, i.e. δn α−1 is a derivation.is a (α2 , α)-derivation.