Multidimensional local skew-fields (792481), страница 16

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. . is conjugate with theautomorphism β̄: β̄(u) = ξ(ᾱ)u + xui(ᾱ) + . . ., where x ∈ k ∗ /k ∗(i(ᾱ)−1) .Lemma 1.3 Let a0 ∈ K̄, ᾱ ∈ Autk (K̄). The linear map T = ᾱ − a0 : K̄ → K̄ has thefollowing property:if ᾱn = Id for some n, then dim(kerT ) = dim(cokerT ) = d, where d = 0 or ∞;if ᾱn = Id and chark = 0, then dim(kerT ) = dim(cokerT ) = d, where d = 0 or 1;if ᾱn = Id and chark = p, then one of the following cases holds:1) dim(kerT ) = dim(cokerT ) = 0 or2) dim(kerT ) = 0, dim(cokerT ) = ∞ or3) dim(kerT ) = 1, dim(cokerT ) = ∞.Proof.

By proposition 1.2 we can assume ᾱ(u) = ξu + xui(ᾱ) + . . ., where ξ is aprimitive n-th root of unity.If ᾱ(u) = ξu, ξ n = 1, then the first claim of lemma is clear, so from now onᾱ(u) = ξu (note that we have proved the first claim in the case chark = 0, because,by corollary 12, any automorphism of a finite order looks like this).Suppose the element a0 satisfy one of the following properties:ν̄(a0 ) = 0 orν̄(a0 ) = 0 but a0 = ξ j for all j ∈ Z.

Let’s study values of the valuation ν̄ on elementsT (ul ) for different l. We have:ᾱ(ul ) − a0 ul = (ξu + xui(ᾱ) + . . .)l − a0 ul = ξ l ul (1 + ξ −1 xui(ᾱ)−1 + . . .)l − a0 ulTherefore:ν̄(T (ul )) = l or l + ν̄(a0 ) if ν̄(a0 ) < 0.So, we can solve any equation ᾱ(y) − a0 y = Y , and the map T is surjective. It isinjective, because the values ν̄(T (ul )) are finite and ν̄(T (ul )) = ν̄(T (ul1 )) if l = l1 .Suppose now a0 = ξ j . Since the injectivity and the projectivity of the map ᾱ−a0 aredefined by the existence and the uniqueness of a solution of the equation ᾱ(y)−a0 y = Yfor any Y ∈ K̄, we cany by yuj and assume that a0 = 1. Then a0 can be writtenreplace∞as the product a0 = j=1 (1 + a0j uj ), a0j ∈ k.Put q = ν̄(a0 − 1). There are two possible case: q < i(ᾱ) − 1 and q ≥ i(ᾱ) − 1.Let q < i(ᾱ) − 1.

Then we can assume n|q. To prove it we have to prove that a0can be written as the product a10 ᾱ(x)for some x, where ν̄(a10 ) = q1 > q, n|q1 .xll)ᾱ(u )= 1+c= 1 + c(ᾱ(ul ) − ul )(1 + cul )−1 , where c is a constant.Note that ᾱ(1+cu1+cul1+culᾱ(ul ) − ul = (ξu + xui(ᾱ) + . . .)l − ul = ξ l ul (1 + ξ −1 xui(ᾱ)−1 + . . .)l − ul73(1.1)From this formula we get the following property:l)if n |l, then ν̄(ᾱ(ul ) − ul ) = l and ᾱ(1+cu= 1 + (ξ l − 1)cul + . . ..

Hence, a0 can be1+culrepresented as the product above, because there exists a constant c such that the valuel)) increases.ν̄(a0 ᾱ(1+cu1+culSo, let we now have: q < i(ᾱ) − 1 and n|q. Let’s study values of the valuation ν̄ onelements T (ul ) for different l. By formula (1.1) we have:ν̄(T (ul )) = l if n |lν̄(T (ul )) = l + q if (l, n) = 1.Therefore, we can solve any equation ᾱ(y) − a0 y = Y , and the map T is surjective.It is injective, because all the values ν̄(T (ul )) are finite and ν̄(T (ul )) = ν̄(T (ul1 )) ifl = l1 .Consider now the case q ≥ i(ᾱ) − 1. As in the first case we can assume that n|q.We divide this case into three cases:q = i(ᾱ) − 1,q > i(ᾱ) − 1 and q is finiteq is infinite, i.e.

a0 = 1.Let q = i(ᾱ) − 1. Let a0 = 1 + wuq + . . .. Note that(ξu + xui(ᾱ) + . . .)lnᾱ(culn )== 1 + nlξ −1 xui(ᾱ)−1 + . . .lnlncuuHence, if w = nlξ −1 x for all l, we can apply the same arguments as in the first caseand get that T is injective and surjective, i.e. d = 0. Otherwise, we can write a0 =a10 ᾱ(uln )/uln , where ν̄(a10 − 1) > q, and reduce this case to the case q > i(ᾱ) − 1.Let chark = 0. We claim that the case q > i(ᾱ) − 1 can be reduced to the case.

We know, thatq = ∞. In this connection it is necessary to show that a0 = ᾱ(A)Aᾱ(1 + cul )= 1 + c(ᾱ(ul ) − ul )(1 + cul )−11 + culand ᾱ(ul ) − ul has the valuation equal to l if (l, n) = 1, and to (i(ᾱ) − 1) + l if n|l,(l, chark) = 1 and l = 0.From here we get the necessary result, because we can multiply a0 sequentially bysuitable elements of the form 1 + cuj + .

. ., each of which can be got from a certainelementof the form 1 + cj uj or 1 + cj uj−(i(ᾱ)−1) . It is clear that the product A =∞j−(i(ᾱ)−1)) converges.j=q (1 + cj uLet now chark = 0 and q = ∞, i.e. a0 = 1. Then we claim that d = 1.

Let us firstfind the dimension of the kernel of the map T . To do that we investigate the values ofthe valuation ν̄ of the elements T (ul ) by different l. We have:ν̄(T (ul )) = l if n |lν̄(T (ul )) = l + (i(ᾱ) − 1) if n|l and l = 0ν̄(T (1)) = ∞ if l = 0, i.e. T (1) = 0.74From this follows that the kernel is one-dimensional and consists of the elements ofthe field k, because all ν̄(T (ul )) are finite if l = 0 and ν̄(T (ul )) = ν̄(T (ul1 )) if l = l1 .On the other hand, we get also that the cokernel is one-dimensional, because we canget an element with any value of valuation except an element with the value (i(ᾱ) − 1),and there exists a pullback of any convergent (to zero) sequence, which is also convergeto zero.Now we must examine the cases, when chark = p.

Let us first consider the casechark = p and q = ∞.We prove that ᾱn = 1 if and only if dimk (ker(ᾱ − 1)) = ∞.Let ᾱn = 1, n = pk m, (p, m) = 1. It is obvious that if exists an element x ∈ K̄,x∈/ k such that (ᾱ − 1)(x) = 0, then dimk (ker(ᾱ − 1)) = ∞. Suppose, that there is nosuch an element. Therefore:ᾱm (u) = u + a1 , a1 ∈ K̄, ν̄(a1 ) > 1, a1 = 0,ᾱ2m (u) = u + 2a1 + a2 , a2 ∈ K̄, ν̄(a2 ) > ν̄(a1 ), a2 = 0,. . .,kᾱp m (u) = u + .

. . + apk , apk ∈ K̄, ν̄(apk ) > ν̄(apk −1 ), apk = 0,and we get a contradiction.Conversely, let dimk (ker(ᾱ−1)) = inf ty. Assume F = ker(ᾱm −1), m = ord(ξ(ᾱ)).It’s clear that F is a field.Let n ∈ N be a minimal positive value of the valuation ν̄ on this field.Then n = pk , k ∈ Z. For, if n = pk l, (l, p) = 1, then there exists an element x ∈ Fwith such a value and, moreover, x = dl , d ∈ K̄. But then d ∈ F , because ξ(ᾱm ) = 1,a contradiction.So, K̄/F is a finite algebraic extension of degree pk , therefore ᾱm is an automorphismof a finite order.

It is easy to see that the order is equal to n, i.e. ᾱ is a generator ofthe cyclic Galois group Gal(K̄/(kerT )).Remark. In particular, we have got a description of a subgroup of elements offinite order in the so-called ”Nottingham” group. See [3], [12], [5] for further detailsabout this group (i.e. the group Autk (K̄), charK̄ = p).Let ᾱ be an automorphism of infinite order. Then kerT = k, dimk (kerT ) = 1. Let(i(ᾱ) − 1, p) = 1. We claim that for any integer N > 0 there exist numbers h(N ) ∈ N,h(N ) > h(N − 1) and x, h(N − 1) < x ≤ h(N ) such that the maximal value of thevaluation on a preimage of arbitrary element with the value x less than −N (or thepreimage is empty).

From this follows that one can construct infinitely many elements,which are not in the image of the map T .For N = 1 it’s clear — h(1) = x = i(ᾱ) − 1. For arbitrary N consider the vectorspace < T (ul ), −N ≤ l ≤ s(k) >, where s(k) = (i(ᾱ)−1)+p(i(ᾱ)−1)+. . .+pk (i(ᾱ)−1).It’s clear that dimk < T (ul ), −N ≤ l ≤ s(k) >≤ (s(k) + N ).

From the other hand side,kν̄(T (up (i(ᾱ)−1) )) = 2pk (i(ᾱ) − 1) > s(k),kk−1ν̄(T (up (i(ᾱ)−1)+p (i(ᾱ)−1) )) = 2pk−1 (i(ᾱ) − 1) + pk (i(ᾱ) − 1) > s(k),75...kν̄(T (up (i(ᾱ)−1)+...+p(i(ᾱ)−1) )) > s(k).So, our property holds for k N . Indeed, assume the converse.

Then < ul , h(N −1) ≤ l ≤ s(k) >⊂< T (ul ), −N ≤ l ≤ s(k) > for all k, and s(k)−h(N −1) ≤ s(k)+N −kfor all k, a contradiction.In the case (i(ᾱ) − 1, p) = 1 we have: ν̄(T (ul )) = ν̄(T (ul1 )) if l = l1 . Therefore, thecokernel of the map T has infinite dimension.Let now chark = p, q > (i(ᾱ) − 1) and q is finite. Note that T is not injective ifand only if a0 = ᾱ(A)/A. Therefore, if T is not injective, this case is equivalent to thecase q = ∞.Let T be injective.Here two cases are possible:1) there exists an integer i ≥ ν̄(a0 − 1) such that for some s1 , s2 ∈ K̄ a0 = s2 ᾱ(s1 )/s1 ,where ν̄(s2 − 1) = i and there are no elements s1 , s2 such that a0 = s2 ᾱ(s1 )/s1 , whereν̄(s2 − 1) > i.2) for any integer i such that i ≥ ν̄(a0 − 1) there exist s1 , s2 ∈ K̄ such that a0 =s2 ᾱ(s1 )/s1 , where ν̄(s2 − 1) ≥ i.For example, the first case takes the place when ν̄(T (a0 − 1)) < (i(ᾱ) − 1), thesecond — when ν̄(T (a0 − 1)) > (i(ᾱ) − 1).

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