Genus change in inseparable extensions of function fields
Genus change in inseparable extensions of function fields
复制标题
功能领域不可分割的扩展中的属变
DOI：
10.1090/s0002-9939-1952-0047631-9
复制
发表时间：
1952
影响因子：
0.8
通讯作者：
J. Tate
中科院分区：
文献类型：
--
作者：
J. Tate
适用于所有tGK。Sa是从K到k的迹的一个特别方便的替代物，它等于0。当然，Sa虽然不是完全任意的，但仍然是非不变的，问题是如果我们用另一个生成元B代替a，Sa如何变换。如果我们回想一下，由于K是一个域，Sa是非平凡的，任何K到k的k-线性映射S都可以表示为S（t）=Sa（，，y），其中y是由S唯一确定的K的某个元素，那么这个问题就可以更精确地表述。因此，我们的问题是：如何根据a和B计算使So（t）= S的元素-y。(tzy) ?答案用派生词来表达最为方便。环中的导子是映射x-？环Dx自身具有性质D（x+y）=D（x）+D（y）和D（xy）=x（Dy）+（Dx）y。如果环是交换的，则规则D（xr）= vx '-'Dx由归纳得出。普通形式微分F（X）-F '（X）是我们的域k上的单字母多项式X的环k [X]中的导子。它将由形式为XP-a的多项式生成的主理想映射到自身，因为（（XPa）F（X））'（XP-a）F'（X）。的
holds for all tGK. Sa is a particularly convenient substitute for the trace from K to k, which is identically 0. Of course Sa, although not completely arbitrary, is nevertheless noninvariant, and the question arises as to how Sa transforms if we replace a by another generator ,B. This question can be more precisely stated if we recall that since K is a field and Sa is nontrivial, any k-linear map S of K into k can be expressed in the form S(t) =Sa(,,y), where y is some element of K uniquely determined by S. Our question is therefore: How does one compute, in terms of a and ,B, the element -y for which So(t) = S.(tzy) ? The answer is most conveniently expressed in terms of derivations. A derivation in a ring is a map x-?Dx of the ring into itself with the properties D(x+y) =D(x)+D(y) and D(xy) =x(Dy)+(Dx)y. The rule D(xr) =vx'-'Dx follows by induction if the ring is commutative. The ordinary formal differentiation F(X)--F'(X) is a derivation in the ring k [X] of polynomials in one letter X over our field k. It maps a principal ideal generated by a polynomial of the form XP-a into itself because ((XP a) F(X)) '(XP-a) F'(X). The