对称分析 * * * Aa + Bb + Cc + .... 忽然意识到这个式子是 非对称 的 (关于根的位置)！即交换任何两个小写字母的位置，则整个式子的值会发生改变。假如直接把根加起来 a + b + c + .... 由于加法交换律，交换任何两个根的位置，则整个式子的值不会发生改变。换句话说，Galois预解式是使得根关于位置不对称的所 ...
《Galois theory》 H.E. p. 53 * * * ?? Now (B) follows immediately from the last corollary because this corollary shows that if t' is any conjugate of t then (set s = t') the substitutions of the roots that result from changing t to one of its conjugetes are the same as ...
《Galois cohomology》 J.P.S. p.73 * * * 13:30 Corollary . Let n be an integer ≥ 1, p rime to the characteristic of k. ---- 令 n 为自然数，并且与 k 的特征互质. ---- 基本域 k 的特征是什么 ? . Let μ n be the group of n-th roots of unity (in k s ). ---- ...
《Galois theory》 H.E. p. 53 * * * ?? Corollary . Let t be a Galois resolvent, let t' be one of its conjugates, and let S be the corresponding substitution of the roots as in the previous corollary. ---- 令 t 为 Galois 预解 ，t' 为其任一共轭，S 为将 t-行 带到 t&# ...
《Galois cohomology》 J.P.S. p.73 * * * 11:10 1.2 First examples . Let G a (resp. G m ) be the additive (resp. multiplicative) group, defined by the relation G a (K) = K (resp. G m (K) = K*). We have (cf. , p. 158): ---- 设 G a 为加法群，由关系 G a( K) = K 定义 ...
《Galois theory》 H.E. p. 53 * * * ?? In order to deduce (B) and (C) from Proposition 1, it is helpful to observe: . Corollary . Let t be a Galois resolvent, let t' be one of its conjugates, and let S be the substitution of the roots which carries the row of the above ta ...
《Galois cohomology》 J.P.S. p.72 * * * 12:20 In particular, we see that two seperable closures of k define cohomology groups H^q(ks/k, A) which correspond bijectively and canonically to each other. ---- k 的两个可分闭包定义上同调群 H^q(ks/k, A)，它们彼此双射、规范地对应. . ...
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