《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'-行的置换.

.

Let s be any other Galois resolvent of the same equation.

---- 令 s 为同一方程的任意其它 Galois 预解.

.

Then there is a conjugate s' of s such that S is equal to the substitution of the roots corresponding to s and s'.

---- 则存在 s 的共轭 s' 使得 S 等于将 s 带到 s' 的置换.

.

评论：此引理是说，S (从而整个 Galois 群) 与 Galois 预解 的选择无关.

.

Proof. Since s is in K(a, b, c, ...), the extension of S defined in the preceding corollary applies to s.

---- 证: 首先 s 属于 K(a, b, c, ...)，于是 S 的扩展可施加于 s.

.

Define s' to be S(s).

---- 定义 s' 为 S(s).

.

评论：将要找的 s' 取作 S(s).

.

If H(x) is an irreducible polynomial of which s is a root,...

---- 若 H(x) 是不可约多项式，s 是它的根...

.

then application of S to H(s) = 0 gives H(s') = 0, so that s' is a conjugate of s.

---- 则施加 S 到 H(s) = 0 得 H(s') = 0，于是 s' 是 s 的共轭.

.

评论：因 S 是自同构，S(H(s)) = H(S(s)) = H(s'). 而 S(0) = 0. 故 H(s') = 0.

.

Now if a = ψa(s), where ψa is a polynomial with coefficients in K, then application of S to both sides gives S(a) = ψa(Ss) = ψa(s').

---- 若 a = ψa(s) 则 S 作用到该式两边，得 S(a) = ψa(Ss) = ψa(s').

.

Similarly, S(b) = ψb(s'), S(c) = ψc(s'), ..., as was to be shown.

---- 对根 b, c, ... 亦可类推. 得证.

.

证明的简记:

..............................H

K(a, b, c, ...) ~ s ~ S ~ s' 

............................ψr 

1. s ∈ K(a, b, c, ...) ==> s':=S(s).     构造

2.              H(s) = 0 ==> H(s') = 0.  共轭

3.                Sψr(s) = ψr(s')                 置换

.

小结：Galois 群 与 Galois 预解的选择无关.

* * *??