Kollár/Smith/Corti 一章, 問題1.8

Let k'/k be a finite Galois extension. The Galois group \mathrm{Gal}(k'/k) acts on \mathbb{A}_{k'}^n = \mathrm{Spec}k'[x_1,\ldots,x_n] coordinatewise. Let X be a closed algebraic set of \mathbb{A}_{k'}^n. Prove that the following are equivalent:

  1. the set X can be defined by polynomials in k[x_1,\ldots,x_n];
  2. the set X is invariant under the \mathrm{Gal}(k'/k)-action.

Proof. Suppose f = \sum a_Ix^I \in k'[x_1,\ldots,x_n] vanishes on X. Let \sigma \in \mathrm{Gal}(k'/k), and set f^\sigma = \sum \sigma^{-1}(a_I)x^I. If X is defined by polynomials in k[x_1,\ldots,x_n] then f = f^\sigma and so X is Galois-invariant.
Conversely, if X is Galois-invariant, then if f vanishes on X then f^\sigma does also. Denoting E_1(f),\ldots,E_d(f) to be the elementary symmetric polynomials of f^\sigma for \sigma \in \mathrm{Gal}(k'/k), each E_i(f) \in k[x_1,\ldots,x_n] and they vanish on X. We claim that these polynomials define X. For, if every E_i(f) vanished at a point P, then every f^\sigma(P) is zero for they are the roots of the polynomial z^d = z^d + \sum_{i=1}^d(-1)^iE_i(f)(P)z^{d-i}. So the E_i(f) define X.