# 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 $X$.