Characterisation of Galois Group with the action of $\sigma \in S_n$ on the roots

Let $f \in K[X]$ be irreducible and separable with roots $x_1,...,x_n$ in a splitting field $L$ of $f$ over $K$. We identify $\text{Gal}(L|K)$ with $\text{Gal}(L|K)\cong G\subset S_n$.

How can I see the equivalence of the following two statements? (which means a characterisation of the galois group with the action of a $\sigma \in G$ on the roots $x_1,...,x_n$)

$(1)$ $\sigma \in G$.

$(2)$ If $P \in K[X_1,...,X_n]$ with $P(x_1,...,x_n)=0$, then for $P(X_{\sigma(1)},...,X_{\sigma(n)})$ it follows that $P(X_{\sigma(1)},...,X_{\sigma(n)})(x_1,...,x_n)=0$.

I prefer using three different notations, say $\sigma$ for the permutation of indices ,$\tau$ for the permutation of roots and $t$ for the field homomorphism extending $\tau$ (when it exists) : thus $\tau(x_i)=x_{\sigma(i)}$.

$(1) \Rightarrow (2)$ If $t$ exists, and $P(x_1,x_2,\ldots,x_n)=0$, we have

$$P(X_{\sigma(1)},...,X_{\sigma(n)})(x_1,...,x_n)= P(x_{\sigma(1)},...,x_{\sigma(n)})= P(\tau(x_1),\tau(x_2),\ldots,\tau(x_n))= t(P(x_1,x_2,\ldots,x_n))=t(0)=0.$$

$(2) \Rightarrow (1)$ The permutation $\tau$ is defined on $\lbrace x_1,x_2, \ldots ,x_n\rbrace$ and we sould like to extend it to the the whole of $L=K[x_1,\ldots,x_n]$. The obvious definition which comes to our mind is

$$t(A(x_1,x_2,\ldots,x_n))=A(x_{\sigma(1)},\ldots,x_{\sigma(n)}) \tag{1}$$

for any $A\in K[X_1,\ldots,X_n]$. The problem with (1) is that it might be an incorrect definition, with two different values set for the same argument. However, if $A(x_1,x_2,\ldots,x_n)=B(x_1,x_2,\ldots,x_n)$ for two polynomials $A,B$, then the polynomial $C=A-B$ satisfies $C(x_1,x_2,\ldots,x_n)=0$, so $C(x_{\sigma(1)},...,x_{\sigma(n)})=0$ by (2), and (1) will therefore yield the same value in both cases.

So $t$ is correctly defined, and it follows immediately from its definition that it is a homomorphism.

Alternatively, you can define $t$ as a "quotient map".

• @prime_dan To see why $t|_{K}=id$, take a constant $A$ in (1) – Ewan Delanoy Jun 1 '14 at 12:34