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 $P(X_{\sigma(1)},...,X_{\sigma(n)})(x_1,...,x_n)=0$.
 A: 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".
