2
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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".

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.