During the solving of a homework problem I had encountered this question. Let $k$ be a base field and $k^{sep}$ its maximal separable extension. I know that the Galois group of $L/k$ – split field of separable irreducible polynomial acts transitively on roots, we can send one root to another and extend the map to $k$-automorphism of $L$, but I have no idea how to prove if it can be extended to $k$-automorphism of $k^{sep}$. And of course I don't know if there is some counterexample. Will be glad for some hints or help.
Absolute Galois group acts transitively on roots of irreducible polynomials.
abstract-algebragalois-theory
Related Solutions
Your argument shows that $\text{Gal}(f)$ is contained in $\text{Gal}(g)\times \text{Gal}(h)$, since it sends roots of $g$ to roots of $g$ and roots of $h$ to roots of $h$. The simplest example that shows that you cannot expect equality in general is $f=g^2$. Clearly, in this case $\text{Gal}(f)=\text{Gal}(g)$. More generally, equality holds if and only if the splitting fields of $g$ and $h$ are disjoint over $\mathbb{Q}$. Note that this is stronger than $g$ and $h$ being coprime.
Exercise: find two distinct irreducible polynomials $g$ and $h$ whose splitting fields are not disjoint over $\mathbb{Q}$.
I seem to be giving lots of answers that depend on very special properties of the supplied example. Here’s an argument, tailored to your polynomial $f(x)=x^3-x-1$. Not the general method you were hoping for at all.
First set $\alpha$ to ba a root of your polynomial, which we all know is irreducible over $\Bbb Q$. It’s not too hard to calculate the discriminant of the ring $\Bbb Z[\alpha]$ as the norm down to $\Bbb Q$ of $f'(\alpha)=3\alpha^2-1$; this number is $23$, surprisingly small for a cubic extension. The fact that it’s square-free implies that $\Bbb Z[\alpha]$ is the ring of integers in the field $k=\Bbb Q(\alpha)$.
Our field $k$ clearly is not totally real, since $f$ has only one real root. So in the jargon of algebraic number theory, $r_1=r_2=1$, one real and one (pair of) complex embedding(s). We can apply the Minkowski Bound $$ M_k=\sqrt{|\Delta_k|}\left(\frac4\pi\right)^{r_2}\frac{n!}{n^n}\,, $$ which for $n=3$ gives a bound less than $2$, so that $\Bbb Z[\alpha]$ is automatically a principal ideal domain.
Let’s factor the number $23$ there: we certainly know that it’s not prime, since $23$ is ramified. Now, we already know a number of norm $23$, necessarily a prime divisor of the integer $23$, it’s $3\alpha^2-1$, and indeed $23/(3\alpha^2-1)=4 + 9\alpha - 6\alpha^2$. But better than that, $23/(3\alpha^2-1)^2=3\alpha^2-4$. This number has norm $23$ (because the norm of $23$ itself is $23^3$). So we’ve found the complete factorization of $23$.
Now let’s look more closely at $f(x)=(x-\alpha)g(x)$, for a polynomial $g$ that we can discover by Euclidean Division to be $g(x)=x^2+\alpha x+\alpha^2-1$. And the roots of $g$ are the other roots of $f$; the Quadratic Formula tells you what they are, and the discriminant of $g$ is $\alpha^2-4(\alpha^2-1)=4-3\alpha^2$. which we already know as $-23/(3\alpha^2-1)^2$. Going back to the Quadratic Formula, our other roots are $$ \rho,\rho'=\frac{\alpha\pm\sqrt\delta}2\>,\>\delta=\frac{-23}{(3\alpha^2-1)^2}\>,\>\sqrt\delta=\frac{\sqrt{-23}}{3\alpha^2-1}\>. $$ And that gives you your roots of this one very special cubic polynomial in terms of one root $\alpha$ and $\sqrt{-23}$.
Best Answer
Since $\DeclareMathOperator{\ksep}{{k}^{\text{sep}}} \ksep \supseteq L \supseteq k$, and $L/k$ is Galois, then by the fundamental theorem of Galois theory we have $$ \DeclareMathOperator{\Gal}{Gal} \Gal(L/k) \cong \frac{\Gal(\ksep/k)}{\Gal(\ksep/L)} \, . $$ The quotient map $\Gal(\ksep/k) \to \frac{\Gal(\ksep/k)}{\Gal(\ksep/L)}$ is surjective, which shows that every element of $\Gal(L/k)$ is the restriction of an element of $\Gal(\ksep/k)$. What this extension looks like is not at all clear, as the proof of this fact uses Zorn's Lemma; see Proposition 7.4 (p. 91) of Milne's Galois theory notes.