Automorphisms of a field extension permute roots of irreducible factors.

Question

Would someone mind confirming (or refuting) the following...

Proposition. Let $K$ be a field and let $f \in K[X]$. Suppose $g \in K[X]$ is an irreducible factor of $f$. Let $L$ be a splitting field for $f$ over $K$. Then Aut$(L/K)$ permutes the roots of $g$.

Proof. Let $x \in L$ be a root of $g$ and let $\sigma \in$ Aut$(L/K)$. As $g \in K[X]$ we have $\sigma g = g$. Clearly $\sigma(x)$ is a root of $\sigma g$ and so the result follows. //

Many thanks :)

Answer 1

This is an expansion of the comment of LUXun.

There is a result that states that splitting fields are unique up to isomorphism.

Put it more precisely, it says the following

Suppose $F, \bar F$ are fields, and $a \mapsto \bar a$ is an isomorphism $\phi : F \to \bar F$. Suppose $g = a_{0} + a_{1} x + \dots + x^{n} \in F[x]$ is a non-constant polynomial, and $\bar g = \bar a_{0} + \bar a_{1} x + \dots + x^{n} \in \bar F[x]$. Let $E_{1}$ be a splitting field of $g$ over $F$, and $E_{2}$ a splitting field of $\bar g$ over $\bar F$. Then there is an isomorphism $\sigma : E_{1} \to E_{2}$ which restricted to $F$ yields $\phi$.

In other words, $\sigma$ is an extension of $\phi$.

In this case you apply this with $E_{1} = E_{2} = L$, $F = K(\alpha)$, $\bar F = K(\beta)$. The fact that $\alpha$ and $\beta$ are two roots of the irreducible polynomial $g \in K[x]$ yields the isomorphism $\phi$ $$ K(\alpha) \cong K[x]/(g) \cong K(\beta) $$ that carries $\alpha$ to $\beta$. By the theorem, $\phi$ extends to an element $\sigma \in Aut(L/K)$ that extends $\phi$, and thus carries $\alpha$ to $\beta$.