Splitting field for primitive element of Galois extension and description of Galois group

I was hoping to get some clarification on the following observation. I'm not sure if it is true or meaningful, but if it is, my way of thinking about it is pretty muddled and I would appreciate a better perspective.

Consider a finite Galois field extension $E/F$ of degree $n$. By the primitive element theorem, there exists $\alpha$ in $E$ such that $E = F(\alpha)$. We can form the minimal polynomial for $\alpha$ in $F[x]$. This polynomial has degree $n$. We will call this polynomial $f$.

We have $Gal(E/F) = Gal(F(\alpha)/F) = G$. It seems to me that $f$ should split in $E = F(\alpha)$ at least if we assume $F$ is perfect (so that $f$ being irreducible implies it is separable) and that the Galois group is induced from replacing $\alpha$ with $\alpha'$ another root of $f$. This gives a bijection between the roots of $f$ and the Galois group $G$.

Why? Let $K$ be the splitting field of $f$. The automorphism of $F(\alpha)$ given by $\alpha \mapsto g \alpha$ for $g \in G$ can probably be extended to an automorphism of $K$. Also, if $g \alpha = g$ then $g$ is the identity. Therefore the action is free so that the set $G\alpha$ has order $n$, so all the roots of $f$ are in $G\alpha \subset F(\alpha)$.

For example, $\mathbb{Q}(\sqrt{2},\sqrt{3}) = \mathbb{Q} (\sqrt{2} + \sqrt{3})$, and the minimal polynomial for $\sqrt{2} + \sqrt{3}$ over $\mathbb{Q}$ is $x^4 - 10 x^2 +1$, which splits into $$(x - \sqrt{2} - \sqrt{3}) (x - \sqrt{2} + \sqrt{3}) (x + \sqrt{2} + \sqrt{3}) (x + \sqrt{2} - \sqrt{3})$$ over $\mathbb{Q}(\sqrt{2} + \sqrt{3})$. The roots are $\pm \sqrt{2} \pm \sqrt{3}$, and replacing one root by another completely determines the action of an element of the Galois group.

My questions is whether this is always true: Given a finite Galois extension $E/F$, let $f$ be the minimal polynomial of the primitive element $\alpha \in E$ for this extension (i.e. $E = F(\alpha)$, and $f$ is the minimal polynomial of $\alpha$). Then $f$ splits over $F(\alpha)$ and any element of the Galois group is determined by $\alpha \mapsto \alpha '$ where $\alpha'$ is another root of the minimal polynomial $f$.

This might also imply the following: an extension is Galois if and only if the minimal polynomial of the primitive element splits by adding one root.

• "My questions is whether this is always true:" My answer is, yes, what you write in that paragraph is always true. – Gerry Myerson May 4 '14 at 12:45