How to describe when a simple extension $F(\alpha)/F$ is Galois in terms of the minimal polynomial of $\alpha$?

I have a question concerning definition in terms of minimal polynomial i.e. if we let $E = F(\alpha)$ be a field extension of $F$ of degree two then how do I describe, in terms of the minimal polynomial for $\alpha$ over $F$ when this field extension is Galois?

Also does there exist a field extension of $\mathbb{Q}$ of degree $3$ that is Galois?

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For your last question, have a look at this: sbseminar.wordpress.com/2008/04/21/… – M Turgeon May 21 '12 at 22:05
Are you still here, Karl? – Gerry Myerson May 27 '12 at 9:34

In general, $F(\alpha)/F$ is Galois if and only if the minimal polynomial of $\alpha$ over $F$ is separable and splits completely in $F(\alpha)$. In the special case of quadratic extensions, it is very easy to see that the "splits completely" part is always satisfied, so it's just a question of separability. If for example $F$ has characteristic 0, then separability is automatic, so in that case, any quadratic extension is Galois.
There do exist Galois extensions of $\mathbb{Q}$ of degree 3. If $\alpha$ has cubic minimal polynomial over $\mathbb{Q}$, then $\mathbb{Q}(\alpha)$ is Galois if and only if the discriminant of this polynomial is a rational square. This is proved in any standard text on Galois theory.
Let $\alpha=e^{2\pi i/7}$, and let $\beta=\alpha+\alpha^{-1}$. You should be able to find the conjugates of $\beta$, prove that they are all in ${\bf Q}(\beta)$, and that that field is Galois of degree 3 over the rationals.