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Let $K=F(\alpha)$. It strikes me that there are two possibilities:

  1. $\alpha \in F$, in which case $K\cong F$
  2. $\alpha\not\in F$, in which case $K$ is the splitting field of $(x-\alpha)$ (and hence Galois).

(I actually think that even #1 might be described as a Galois extension - $|G(F/F)|=[F:F]$, for example.)

In any case, what does it mean to say that an extension is not Galois?

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    It's true that $K$ is the splitting field of $x-\alpha$... *over $F(\alpha)=K$*. So that tells you that $K$ is Galois over $K$. Alas, you are interested in knowing whether $K$ is Galois *over $F$*, not over itself. Every field is Galois over itself.2011-09-22

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No. Let $\alpha$ be a cube root of $2$. Then $\mathbb{Q}(\alpha)$ is not a Galois extension. Indeed, note that $\mathbb{Q}(\alpha)$ can be embedded in $\mathbb{R}$, so how could it possibly be the splitting field of $X^3 - 2$?

In the general case, we define a Galois extension to be an algebraic field extension $F / K$ such that the fixed field of $\mathrm{Aut}_K(F)$, the group of $K$-automorphisms of $F$, is not $K$. Notice that $\mathbb{Q}(\alpha)$ has no non-trivial $\mathbb{Q}$-automorphisms: thus it cannot possibly be a Galois extension. For finite extensions, by Dedekind's theorem on the linear independence of $K$-automorphisms, it suffices to check that $[F : K] = | \mathrm{Aut}_K(F) |$.

If a finite extension is not Galois, then that can only be because there are not enough automorphisms: Dedekind's theorem mentioned above in fact guarantees that $[F : K] \ge | \mathrm{Aut}_K(F)|$. In the separable case, if $F / K$ does not have enough automorphisms, that essentially means that $F$ does not contain enough roots.

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Consider $K=\mathbb Q[\sqrt[3]{2}]$ over $F=\mathbb Q$.

$K$ is certainly a finite, algebraic extension. However the minimal polynomial, $x^3-2$, has two more roots which are not in $K$.

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I think the key point you're missing is that the label "Galois" isn't something that applies to the extension field -- it is something that applies to the extension.

It doesn't make sense to say that $K$ is Galois -- instead, what you observed in your question is that the extension $K/K$ is a Galois one, because $K$ is the splitting field of a polynomial with coefficients in $K$.

But $K/F$ need not be Galois, because $K$ might not be a splitting field of a polynomial with coefficients in $F$.