Are there (finite) field extensions which aren't Galois? Let $K=F(\alpha)$. It strikes me that there are two possibilities:


*

*$\alpha \in F$, in which case $K\cong F$

*$\alpha\not\in F$, 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?
 A: 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.
A: 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 $K$.
A: 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 $F$.
