Continuation on: Let $K = \mathbb{Q}(i2^{1/3}, 3^{1/4})$. Is this a Galois extension? Let $K = \mathbb{Q}(i2^{1/3}, 3^{1/4})$. Is this a Galois extension?
People seem pretty convinced this is Galois.  I present an issue I am having though.
So the monic polynomial p(x) in $\mathbb{Q}[x]$ of least degree whose root field over $\mathbb{Q}$ is $K$ 
($x^2$+1)($x^6$+4)($x^4$-3) = $x^{12}+x^{10}+4x^6+4x^4-3x^8-3x^6-12x^2-12$
Is the fact that this is degree 12 contradict that $K$ is Galois since the extension K has degree 24?
 A: No, it does not contradict the fact.  Here is an example of a Galois extension $K$ such that $[K:F] > \deg(p(x))$:
Let $K = \mathbb{Q}[\sqrt[4]{2}, i]$.  You can check that this is a splitting field of the irreducible polynomial $f(x) = (x^4 - 2)$.  Note that $\deg(f) = 4$.  However, $[K:\mathbb{Q}] = 8$.  The important thing is that $K$ is a Galois extension since it is the splitting field of a polynomial over the rationals.
To conclude, the following statements are equivalent to saying that an extension $K$ over a field $F$ is Galois (think if and only if):


*

*$K$ is a separable extension, and it is a splitting field of some polynomial $F[x]$.  Since $\mathbb{Q}$ and $\mathbb{R}$ are perfect fields, all finite extensions  of them are separable.  Therefore, we need only check that extensions over $\mathbb{Q}$ or $\mathbb{R}$ are splitting fields.

*$[K:F] = |Aut(K/F)|$



Returning to the example above, we can check the second condition as well.  If we take two "base" $\mathbb{Q}$-automorphisms defined as follows: $\phi(\sqrt[4]{2}) = i\sqrt[4]{2}$ and $\psi(i) = -i$, then we can construct an automorphism group of order $8$ isomorphic to $D_4$ by composing $\psi$ with powers of $\phi$.  
A: It might be worth mentioning that if $f(x)$ is a typical degree $n$ irred. polynomial over $\mathbb Q$, then the splitting field of $f$ will have Galois group
$S_n$, and in particular be of degree $n!$.  So it is not surprising that the degree
of the field extension is greater than the degree of the polynomial of which it is a splitting field; indeed, such behaviour is typical.
[Added in response to comments: Here, by "typical" I mean "random".  See the links in the comments below for more details about this.]

One theoretical way to think about it is as follows: if $L/K$ is Galois with Galois group $G$, then the minimal degree of a poly. for which it is a splitting field is equal to the minimal degree of a faithful permutation rep'n of $G$.
(If $L$ is the splitting field of a polynomial $f \in K[x]$, then $G$ acts faithfully on the roots of $f$.  Conversely, every permutation rep'n of $G$
arises in this way.)
Any group $G$ has a faithful perm. rep'n of degree $|G|$, namely its regular
rep'n on itself, but this it typically not minimal.

In your example, the Galois group is an index $2$ subgroup of the direct product $D_6 \times D_8$.   Thus it has a faithful permutation rep'n of degree $7$:  act 
on $3$ elements via the projection to $D_6$ and the isomorphism $D_6 \cong S_3$,
and act on $4$ elements via the projection to $D_8$ and the embedding
$D_8 \subset S_4$.  
Concretely, this corresponds to the degree $7$ polynomial $(x^4 - 3)(x^3 - 2)$.  
(So incidentally, your claim about $12$ being the minimal degree is false.)
