My understanding is that for $\mathbb{Q}(3^{1/3},3^{1/4})$ to be a Galois extension of $\mathbb{Q}$ there must exist a polynomial over $\mathbb{Q}$ with roots $3^{1/3},3^{1/4}$ that splits in $\mathbb{Q}(3^{1/3},3^{1/4})$. Is it true then that this field is not a Galois extension in this case as the field does not contain $i$ (since some roots of $x^3-3$ and $x^4-3$ are complex) or am I missing the point?
Thanks.
You're right on the point: some roots of $x^3-3$ are complex.
More precisely:
A Galois extension contains all the algebraic conjugates of its elements.
$3^{1/3}$ has algebraic conjugates that are complex, such as $3^{1/3}\omega$, where $\omega$ is a primitive cubic root of unity.
$\mathbb{Q}(3^{1/3},3^{1/4}) \subseteq \mathbb{R}$.
Therefore, $\mathbb{Q}(3^{1/3},3^{1/4})$ does not contain all the algebraic conjugates of $3^{1/3}$ and so is not a Galois extension of $\mathbb{Q}$.