Is $\mathbb{Q}(\sqrt{2}, 2^{1/3})$ a Galois extension of $\mathbb{Q}$?

Question

I am wondering about the following question: Is $\mathbb{Q}(\sqrt{2}, 2^{1/3})$ a Galois extension of $\mathbb{Q}$ ?

For a finite extension $E/F$, the statement "$E/F$ is Galois" is equivalent to the statement "$E$ is a splitting field of a separable polynomial with coefficients in $F$".

Per the above statement, I believe that $\mathbb{Q}(\sqrt{2}, 2^{1/3})$ is not a Galois extension of $\mathbb{Q}$. This is because, it doesn't seem to me that $\mathbb{Q}(\sqrt{2}, 2^{1/3})$ is even a splitting field of any polynomial with coefficients in $\mathbb{Q}$.

Am I correct? If so, how can I make that last sentence above more rigorous, and actually show it's not a splitting field of any polynomial with coefficients in $\mathbb{Q}$ ?

Thanks!

Answer 1

Hint: It isn't a normal extension, as it doesn't have all the roots of $x^3-2$.

Answer 2

Yes, you are correct

Galois means algebraic, normal and separable extension. Equivalent conditions to normality of the algebraic extension $E|F$ are:

$\bullet$ For each homomorphism $\sigma: E \to L$, where $L$ is the algebraic closure of $F$, such that $\sigma \vert_{F} = id_F$, $\sigma(E) = E$;

$\bullet$ $E$ is the splitting field of some collection of polynomials in $F[X]$;

$\bullet$ For each irreducible polynomial $f \in F[X]$, if $f$ has a root in $E$, then it splits in $E$.

Take $X^3 - 2 \in \mathbb{Q}[X]$, which is irreducible. It has one root in $\mathbb{Q}(2^{1/3}, \sqrt{2})$, but doesn't split in this field. To work in your definition, it might be better if you assume that $\alpha \in \mathbb{Q}(2^{1/3}, \sqrt{2})$ is a primitive element, and work from there.