I believe the extension $\mathbb Q(16^{\frac{1}{3}}):\mathbb Q$ is not Galois, and I'm trying to prove why. Is it enough to say that $16^{\frac{1}{3}}$ is a root of the polynomial $x^3-16$ over $\mathbb Q$ but it also contains complex roots, thus $\mathbb Q(16^{\frac{1}{3}})$ does not contain all the algebraic conjugates of $16^{\frac{1}{3}}$ so is not a Galois extension of $\mathbb Q$?
Is this correct? If so, is there a neater way to say this?
Actually $\Bbb Q(16^{1/3})=\Bbb Q(2^{1/3})$. A Galois extension of $\Bbb Q$ must contain all conjugates of any of its elements over $\Bbb Q$. One of the conjugates of $2^{1/3}$ is $2^{1/3}\exp(2\pi i/3)$ which is not real, and so cannot lie in $\Bbb Q(2^{1/3})$.