Proving that a field is not a splitting field of any polynomial I am trying to prove that $Q(2^{1/3})$ is not a splitting field of any polynomial $p(x) \in Q[x]$. 
I know I need to show that there does not exist any polynomial in $Q[x]$ that has all of its roots in $Q(2^{1/3})$. I know that a basis for $Q(2^{1/3})$ over $Q$ is $\{1, 2^{1/3}, 2^{2/3}\}$.  But I can't figure out how to start. Can someone give me a hint?
 A: Splitting fields are normal, so if it is a splitting field, any polynomial with a root has all the roots in the field. But $x^3-2$ is such a polynomial, and clearly does not contain all its roots since $\Bbb Q(\sqrt[3]{2})$ is a purely real field.
Edit (since the op doesn't know about normality): The other option is ad hoc, but works, if it's a splitting field, since it has degree $3$ it has no proper, non-trivial subfields, hence it must be the splitting field for a third-degree polynomial with all real roots. But then it is a totally real field, i.e. it has no embeddings into $\Bbb C$ which do not end in $\Bbb R$. However, since $K\cong \Bbb Q[x]/(x^3-2)$ this is a contradiction since the map which sends $\overline{x}$ in the quotient to $\zeta_3\sqrt[3]{2}$ is a map which does not land in a subset of $\Bbb R$.
A: Suppose we have a finite extension $\mathbb{Q}(\alpha):\mathbb{Q}$ for some $\alpha$. The condition 'finite' means that $\alpha,\alpha^2,\alpha^3,...$ is dependent for some $n\in\mathbb{N}$. Now pick the smallest such $n$, then you can write $\alpha^n = ...$ with '...' some combination of $1,\alpha,\alpha^2,...\alpha^{n-1}$ over $\mathbb{Q}$. Now I claim that this gives you the minimal polynomial of $\alpha$! 
For your example, with $\alpha=2^{1/3}$, you already know that $\alpha^3$ is dependent on $1,\alpha,\alpha^2$. Yes: $\alpha^3=2*1$. Thus $\alpha$ is a root of $X^3-2$ and this polynomial is the minimal polynomial of $\alpha$.
Now to your problem. If $\mathbb{q}(\alpha)$ is the splitting field of some polynomial, then some irreducible polynomial in $\mathbb{Q}[X]$ splits in $\mathbb{Q}(\alpha)$. Thus it has its roots in $\mathbb{Q}(\alpha)$. Thus we can  check the elements of $\mathbb{Q}(\alpha)$ and see if it their irreducible polynomials in $\mathbb{Q}$ generate $\mathbb{Q}(\alpha)$. We only have to check $\alpha$ (Why?): It has minimal polynomial $X^3-2$ in $\mathbb{Q}$. Now you have to check that $\mathbb{Q}(\alpha)$ does not contain all roots of $X^3-2$. It does contain $\alpha$ so divide $X^3-2$ by $X-\alpha$ and then look at the resulting polynomial $X^2+\alpha X+\alpha^2$. It has roots $b\alpha$ with $b$ some element satisfying $b^2+b+1=0$. Do you know about roots of unity? Otherwise just show that no such element is in $\mathbb{Q}(\alpha)$.
