# Find an extension $K$ of $\mathbb Q$ having all roots of $x^3-2$ such that $[K:\mathbb Q]=6$.

I'm trying to solve this problem:

Show that $p(x)=x^3-2 \in \mathbb Q[x]$, is irreducible over $\mathbb Q$. Find an extension $K$ of $\mathbb Q$ having all roots of $p(x)$ such that $[K:\mathbb Q]=6$.

The easiest part is to show that $p(x)$ is irreducible, it suffices to use the Eisenstein criterion.

By Kronecker's theorem there exists a root $u$ of $p(x)$ in an extension $E$ of $\mathbb Q$, then we can write $p(x)$ as $p(x)=(x-u)q(x)$, where $q(x)\in \mathbb Q[x]$ has degree 2. I know also there is a root $v$ of $q(x)$ in an extension $G$ of $E$.

I noticed if I prove $[\mathbb Q(u,v):\mathbb Q(u)]=2$, where $\mathbb Q(u,v)=\mathbb Q(u)(v)$ I solve the problem, but I don't know how to prove this. This $[\mathbb Q(u,v):\mathbb Q(u)]$ in this context seems very weird to me, I really need help

Thank you

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You know the (complex) roots of $p(x)$. Use that knowledge to find a simple description of $\mathbb{Q}(u,v)$ and its relation to $\mathbb{Q}(u)$. Once you have that, decide if $[\mathbb{Q}(u,v):\mathbb{Q}]=6$, or if $=3$ and you have to construct $K$ as an (arbitrary) quadratic extension. – Hurkyl Oct 25 '12 at 0:58

You could simply calculate all the roots of $X^3-2$, pick $u=\sqrt[3]{2}$, and then prove that the second root is not in $\mathbb Q(u)$.
Alternately, since $v$ is the root of a quadratic polynomial over $\mathbb Q[u]$, you know that $[\mathbb Q(u,v):\mathbb Q(u)] \leq 2$. To prove that this is 2, you only need show that $v \notin \mathbb Q(u)$ (since it follows that the degree cannot be 1). This can easily be shown by finding explicitly $u, v$, but can probably be also proven without calculating them:
Assume by contradiction that $v \in \mathbb Q[u]$, then $v=au^2+bu+c$ for some rationals $a,b,c$. Use now that $1,u,u^2$ are independent over $\mathbb Q$, that $v^3=2$ and $u^3=2$ and of course $v \neq u$..After many computations and many tears, you should get a contradiction.
this is my problem, why $v$ is the root of a polynomial over $\mathbb Q[u]$? – user42912 Oct 25 '12 at 1:12
@user42912 Because $X^3-2: (X-u)$ is a division which can be done there. – N. S. Oct 25 '12 at 5:01