TITLE: Describe the subfields of $\mathbb{C}$ of the form: $\mathbb{Q}(\alpha)$ where $\alpha$ is the real cube root of $2$.
QUESTION [6 upvotes]: Describe the subfields of $\mathbb{C}$ of the form: $\mathbb{Q}(\alpha)$ where $\alpha$ is the real cube root of $2$.
Let $\alpha$ be the real cube root of $2$, and consider $\mathbb{Q}(\alpha)$. As well as $\alpha$, the subfield $\mathbb{Q}(\alpha)$ must contain $\alpha^2$. We show that $$\alpha^2\neq j+k\alpha \text{ for } j,k \in \mathbb{Q}.$$ For a contradiction, suppose that $\alpha^2=j+k\alpha$. Then $$2=\alpha^3=\alpha(j+k\alpha)=j \alpha + k \alpha^2=j\alpha + k(j+k\alpha)=j\alpha+jk+k^2\alpha=jk+(j+k^2)\alpha.$$ Therefore $(j+k^2)\alpha=2-jk$. Since $\alpha$ is irrational, 
$j+k^2=0=2-jk.$ Note that $j+k^2=0 \iff -j=k^2$, so $$j+k^2=0=2-jk \iff k^3=2,$$ which is a contradiction because $k\in \mathbb{Q}$.
\null
In fact, $\mathbb{Q}(\alpha)$ is precisely the set of all elements of $\mathbb{R}$ of the form $$p+q\alpha + r\alpha^2, \text{ where } p,q,r\in \mathbb{Q}.$$ To show this, we prove that the set of such elements is a subfield. We will show that every element of $\mathbb{Q}(\alpha)$ can be expressed in this way. Set 
$$X=\{p+q\alpha+r\alpha^2 | p,q,r \in \mathbb{Q} \}.$$

$X$ is a subgroup of the additive group $(\mathbb{Q}(\alpha), +)$.
$1\in X$ is an identity element for multiplication.
Multiplication between to elements: 
$$ (p+q\alpha+r\alpha^2)(p'+q'\alpha+r'\alpha^2) = p'p+(p'q+pq')\alpha+(p'r+pr'+qq')\alpha^2 +(r'q+rq')\alpha^3+rr'\alpha^4 $$

I know that I can't have $\alpha^4$, so I need to rewrite it. How would I do that? 
Answer: $$\alpha^4=2\alpha$$
How would I approach find the inverse of $p+q\alpha+r\alpha^2$?

REPLY [3 votes]: The equation $X^3 - 2 = 0$ has three roots, namely $\alpha$, $j\alpha$ and $j^2\alpha$, where $j$ is a cubic root of unity. Therefore $p+q\alpha+r\alpha^2$ has two conjugates: $p+qj\alpha+rj^2\alpha^2$ and $p+qj^2\alpha+rj\alpha^2$. The product of the three will be rational. The inverse can be written:
$$\frac{1}{p+q\alpha+r\alpha^2}
= \frac{(p+qj\alpha+rj^2\alpha^2)(p+qj^2\alpha+rj\alpha^2)}
{(p+q\alpha+r\alpha^2)(p+qj\alpha+rj^2\alpha^2)(p+qj^2\alpha+rj\alpha^2)}
= \frac{(p^2-2qr)+(2r^2-pq)\alpha+(q^2-pr)\alpha^2}
{p^3+2q^3+4r^3-6pqr}$$
If you want to know more, you should read a course on Galois theory, for example Milne's lecture notes which are available online: http://www.jmilne.org/math/CourseNotes/FTe6.pdf