Describe the subfields of $\mathbb{C}$ of the form: $\mathbb{Q}(\alpha)$ where $\alpha$ is the real cube root of $2$. 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 $k\in \mathbb{Q}$.
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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 $p+q\alpha+r\alpha^2$?
 A: Since $\alpha^3=2$, $\alpha^4=\alpha^3\cdot\alpha=2\alpha$.
Inverses require a lot more cleverness.  Here's one possible approach.  Note that $X$ is a finite-dimensional vector space over the field $\mathbb{Q}$, and that for any $x\in X$, the map $\mu_x(y)=xy$ is a $\mathbb{Q}$-linear map $X\to X$.  If $x\not=0$, furthermore, $\mu_x$ is injective.  But any injective linear map from a finite-dimensional vector space to itself is also surjective.  It follows that $1$ is in the image of $\mu_x$, which says exactly that $x$ has an inverse.
(In principle, using Cramer's rule to compute the inverse of the linear map $\mu_x$, you can use this argument to explicitly write down a formula for $x^{-1}$, but it will be quite complicated!)
A: 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
A: Let me first reformulate precisely your question: if a is a real cubic root of 2, you know that every element x of the field Q(a) can be written uniquely as x = p + qa + ra^2, with p, q, r in Q, and you want to compute an analogous "polynomial" expression for the inverse x^-1 ?
The answer is very simple if we go back to the classical "polynomial" description of the field Q(a), which is the beginning of the theory of algebraic extensions of a field. I recall it here. Denote as usual by Q[a] the ring generated by Q and a. By the universal property of the ring of polynomials Q[X], we have a ring homomorphism h from Q[X] to Q[a] which sends X to a. By construction, h is surjective, and its kernel is obviously the principal ideal generated by X^3 - 2, because this polynomial is irreducible over Q (the shortest proof uses Eisenstein's criterion; a longer proof is the one you gave for the "polynomial" description of the elements of Q(a)). Thus we have that Q[a] is isomorphic to the quotient Q[X]/(X^3 - 2), which is a field since the ideal (X^3 - 2) is maximal because  X^3 - 2 is irreducible. So Q[a] = Q(a). Note that this gives again the aforementioned "polynomial" description of the elements of Q(a). But it gives more if we detail the proof of the existence of inverses (maximality of (X^3 - 2), etc.) : the ring Q[X] being a UFD, the Bézout identity is available, i.e.  polynomials f(X) and g(X) are coprime iff there exists u(X) and v(x) such that : u(X)f(X) + v(X)g(X) = 1. Here take f(X) = p + qX + rX^2, which is coprime to g(X) = X^3 - 2. Applying h , we get u(a).f(a) = 1 , i.e. u(a)is the inverse we are looking for. 
Concretely, given f(X), we must find u(X): this is the euclidian algorithm applied to the computation of the Bézout identity, see e.g. www.math.uconn.edu/~kconrad/blurbs/.../divgcd.pdf               ¤
