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 $p+q\alpha+r\alpha^2$?
Best Answer
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!)