How to show that $\mathbb{Q}(\alpha) = \left\{ p+q\alpha+r\alpha^2 \mid p, q, r\in \mathbb{Q} \right\}$, where $\alpha$ is the real cube root of $2$?

Question

Context: Galois Theory by Ian Stewart, 4th edition, Sec. 4.1 in Chap. 4

Let $\alpha$ be the real cube root of $2$. Let $\mathbb{Q}(\alpha)$ denote the smallest subfield of the field $\mathbb{R}$ of real numbers that contains $\mathbb{Q} \cup \{ \alpha \}$. Of course, every subfield of $\mathbb{C}$ necessarily contains $\mathbb{Q}$.

Then how to show that $$ \mathbb{Q}(\alpha) = \left\{ \ p+q\alpha+r\alpha^2 \, \vert \, p, q, r \in \mathbb{Q} \ \right\}?$$

My Attempt:

As $\mathbb{Q}(\alpha)$ must contain $\mathbb{Q}$ and $\alpha$, so it must contain all real numbers of the form $p + q\alpha$ for any $p, q \in \mathbb{Q}$.

But, as has been shown by Ian Stewart in Example 4.9 (3), the number $\alpha^2$ cannot be expressed in the form $j+k\alpha$ for any $j, k \in \mathbb{Q}$. So $\mathbb{Q}(\alpha)$ must contain all numbers of the form $p + q\alpha + r\alpha^2$ for any $p, q, r \in \mathbb{Q}$. Let $L$ be the set of all such real numbers.

Then $L$ is an abelian group under addition and also contains $1 = 1 + 0\alpha + 0\alpha^2$.

Moreover, if $p_1 + q_1 \alpha + r_1 \alpha^2$ and $p_2 + q_2 \alpha + r_2 \alpha^2$ are any two elements of $L$, then we find that $$ \begin{align} & \qquad \left( p_1 + q_1 \alpha + r_1 \alpha^2 \right) \left( p_2 + q_2 \alpha + r_2 \alpha^2 \right) \\ &= p_1 p_2 + \left( p_1q_2 + q_1 p_2 \right)\alpha + \left( p_1 r_2 + r_1 p_2 \right) \alpha^2 + \left( q_1 r_2 + r_1 q_2 \right) \alpha^3 + r_1 r_2 \alpha^4 \\ &= \left( p_1p_2 + 2q_1r_2 + 2q_2 r_1 \right) + \left( p_1q_2 + q_1 p_2 + 2r_1 r_2 \right)\alpha + \left( p_1 r_2 + r_1 p_2 \right) \alpha^2. \end{align} $$ Thus $L$ is also closed under multiplication.

In fact, $L$ is a commutative ring with unity.

Now let $x = p+q\alpha + r\alpha^2$ be any non-zero element of $L$. We need to show that $\frac{1}{x}$ also is in $L$.

As $x \neq 0$, so $(p, q, r) \neq (0, 0, 0)$.

What next? How to proceed from here?

Answer 1

Let $L$ be the $\Bbb Q$-span of $1$, $\alpha$ and $\alpha^2$. Then $L$ is a ring, and also a three-dimensional $\Bbb Q$-vector space. If $\beta=p+q\alpha+r\alpha^2$ is a nonzero element of $L$, then $f:u\mapsto\beta u$ is a $\Bbb Q$-linear map from $L$ to $L$. As the real numbers form a field, $f$ is injective. By rank-nullity, $f$ is surjective, so there is $u\in L$ with $f(u)=1$. Then $u=1/\beta$.