If $\alpha=\sqrt[3]{2}$ and $p,q,r\in\mathbb{Q}$ then show $p+q\alpha+r\alpha^2$ is a subfield of $\mathbb{C}$.
For context, this is number $5$ in Chapter $1$ of Ian Stewart's Galois Theory. At this point in the text, we have only learned how to solve cubics and quartics, while introducing subring and subfield language.
First to show $$R=\{p+q\alpha+r\alpha^2: p,q,r\in\mathbb{Q} \wedge \alpha=\sqrt[3]{2}\}$$ is a subfield we show $R$ is a subring of $\mathbb{C}$ and then finish by showing $\forall x\in R, \exists x^{-1}\in R$.
Note that clearly $R\subset\mathbb{C}$ since $p+q\alpha+r\alpha^2$ is a real number for all rational $p,q,r.$ Take $p=1,q=0,r=0$ to see $1\in R$.
If $p_1+q_1\alpha+r_1\alpha^2\in R$ and $p_2+q_2\alpha+r_2\alpha^2\in R$, then $$\left(p_1+q_1\alpha+r_1\alpha^2\right)+\left(p_2+q_2\alpha+r_2\alpha^2\right)=\left(p_1+p_2\right)+\left(q_1+q_2\right)\alpha+\left(r_1+r_2\right)\alpha^2\in R$$ $$-(p_1+q_1\alpha+r_1\alpha^2)=-p_2-q_2\alpha-r_2\alpha^2\in R$$ $$\left(p_1+q_1\alpha+r_1\alpha^2\right)\left(p_2+q_2\alpha+r_2\alpha^2\right)$$ $$=\left(p_1p_2+2q_1r_2+2q_2r_1\right)+\left(p_1q_2+p_2q_1+2r_1r_2\right)\alpha+\left(p_1r_2+q_1q_2+p_2r_1\right)\alpha^2\in R$$
The preceding argument follows from the facts that the rationals are closed under addition and multiplication. The above also shows $R$ is a subring of $\mathbb{C}$. To complete the proof that $R$ is a subfield, we find an expression for the inverse $$(p_1+q_1\alpha+r_1\alpha^2)^{-1}$$
Here is where I run into issues. My first thought was to set the product $\left(p_1+q_1\alpha+r_1\alpha^2\right)\left(p_2+q_2\alpha+r_2\alpha^2\right)$ equal to $1$: $$\left(p_1+q_1\alpha+r_1\alpha^2\right)\left(p_2+q_2\alpha+r_2\alpha^2\right)=1\implies$$ $$p_1p_2+2q_1r_2+2q_2r_1=1$$ $$p_1q_2+p_2q_1+2r_1r_2=0$$ $$p_1r_2+q_1q_2+p_2r_1=0$$ which is equivalent to $$\begin{pmatrix}
p_2&2r_2&2q_2\\ q_2&p_2&2r_2\\r_2&q_2&p_2\end{pmatrix}\begin{pmatrix}p_1\\q_1\\r_1\end{pmatrix}=\begin{pmatrix}1\\0\\0\end{pmatrix}$$
If I could find an explicit inverse for the above $3\times3$ matrix, the problem would be solved yielding exact expressions for $p_1,q_1, r_1$ in terms of $p_2,q_2,r_2$.
However, I am not seeing a way to guarantee the determinant is nonzero for as long as $p_2,q_2,r_2\neq0$.
I noticed the matrix is Toeplitz, but I don't know if that tells us anything about invertibility.
Any help with finding the inverse element here without resorting to high power machinery in these answers Describe the subfields of $\mathbb{C}$ of the form: $\mathbb{Q}(\alpha)$ where $\alpha$ is the real cube root of $2$. and 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 $2$? is much appreciated.
Best Answer
Presumably, you know how to calculate the multiplicative inverse of complex numbers. This uses a similar idea to that, with rationalising the denominator. However, in this case there are three terms to the denominator, so it's a bit more tricky to find the exact term that works.
Note that $$ (x+y+z)(x^2+y^2+z^2-xy-xz-yz)=x^3+y^3+z^3-3xyz $$ Using this, we get $$ (p+q\alpha+r\alpha^2)(p^2+q^2\alpha^2+2r^2\alpha-pq\alpha-pr\alpha^2-2qr)\\ =p^3+2q^3+4r^3-6pqr $$ Now consider $\frac1{p+q\alpha+r\alpha^2}$ for some non-zero $p+q\alpha+r\alpha^2$, and expand this fraction according to the above. You now have a fraction with a rational number in the denominator, so it's in your ring.
Final piece: Showing that what we expand by is non-zero. We have that $$ (x-y)^2+(x-z)^2+(y-z)^2\geq0\\ x^2+y^2+z^2- xy-xz-yz\geq0 $$ holds for any real $x,y,z$ with equality iff $x=y=z$. However, in our case that would mean $$ p=q\alpha=r\alpha^2 $$ which by irrationality of $\alpha$ would imply $p=q=r=0$, which is not the case.