Prove the ring $a+b\sqrt[3]{2}+c\sqrt[3]{4}$ has inverse and is a field

Question

How can I prove that $\frac{1}{a+b\sqrt[3]{2}+c\sqrt[3]{4}}$ is of the form $a+b\sqrt[3]{2}+c\sqrt[3]{4}$ (i.e. that $a+b\sqrt[3]{2}+c\sqrt[3]{4}$ is a field) for all rational $a,b,c$ and $a+b\sqrt[3]{2}+c\sqrt[3]{4} \neq 0$?

Answer 1

This is very standard. Let $\phi(a,b,c)=a+b\sqrt[3]{2}+c\sqrt[3]{4}$. Then you have

$$ \phi(a,b,c)+\phi(a',b',c')=\phi(a+a',b+b',c+c') $$

$$ \phi(a,b,c)\phi(a',b',c')=\phi(aa'+2bc'+2cb',ab'+ba'+2cc',ac'+bb'+ca') $$

$$ \phi(a,b,c)^{-1}=\phi(\frac{a^2-2bc}{D},\frac{2c^2-ab}{D},\frac{a^2-2bc}{D}), \ D=a^3+2b^3+4c^3-6abc $$

Note that $a^3+2b^3+4c^3-6abc \neq 0$ whenever one of $a,b$ or $c$ is nonzero.

Answer 2

Note that $\sqrt[3]4$ is $\sqrt[3]2^2$, so what you're looking at is the set of all polynomials in $\sqrt[3]2$ with rational coefficients, or to call it by its name, $\Bbb Q[\sqrt[3]2]$. What you want to show is a special case of a basic result in field theory that $\Bbb K[x]=\Bbb K(x)$.

The proof makes essential use of the minimal polynomial of $x$ over $\Bbb K$ and of Bézout's theorem for polynomials. In your case, you need to express $1/P(\sqrt[3]2)$ as a polynomial, with $P$ some polynomial, so let $M$ be the minimal polynomial of $\sqrt[3]2$ over $\Bbb Q$. Now you need to combine that with Bézout's theorem to show that $P$ has an inverse, that is there exists $Q$ such that $P(X)Q(X)=1$, and finally use that to express your fraction as a polynomial.