Multiplicative Inverse Element in $\mathbb{Q}[\sqrt[3]{2}]$

Question

So elements of this ring look like $$a+b\sqrt[3]{2}+c\sqrt[3]{4}$$ If I want to find the multiplicative inverse element for the above general element, then I'm trying to find $x,y,z\in\mathbb{Q}$ such that $$(a+b\sqrt[3]{2}+c\sqrt[3]{4})(x+y\sqrt[3]{2}+z\sqrt[3]{4})=1$$ I can see that expanding gives me the system $$ax+2cy+2bz=1$$ $$bx+ay+2cz=0$$ $$cx+by+az=0$$ I don't want to solve this using matrices because I know it will turn ugly. Is there a more elegant way to approach the inverse calculation to avoid the ugly calculation? The only thing I thought of was setting the bottom two equations equal to each other $$bx+ay+2cz=cx+by+az$$ Which seems to indicate that $$a=b, b=c, a=2c$$ but this would make me think $a=b=c=0$ and thus a multiplicative inverse does not exist.

Answer 1

First let me remark that there is a general abstract argument which shows that if $L/K$ is a field extension and $a \in L$ is algebraic over $K$, then the $K$-algebra $K[a]$ is a field. Namely, $K[a]$ is an integral domain which is also a finite-dimensional vector space over $K$. This implies for $0 \neq x \in K[a]$ that the linear map $K[a] \to K[a]$, $y \mapsto x \cdot y$ is surjective, since it is injective, which means that $x$ is invertible. To get a constructive proof, we just have to sit down and make the linear algebra argument here explicit.

The underlying vector space of $\mathbb{Q}[\sqrt[3]{2}]$ has basis $1,\sqrt[3]{2},\sqrt[3]{4}$. For some fixed non-zero element $x=a + b \sqrt[3]{2} + c \sqrt[3]{4}$, let's write down the linear map $y \mapsto x \cdot y$ in terms of this basis:

$~~~\,1 \mapsto a + b \sqrt[3]{2} + c \sqrt[3]{4}$

$\sqrt[3]{2} \mapsto 2c + a \sqrt[3]{2} + b \sqrt[3]{4} $

$\sqrt[3]{4} \mapsto 2b+2c\sqrt[3]{2} + a \sqrt[3]{4} $

The corresponding matrix is:

$$\begin{pmatrix} a & 2c & 2b \\ b & a & 2c \\ c & b & a \end{pmatrix}$$

From linear algebra we know how to invert matrices, for example via Cramer's rule. In this case, we get:

$$\frac{1}{a^3 - 6abc + 2b^3 + 4c^3} \cdot \begin{pmatrix} a^2 - 2bc & 2b^2 - 2ac & 4c^2 - 2ab \\ 2c^2 - ab & a^2 - 2bc & 2b^2 - 2ac \\ b^2 - ac & 2c^2 - ab & a^2 - 2bc \end{pmatrix}$$

The determinant $a^3 - 6abc + 2b^3 + 4c^3$ has been computed via the Rule of Sarrus and the cofactors have been computed by the usual formula for $2 \times 2$-determinants.

This matrix represents the linear map $y \mapsto x^{-1} \cdot y$ with respect to our basis. Thus, to get $x^{-1}$, we just have to evaluate at $1$, and we get:

$$x^{-1} = \frac{1}{a^3 - 6abc + 2b^3 + 4c^3} \cdot ((a^2 - 2bc) + (2c^2 - ab) \sqrt[3]{2} + (b^2 - ac) \sqrt[3]{4})$$

Of course, this method works quite generally. For example, for $x=a + b \sqrt[3]{p} + c \sqrt[3]{p}^2$ we have:

$$x^{-1} = \frac{1}{a^3 - 3pabc + pb^3 + p^2 c^3} \cdot ((a^2 - pbc) + (pc^2 - ab) \sqrt[3]{p} + (b^2 - ac) \sqrt[3]{p}^2)$$

Answer 2

Try these:
Rationalizing the denominator 3
Deradicalization of denominators

Answer 3

Not a great method if you'd like larger matrices, but I used Maple to get:

$$ x=\frac{a^2-2bc}{a^3-6abc+2b^3+4c^3}\qquad y=-\frac{ab-2c^2}{a^3-6abc+2b^3+4c^3}\qquad z=-\frac{ac-b^2}{a^3-6abc+2b^3+4c^3}. $$

I didn't check that I typed everything correctly, but it's not too bad.