How to find the multiplicative inverse of $a+b\sqrt{2}$ in $\mathbb{Q}({\sqrt{2}})$

Question

I am struggling with this a lot. I set the product of 2 elements in the field equal to 1 but cannot find a way to write the elements of one element in terms of the other.

Instead of just knowing the answer I’d like to understand how to find it so I can apply this again.

Answer 1

You can use a method that works in every field of the form $F(r)$, with $r$ of degree $n$ over $F$.

If you are given $t=a_0+a_1r+\dots+a_{n-1}r^n\in F(r)$, $t\ne0$, then you know that the map $u\mapsto tu$ is invertible. Then you can find its matrix with respect to the basis $\{1,r,\dots,r^{n-1}\}$ by taking $t,tr,\dots,tr^{n-1}$ (using the minimal polynomial to reduce higher powers). Find the matrix inverse and you're done, because the inverse is the image of the element $1$, whose coordinate vector is $\left[\begin{smallmatrix}1\\0\\0\end{smallmatrix}\right]$, so you just take the first column of the matrix.

Example for your case: the minimal polynomial is $x^2-2$; then the elements you need are $(a+b\sqrt{2})1$ and $(a+b\sqrt{2})\sqrt{2}=2b+a\sqrt{2}$. Thus the matrix is \begin{bmatrix} a & 2b \\ b & a \end{bmatrix} whose inverse is $$ \frac{1}{a^2-2b^2}\begin{bmatrix} a & -2b \\ -b & a \end{bmatrix} $$ and the first column will give the required inverse, namely $$ \frac{a}{a^2-2b^2}-\frac{b}{a^2-2b^2}\sqrt{2} $$ The same would do for $r=\sqrt[3]{2}$ and $t=a+br+cr^2$. The elements to compute are $$ a+br+cr^2,\quad (a+br+cr^2)r=2c+ar+br^2,\quad (a+br+cr^2)r^2=2b+2cr+ar^2 $$ so the matrix is $$ \begin{bmatrix} a & 2c & 2b \\ b & a & 2c \\ c & b & a \end{bmatrix} $$ The inverse matrix is $$ \frac{1}{a^3+2b^3+4c^3-6abc} \begin{bmatrix} a^2-2bc & 2b^2-2ac & 4c^2-2ab \\ 2c^2-ab & a^2-2bc & 2b^2-2ac \\ b^2-ac & 2c^2-ab & a^2-2bc \end{bmatrix} $$ so the inverse is $$ \frac{a^2-2bc}{a^3+2b^3+4c^3-6abc}+ \frac{2c^2-ab}{a^3+2b^3+4c^3-6abc}\sqrt[3]{2}+ \frac{b^2-ac}{a^3+2b^3+4c^3-6abc}\sqrt[3]{4} $$ Of course this is easier if you have explicit $a,b,c$.

By the way, this proves that if $t=a+br+cr^2\ne0$, then $a^3+2b^3+4c^3-6abc\ne0$ (this is the norm function).

In the $\mathbb{Q}(\sqrt{2})$ the norm function on $a+b\sqrt{2}$ is $a^2-b^2$.

Answer 2

Since $(a+b\sqrt{2})(a-b\sqrt{2})=a^2-2b^2$ is non-zero because $\sqrt{2}$ is irrational, the inverse is$$\frac{a-b\sqrt{2}}{a^2-2b^2}=\frac{a}{a^2-2b^2}+\frac{-b}{a^2-2b^2}\sqrt{2}.$$

Answer 3

Set up a system of equations. Say $(a + b \sqrt{2})^{-1} = u + v \sqrt{2}$, multiply and equate:

$\begin{align*} (a + b \sqrt{2}) \cdot (u + v \sqrt{2}) &= (a u + 2 b v) + (a v + b u) \sqrt{2} \end{align*}$

This gives a system of equations:

$\begin{align*} a u + 2 b v &= 1 \\ b u + a v &= 0 \end{align*}$

This is a linear system of equations for $u, v$.

Or compute directly, rationalizing:

$\begin{align*} \frac{1}{a + b \sqrt{2}} &= \frac{a - b \sqrt{2}}{(a - b \sqrt{2}) (a + b \sqrt{2})} \\ &= \frac{a - b \sqrt{2}}{a^2 - 2 b^2} \end{align*}$

Unless $a = b = 0$, the denominator is never zero as long as $a, b \in \mathbb{Q}$.

Answer 4

Here is a different take:

Let $\beta=a+b\sqrt{2}$. Then $\beta^2-2a\beta+a^2-2b^2=0$. Therefore $ \beta(\beta-2a)=-(a^2-2b^2) $ and so $$ \beta\frac{2a-\beta}{a^2-2b^2}=1 $$ Thus, the fraction above is the inverse of $\beta$. (Note that $a^2-2b^2=0$ iff $\beta=0$.)

This approach is systematic and works for every nonzero algebraic number since the independent term in the minimal polynomial is not zero.