The group of units of $\{a+bi\sqrt{2}:a,b\in\mathbb{Z}\}$ is $\{1,-1\}$.

Question

The exercise is to show the group of units of the ring $\{a+bi\sqrt{2}:a,b\in\mathbb{Z}\}$ is $\{1,-1\}$. I don't understand the beginning of the book's solution:

Suppose $a+bi\sqrt{2}$ has an inverse $x+yi\sqrt{2}$. Then $|a+bi\sqrt{2}|^{2}|x+yi\sqrt{2}|^{2}=1$. That is, $(a^{2}+2b^{2})(x^{2}+2y^{2})=1$.

First, why the absolute value instead of just $(a+bi\sqrt{2}) (x+yi\sqrt{2})=1\implies(a+bi\sqrt{2})^{2}(x+yi\sqrt{2})^{2}=1?$

Second, how did they get $(a^{2}+2b^{2})(x^{2}+2y^{2})=1?$

When I did this: $$\begin{align} |a+bi\sqrt{2}|^{2}|x+yi\sqrt{2}|^{2}&=|(a+bi\sqrt{2})^{2}(x+yi\sqrt{2})^{2}|\\ &=|((a^{2}-2b^{2})+2abi\sqrt{2})((c^{2}-2d^{2})+2cdi\sqrt{2})|\\ &=|(a^{2}-2b^{2})(x^{2}-2y^{2})-8abxy+\text{other stuff with }i|. \end{align}$$

What about the $-8abxy$? Doesn't that have to be set $=1$, too?

Thanks in advance.

Answer 1

There is a norm with values in $\mathbb{N}$ on that ring. It is defined by $N(a+i\sqrt{2}b)=a^2+2b^2$. This norm is multiplicative, you can check that. So if $x$ and $y$ are inverses of each other $i.e.$ $xy=1$ then $1=N(1)=N(xy)=N(x)N(y) $ which implies $N(x)=N(y)=1$. From this you get that the only invertible elements are $\pm 1$.

Answer 2

You are working with complex numbers here, and the absolute value of a complex number is defined by $|a+bi| = \sqrt{a^2+b^2}$ ($a$ and $b$ being the real and imaginary parts), the length of the vector from (0,0) to $(a,b)$.

You seem to be saying that $|z|^2 = z^2$, which is incorrect; that only works with real numbers.