# The ring $\{a+b\sqrt{2}\mid a,b\in\mathbb{Z}\}$

The set $\{a+b\sqrt{2}\mid a,b\in\mathbb{Z}\}$ spans a ring under real addition and multiplication. Which elements have multiplicative inverses?

This is part of an exercise from an introductory text to algebraic structures. The answer is that an element has a multiplicative inverse if and only if $a^2 - 2b^2 = \pm 1$. It is evident that elements verifying the condition are units but I fail to see that it is the only possible solution. Any one can shed some light?

-
Did you mean $\impliedby$ rather than $\implies$? Written with $\implies$ it looks like you are saying you understand why units have that form. – rschwieb May 8 '12 at 16:49
I updated it to make it clear. – Euclean May 8 '12 at 16:57

## 3 Answers

Hint $\$ Use either the multiplicativity of the norm, or rationalize denominators. Since the former is well-known, but the latter is not, I'll elaborate on that. Let $\rm\: d = gcd(a,b) = 1.\:$ Then

$$\rm \frac{1}{a+b\sqrt{2}} = \frac{a-b\sqrt{2}}{a^2\!-2b^2}\in \mathbb Z[\sqrt{2}]\:\Rightarrow\: c = a^2\!-\!2b^2\:|\:a,b\:\Rightarrow\: d^2\:\!|\:c\:|\:d\:\Rightarrow\: d = 1\:\Rightarrow\: c=\pm1$$

-
Thanks for the answer, I follow all your reasoning except for the assumption that $\textrm{gcd}(a,b)=1$ for every unit. I understand that if $\textrm{gcd}(a,b)\neq 1$ then there could be units beyond those that verify $a^2-2b^2=\pm 1$ – Euclean May 8 '12 at 16:46
[Note: This is not used in the current simpler answer version] In $\rm\:R = \mathbb Z[\sqrt{2}],\:$ $\rm\:n\:|\:\alpha\:|\:1\:\Rightarrow\: n\:|\:1\:$ in R, so $\rm\:1/n \in R\cap\mathbb Q = \mathbb Z,\:$ i.e. we are exploiting the fact that no nonunit integers become units in R. But be sure to understand the norm approach too - it is very important. – Bill Dubuque May 8 '12 at 17:03
I see now both approaches. Thanks a lot! – Euclean May 8 '12 at 17:20
Note that you may already know a special case of norm multiplicativity, viz. the formula for composition of sums of squares (Brahmagupta-Fibonacci identity) which arises from norm multiplicativity in $\mathbb Z[{\it i}\:\!]$. – Bill Dubuque May 8 '12 at 17:33
@BillDubuque I admire your succinctness in answers. – Pedro Tamaroff May 8 '12 at 17:46

The easiest way to see this is to note that $f(a+b\sqrt{2}) = a^2 - 2b^2$ is multiplicative: $f((a+b\sqrt{2})(c+d\sqrt{2})) = f(a+b\sqrt{2})f(c+d\sqrt{2})$ (why?). Therefore, if $a+b\sqrt{2}$ has a multiplicative inverse $c + d \sqrt{2}$, we have $f(a+b\sqrt{2}) \cdot f(c+d\sqrt{2}) = f(1) = 1$. It's also easy to see that $\operatorname{im} f \subseteq \mathbb Z$, from which the claim follows.

-
Ok, thanks for the answer. If I understand correctly, the same technique can be used to prove that $\mathbb{Z}[\sqrt{2}]$ is an integral domain: $f(a+b\sqrt{2})f(c+d\sqrt{2})=f(0)=0$. And since $\textrm{im}f\subseteq\mathbb{Z}$ the claim follows. – Euclean May 9 '12 at 9:04
Yes, this argument is valid. – Johannes Kloos May 9 '12 at 9:19

If

$$(a+b\sqrt{2})(c+d\sqrt{2})=1$$

then

$$(a-b\sqrt{2})(c-d\sqrt{2})=1$$

By multiplying them together you get the desired result.

P.S. This is basically the same solution as Johannes's.

-