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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?

Euclean
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3 Answers3

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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.

Johannes Kloos
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  • 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 09 '12 at 09:04
  • Yes, this argument is valid. – Johannes Kloos May 09 '12 at 09:19
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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

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

Bill Dubuque
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    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 08 '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 08 '12 at 17:03
  • I see now both approaches. Thanks a lot! – Euclean May 08 '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 08 '12 at 17:33
  • @BillDubuque I admire your succinctness in answers. – Pedro May 08 '12 at 17:46
  • @BillDubuque thanks indeed, the extra insight is greatly appreciated. – Euclean May 09 '12 at 08:53
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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.

N. S.
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