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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?

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    I updated it to make it clear.2012-05-08

3 Answers 3

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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.

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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.

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    Yes, this argument is valid.2012-05-09
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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

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

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    @BillDubuque thanks indeed, the extra insight is greatly appreciated.2012-05-09