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Prove that $N(\gamma) = 1$ if, and only if, $\gamma$ is a unit in the ring $\mathbb{Z}[\sqrt{n}]$

Where $N$ is the norm function that maps $\gamma = a+b\sqrt{n} \mapsto \left | a^2-nb^2 \right |$

I have managed to prove $N(\gamma) = 1 \Rightarrow \gamma$ is a unit (i think), but cannot prove $\gamma$ is a unit $\Rightarrow N( \gamma ) = 1$

Any help would be appreciated, cheers

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I expect there is a typo, where you have $|a^2-nb^2|$, probably $|a^2-2b^2|$ is intended. –  André Nicolas May 26 '12 at 1:07
    
thanks for the spot, the $\sqrt{2}$ was actually the typo, I meant $\sqrt{n}$ I realise now that the title is therefore slightly wrong.. whoops –  Sam May 26 '12 at 1:10

4 Answers 4

up vote 2 down vote accepted

If $\,\gamma\in\mathbb{Z}[\sqrt{n}]\,$ is a unit, then $\exists\beta\in\mathbb{Z}[\sqrt{n}]\,\,s.t.\,\,\gamma\beta=1\,\Longrightarrow 1=N(1)=N(\gamma\beta)=N(\gamma)N(\beta)$ ...

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Hint $\rm\ \ unit\ \alpha\iff \alpha\:|\: 1\iff \alpha\alpha'\:|\:1 \iff unit\ \alpha\alpha',\ $ since $\rm\:\alpha\:|\:1\iff\alpha'\:|\:1' = 1$

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Thanks for the hints. I have come up with a proof that only uses the facts that

$$ N(\alpha\beta) = N(\alpha)N(\beta) $$ and $$ N(1) = 1 $$

Let $\gamma$ be a unit. Then $\gamma\beta = 1$ for some $\beta\in\mathbb{Z}[\sqrt{n}]$

$N(1) = N(\gamma\beta) = N(\gamma)N(\beta) = 1$

As both $N(\gamma)$ and $N(\beta)$ are integers, they must both equal 1, hence $N(\gamma) = N(\beta) = 1$

See anything wrong with this proof?

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That's fine, except that generally an integer unit can be $\pm 1$. The way I hinted is closely related, viz. $\rm\:a\:|\:b\:\Rightarrow ac = b\:\Rightarrow\:a'c' = b'\:\Rightarrow\:a'\:|\:b'.\:$ Thus $\rm\:a\:|\:1\:\Rightarrow\:a'\:|\:1\:\Rightarrow\:aa'\:|\:1\cdot 1 = 1.$ So this is really just a special case of the fact that units are closed under multiplication and divisors (and, here, conjugation). $\ \ $ –  Bill Dubuque May 26 '12 at 1:43

Hint: The Brahmagupta Identity $$(a^2-nb^2)(c^2-nd^2)=(ac+nbd)^2-n(ad+bc)^2$$ is easy to verify, and useful. It implies that the norm of a product is the product of the norms.

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