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In the ring $\mathbb{Z}[\sqrt{2}]$, how do I prove that an element $\alpha$ is a unit if and only if $N(\alpha) = 1$?

We are told that $N(a+b\sqrt{2}) = a^2-2b^2$.

I've shown that $N(\alpha\beta)=N(\alpha)N(\beta)=1$, but since $N(\alpha) \text{and} N(\beta)$ are both in $\mathbb{Z}$, they could be $1$ or $-1$? That is the problem I'm having.

2 Answers2

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You have already shown that it is necessary that a unit have norm of (plus or minus) one. To show sufficiency, assume that $a + b\sqrt{2}$ is such that $N(a+b\sqrt{2})=a^2 -2 b^2=\pm 1$. Then notice that $(a+b\sqrt{2})(a-b\sqrt{2})=a^2-2b^2=\pm 1 $. Hence $a-b\sqrt 2$ is the desired multiplicative inverse, and so $a+b\sqrt 2 $ is a unit.

mwalth
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  • I'm supposed to prove that it must be +1 only though...is that possible? –  Apr 26 '15 at 01:17
  • As @Rolf pointed out, the element $1+\sqrt{2}$ is a unit whose norm is $-1$. So unless you're using a different norm, i.e. $N(a+b\sqrt{2})=\left |a^2-2 b^2 \right |$, then the you need to include both positive and negative one. – mwalth Apr 26 '15 at 01:33
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Note that units can easily have norm $-1$. For instance, if you look at $1 + \sqrt{2}$, the norm is $1^2-2(1^2) = -1$, but $(1+\sqrt{2})(-1+\sqrt{2}) = 1$. You should instead prove that the units are precisely the elements with norm $\pm 1$.

Rolf Hoyer
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  • I'm trying to show that $\alpha$ is a unit if and only if $N(\alpha)=1$, though. And it seems that units exist with norm -1. –  Apr 26 '15 at 01:09
  • So it is impossible to prove that $\alpha$ is a unit if and only if the norm is positive 1? –  Apr 26 '15 at 01:23
  • Since there exist counterexamples to the only if part, any such proof would show mathematics to be inconsistent. – Rolf Hoyer Apr 26 '15 at 01:27
  • Arggh...that is super annoying. However, if the norm function had the absolute value around it, then it would work, right? –  Apr 26 '15 at 01:33
  • The result would be true if you defined the norm that way, and the proof would be identical to showing that the units are those elements are those with norm $\pm 1$ under the usual norm. – Rolf Hoyer Apr 26 '15 at 01:35