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If we define the norm on $\mathbb{Z}[\sqrt{-3}]$ to be $N(\alpha)=a^2+3b^2$, then how do we use this norm to find all the units in $\mathbb{Z}[\sqrt{-3}]$.

I know what a unit is, so we are looking for all the invertible elements in $\mathbb{Z}[\sqrt{-3}]$ e.g.

$(a+b\sqrt{-3})(c+d\sqrt{-3})=1$ and so we have $(a-b\sqrt{-3})(c-d\sqrt{-3})=1$ which gives: $$(a^2+3b^2)(c^2+3d^2)=1$$ so the only units are $1,-1$ but I'm not sure how to use the norm to show this (or have I as used it without realising?)

Thanks very much for any help

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Perhaps it is usually done by showing that the norm of any unit will be 1. –  mixedmath May 10 '12 at 16:18
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Notice that $N(a+b\sqrt{-3})=(a+b\sqrt{-3})(a-b\sqrt{-3})$. So yes, you have used it without realizing. –  Micah May 10 '12 at 16:19

2 Answers 2

up vote 8 down vote accepted

The norm is multiplicative: $N(\alpha\beta) = N(\alpha)N(\beta)$.

In particular, if $\alpha$ is a unit, then there exists $\beta$ such that $\alpha\beta=1$, so $1= N(1) = N(\alpha\beta) = N(\alpha)N(\beta)$. Thus, if $\alpha$ is a unit, then $N(\alpha)=\pm 1$.

Conversely, if $N(\alpha)=\pm 1$, then $\alpha\overline{\alpha}=1$ or $\alpha(-\overline{\alpha}) = 1$, so $\alpha$ is a unit.

(In fact, since the norm is always positive you can ignore one of the possibilities, but notice that I never used that we are working in $\mathbb{Z}[\sqrt{-3}]$. All we are using is that we have a multiplicative map $N\colon R\to\mathbb{Z}$ to get the necessity; and the sufficiency from the fact that we can evaluate this map $N$ via a unary operator, in this case the map $a+b\sqrt{-3}\mapsto a-b\sqrt{-3}$. So the argument easily generalizes to any $\mathbb{Z}[\sqrt{d}]$ with $d$ squarefree, and the ring of integers of any Galois extension of $\mathbb{Q}$)

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You implicitly use multiplicativity of the norm. Essentially the proof amounts to the fact that multiplicative maps preserve divisibility, so if they preserve $1$ then they preserve its divisors (= units). I prove this generally below. Put $\rm\: \bar x = xx' =$ norm $\rm x,\:$ and $\rm\: e = 1\:$ to get your case, viz.

$$\rm unit\ \alpha\iff \alpha\:|\: 1\iff \alpha\alpha'\:|\:1 \iff unit\ \alpha\alpha'$$

Theorem $\ $ If $\rm\:\overline{xy} = \bar x\:\bar y,\:$ and $\rm \: x\:|\:\bar x,\:$ and $\rm\:\bar e = e\:$ then

$$\rm\:a\:|\:e \iff \bar a\:|\:e $$

Proof $\rm\ \ (\Rightarrow)\ \ \ a\:|\:e\:\Rightarrow\: e = ab\:\Rightarrow\: e = \bar e = \overline{ab} = \bar a\:\bar b\:\Rightarrow\: \bar a\:|\:e.\:$

$\rm (\Leftarrow)\ \ \ a\:|\:\bar a\:|\:e\:\Rightarrow\:a\:|\:e.\ \ $ QED

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