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I am trying to show that, using the norm map $N : \mathbb{Z}[\sqrt{-5}] \to \mathbb{Z}$ defined by:

$N(x+y\sqrt{-5}) = (x+y\sqrt{-5})(x-y\sqrt{-5}) = x^2 + 5y^2$, the units in $\mathbb{Z}[\sqrt{-5}]$ are $\pm$1.

I am unsure how to proceed but am thinking that the $x^2 + 5y^2$ will have to be equal to $\pm$1 as this will form a unit circle, is this right?

Thanks for the help.

Zombiegit123
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  • Note that complex conjugation defines an automorphism of $\mathbb Z[\sqrt{-5}]$ and $N(z) = z\overline z$. You should be able to use that to deduce that $z$ is a unit if and only if $N(z)$ is a unit in $\mathbb Z$. Then it's just a matter of solving the equation $N(z) = \pm 1$. – Dustan Levenstein Dec 05 '17 at 22:45

2 Answers2

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Let $a\in\mathbb{Z}\left[\sqrt{D}\right]$. Then $a$ is a unit iff $N\left(a\right)=\pm1$.

To prove this, remember that $N\left(ab\right)=N\left(a\right)N\left(b\right)$. Thus, if $a|b$ also $N\left(a\right)|N\left(b\right)$. A unit $a$ satisfies both $1|a$ and $a|1$ and therefore $N\left(1\right)|N\left(a\right)$ and $N\left(a\right)|N\left(1\right)$. But $N\left(1\right)=1$ and this happens in $\mathbb{Z}$, forcing $N\left(a\right)=\pm 1$. The other direction follows immediately from the fact that $N\left(a\right)=a\bar{a}=\pm 1$ so $a^{-1}=\pm\bar{a}$.

Now, for a unit $a\in\mathbb{Z}\left[\sqrt{-5}\right]$. If one writes $a=x+y\sqrt{-5}$, then you have to require

$$N\left(a\right)=x^2+5y^2=1$$

If $y\neq 0$, you can clearly see that $N\left(a\right)\geq 5y^2\geq5$. Thus for this equation to hold you must have $y=0$ and then it is easy to see that $x=\pm 1$.

eranreches
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Hint: With $z_k=x_k+y_k\sqrt{-5}$ for $k=1,2$, we have $$N(z_1z_2)=N(z_1)N(z_2)$$ So if $z_1$ is a unit, $N(z_1)\,|\,1$.

Berci
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