# Find all units in the ring $R$ as defined.

$$R = \mathbb{Z}\left[\sqrt{-7}\right] \le \mathbb{C}$$

In other words, $$R$$ is the set of all integers and integers multiplied by the square root of $$-7$$. I believe this might be called $$\mathbb{Z}$$ adjoin root $$-7$$ but I am not completely sure.

I am asked to find all units in $$R$$. In $$\mathbb{Z}$$, the only units are $$1$$ and $$-1$$ which are obviously units of $$R$$ as well. I know $$\mathbb{Z}[i]$$ has units $$1, -1, i$$, and $$-i$$ but the latter two are not in $$R$$.

Are there any non-trivial units in this ring? All I can find is $$1$$ and $$-1$$, but that seems too simple.

• The set $\mathbb Z[\sqrt{-7}]$ is the set of all complex numbers of the form $a+b\sqrt{-7}$, with $a,b\in \mathbb Z$. – lhf Apr 2 '15 at 1:24

Hint:

• Consider $N(a+b\sqrt{-7}) = a^2+7b^2$, which coincides with the norm or absolute value of the complex number $a+b\sqrt{-7}$.

• Conclude that $N$ is multiplicative.

• Prove that $N(\alpha)=\pm1$ iff $\alpha$ is a unit in $\mathbb Z[\sqrt{-7}]$.

• Conclude that you need to solve $a^2+7b^2=\pm 1$ with $a,b\in \mathbb Z$.

• Solve it.

• Thank you, but that suggests to me that my trivial answer was right after all. If I have an element of R written as a + b*sqrt(-7), then its norm is a^2 + 7b^2, which is clearly an integer as a and b are both integers. So say I have two elements of R, x and y, and they are both units. Then abs(xy) = abs(x)*abs(y) = 1. So I need two integers abs(x) and abs(y) such that their product is 1. The only solutions are abs(x) = abs(y) = 1. And if b cannot 0 this would force the norm to be greater than 1, b must be zero. Thank you for reminding me of the technique of using norms to find units. – ocyeung Apr 2 '15 at 1:44
• Yes, imaginary quadratic rings have only trivial units, except for two cases ($-1$ and $-3$ instead of $-7$). – lhf Apr 2 '15 at 2:36