# Math Help - Find units in ring of integers

1. ## Find units in ring of integers

Let R be the ring of integers in $\mathbb {Q} [ \sqrt {-3} ]$. Determine the units in R.

Let $x = a + b \sqrt {-3}, \ \ \ a,b \in \mathbb {Z}$ be a unit of R, then the norm of x, that is, $N_{R} (x) = 1,-1$

$N_{R} (x) = N_{R} (a+b \sqrt {-3} ) = a^2 + b^2 (-3) = =1, -1$

I can only think of +1 and -1 being units in R, anything more?

2. A unit in a ring with multipicative identity 1 is a divisor of 1; if the ring is also a normed vector space, then the norm of a unit must be 1.

So $a+b\sqrt{-3}$ is a unit in $\mathbb{Q}\left[\sqrt{-3}\right]\ \Rightarrow\ a^2+3b^2=1$..

Conversely, any element of the form $a+b\sqrt{-3}$ where $a^2+3b^2=1$ is a unit in $\mathbb{Q}\left[\sqrt{-3}\right]$, its mutliplicative inverse being $a-b\sqrt{-3}$:

$\left(a+b\sqrt{-3}\right)\left(a-b\sqrt{-3}\right)\ =\ 1$

Hence the units of $\mathbb{Q}\left[\sqrt{-3}\right]$ are $\left\{a+b\sqrt{-3}:a,b\in\mathbb{Q},a^2+3b^2=1\right\}$.

By the way, there are units other than ±1: for example, $\left(\frac{1+\sqrt{-3}}{2}\right)\left(\frac{1-\sqrt{-3}}{2}\right)=1$.