If $K = \mathbb Q(\sqrt{-d})$, then $\mathcal O_K^* = \{\pm 1\}$ for $d \neq 1,3$ (squarefree) I'd like to know why if $K = \mathbb Q(\sqrt{-d})$, then $\mathcal O_K^* = \{\pm 1\}$ for $d \neq 1, 3$. 
Dirichlet's unit theorem tells us that the only units in $\mathcal O_K$ are the roots of unity contained in $K$. Why does $\mathbb Q(\sqrt{-d})$ not contain any roots of unity other than $1,-1$ for the specified $d$?
Thanks
 A: For $d$ not congruent to $3$ modulo $4$, the elements of $\mathcal{O}_K$ are of the form $a+b\sqrt{-d}$ with $a,b\in\mathbb{Z}$; the map $a+b\sqrt{-d}\mapsto (a+b\sqrt{-d})(a-b\sqrt{-d}) = a^2+db^2$ is multiplicative and has positive integers as images, hence $a+b\sqrt{-d}$ is a unit if and only if its image is $1$. But $a^2+db^2=1$ with $d\gt 1$ requires $b=0$ and $a=\pm 1$. If $d=1$, then we also get $a=0$, $b=\pm 1$, i.e., $i$ and $-i$. 
For $d\equiv 3 \pmod{4}$, the element sof $\mathcal{O}_k$ are of the form $$\frac{a+b\sqrt{-d}}{2}$$
with $a,b\in\mathbb{Z}$ of the same parity. The map
$$\frac{a+b\sqrt{-d}}{2} \longmapsto \left(\frac{a+b\sqrt{-d}}{2}\right)\left(\frac{a-b\sqrt{-d}}{2}\right) = \frac{a^2+db^2}{4}$$
is multiplicative and has positive integers as their images, hence an element of $\mathcal{O}_K$ is a unit if and only if $a$ and $b$ are of the same parity and $a^2+db^2=4$. If $d\gt 3$ then $d\gt 7$, so this requires $b=0$ and $a=\pm 2$, hence the element in question is $\pm 1$. 
If $d=3$, then we also have the solution $a=\pm 1$, $b=\pm 1$, which gives 
$$\pm\frac{1+\sqrt{-3}}{2},\quad \pm\frac{1-\sqrt{-3}}{2}.$$
A: You can either just solve the norm equation, as Qiaochu suggests, in which case you don't need Dirichlet's theorem; alternatively, you can show that if $\zeta_n$ is a primitive $n$-th root of unity, then $\mathbb{Q}(\zeta_n)$ has degree $\phi(n)$, the Euler totient function. From here, you will quickly find that the only $n$ for which the degree is two are 3, 4, 6. Thus, you cannot have any other roots of unity defined over a degree two field.
