# 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

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Consider norms. –  Qiaochu Yuan May 24 '12 at 23:57
The degree of an $n$-th root of unity over the rationals is $\varphi(n)$, so to be in a quadratic field we need $\varphi(n) \leq 2$, and the only such $n$ are 1, 2, 3, 4, and 6. The cases $n = 1$ and 2 don't take you beyond the rationals, and for the others we have ${\mathbf Q}(\zeta_3) = {\mathbf Q}(\zeta_6) = {\mathbf Q}(\sqrt{-3})$, ${\mathbf Q}(\zeta_4) = {\mathbf Q}(i)$. This approach, though, is overkill. A unit in ${\mathcal O}_K$ has norm $1$ and if you look at the norm formula for algebraic integers other than 0 or $\pm 1$ you'll see the norm is at least 2 unless $d$ is 1 or 3. –  KCd May 25 '12 at 0:10

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 $$\pm\frac{1+\sqrt{-3}}{2},\quad \pm\frac{1-\sqrt{-3}}{2}.$$

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I think you mean to say "$d$ not congruent to $1$ modulo $4$" –  Daniel Freedman May 31 '12 at 23:47
@Daniel: No, because we are looking at $\mathbb{Q}(\sqrt{-d})$; so we care about $-d\equiv 1\pmod{4}$, which is equivalent to $d\equiv 3\pmod{4}$. –  Arturo Magidin May 31 '12 at 23:50
Oops. This isn't the first time I've made a mistake like that... –  Daniel Freedman May 31 '12 at 23:52

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.

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