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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.2012-05-24
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    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 3.2012-05-25

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