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Let $K = \mathbb{Q}(\sqrt{-d})$ be an imaginary quadratic field of class number one (i.e. every ideal in $\mathcal{O}_K$ is principal, i.e. $\mathcal{O}_K$ is a principal ideal domain). Let $d_K$ be the discriminant of $K$.

How does one prove that all primes less than $\frac{1 + |d_K|}{4}$ are inert in $K$?

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Let $p$ a prime. If ${\frak P}$ is a prime ideal of $O_K$ above $p$, then the assumption on the class number implies that ${\frak P}=\langle \alpha\rangle$ for some $\alpha\in O_K$. Let $\alpha=a+b\theta$, where $\theta$ is either $\sqrt{-d}$ or $(1+\sqrt{-d})/2$ depending. Show that if $\alpha$ is not a rational integer, then $N(\alpha)\ge(1+|d_K|)/4$.

If $p$ splits, then we must have $N(\alpha)=p$. This is clearly impossible for a rational integer $\alpha$. So the above result shows that $p\ge(1+|d_K|)/4$.

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  • $\begingroup$ Ah, that is it! :) $\endgroup$ – Bruno Joyal Nov 22 '13 at 14:20
  • $\begingroup$ I'm a bit worried about the hole formed by the case of $d$ being a prime $\equiv1\pmod4$. In that case $N(\sqrt{-d})=|d_K|/4$ is smaller than I promised, and the prime $d$ is ramified. In those cases we need the extra bit that primes above $p=2$ are not principal, so those cases are excluded by the class number condition. $\endgroup$ – Jyrki Lahtonen Nov 28 '13 at 16:17
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Well, according to Stark-Heegner, there are only 9 such fields, corresponding to the values of $-d$ $$−1, −2, −3, −7, −11, −19, −43, −67, −163.$$ Thus, the statement can be verified by a finite computation (left to the reader ;)).

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