What is known about the computational complexity of primality testing in $\mathcal{O}_{\mathbb{Q}(\sqrt{d})}$ where $d$ is a square-free number? For what values of $d$ is primality testing easy (i.e., can be determined in polynomial time)?
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$\begingroup$ You mean in the ring of integers of a quadratic number field? $\endgroup$– Qiaochu YuanDec 19, 2011 at 17:35
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$\begingroup$ Yes, sorry. I have edited my question—hopefully it is clearer now. $\endgroup$– Zach LangleyDec 19, 2011 at 17:39
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1$\begingroup$ $\mathbb{Z}[\sqrt{d}]$ isn't the full ring of integers if $d \equiv 1 \bmod 4$. $\endgroup$– Qiaochu YuanDec 19, 2011 at 17:43
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1 Answer
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For $K = \mathbb{Q}(\sqrt{d})$, primality testing in $\mathcal{O}_K$ reduces to primality testing of integers. It's known that the prime elements $\alpha \in \mathcal{O}_K$ are either
- elements whose norms $N(\alpha)$ are prime, or
- elements whose norms $N(\alpha)$ are squares of a prime $p$ such that $\left( \frac{d}{p} \right) = -1$.
(A slight modification is necessary in the case that $d \equiv 1 \bmod 4$ and $N(\alpha) = 4$. In that case $\alpha$ is prime if and only if $\frac{d-1}{4}$ is odd.)
The first condition can be tested for in polynomial time by the AKS primality test. The second condition can be tested for in polynomial time by AKS and quadratic reciprocity.
answered Dec 19, 2011 at 17:42
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$\begingroup$ If you don't mind me asking, is there a result stating this? Could you perhaps point me in the right direction? Thanks in advance. $\endgroup$– tc1729Jan 28, 2014 at 1:55
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1$\begingroup$ @Siddharth: probably this result can be found in most textbooks on algebraic number theory, possibly as an exercise. $\endgroup$ Jan 28, 2014 at 2:15