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[数论] $ℚ[\sqrt{-23}]$,计算$IJ$

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hbghlyj posted 2024-5-27 03:42 |Read mode
$K=ℚ[\sqrt{-23}]$,$𝒪_K=ℤ[\frac{1+\sqrt{-23}}2]$

$ I=2𝒪_K+\dfrac {\sqrt {-23}-1}{2}𝒪_K$

$ J=4𝒪_K+\dfrac {\sqrt {-23}+3}{2}𝒪_K$

如何证明
$ IJ=\dfrac {3+\sqrt {-23}}{2}𝒪_K$

验证:sagecell.sagemath.org/?z=eJwLDNSzSbRTsFUILE1M … cts=eJyLjgUAARUAuQ==

小尝试:
$ IJ=8𝒪_K+4\dfrac {\sqrt {-23}-1}{2}𝒪_K+2\dfrac {\sqrt {-23}+3}{2}𝒪_K+\dfrac {\sqrt {-23}-1}{2}\dfrac {\sqrt {-23}+3}{2}𝒪_K=8𝒪_K+\dfrac {\sqrt {-23}+3}{2}𝒪_K$
然后没法继续化简了

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original poster hbghlyj posted 2024-5-27 03:52
hbghlyj 发表于 2024-5-26 19:42
然后没法继续化简了
我知道了:$\dfrac{8}{\frac{\sqrt{-23}+3}2}=\dfrac{3-\sqrt{-23}}2\in𝒪_K$

$8\in\dfrac{\sqrt{-23}+3}2𝒪_K$

然后就只剩下$\dfrac{\sqrt{-23}+3}2𝒪_K$

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