$ℚ[\sqrt{-23}]$，计算$IJ$

hbghlyj

$K=ℚ[\sqrt{-23}]$，$𝒪_K=ℤ[\frac{1+\sqrt{-23}}2]$

$ 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=eJwLDNSzSbRTsFUILE1MKUosyUx2y0zNSdHQNTL ... 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

hbghlyj

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$\dfrac{\sqrt{-23}+3}2𝒪_K$