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Jan 23, 2021 at 14:30 history edited GH from MO
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Jan 23, 2021 at 14:19 answer added Davood Khajehpour timeline score: 1
Sep 28, 2011 at 12:49 comment added Cam McLeman For your second question, the answer is no by genus theory: The more primes that divide d, the bigger the class group.
Sep 28, 2011 at 12:37 comment added Tommaso Centeleghe Observe that the ring of Integers of $Q(\sqrt{-d})$ when $d\equiv 1$ mod $4$ is not a UFD because the only ideal lying above $2$, which is ramified, cannot be principal (the integer $2$ is not a norm). The theory of complex multiplication (in case you did not know) relates the Hilbert class field of an imaginary quadratic filed K to the j-invariant of any elliptic curve with CM by the integers of K (one can learn this in "Primes of the form x^2+ny^2" by D.cox)
Sep 28, 2011 at 12:21 history edited Sina CC BY-SA 3.0
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Sep 28, 2011 at 12:18 comment added Emerton It is not true that $\mathbb Q(\sqrt{d},i)$ is the Hilbert Class Field of $\mathbb Q(\sqrt{-d})$ in general when $4 | (d-1)$, just that it is contained in the Hilbert Class Field.
Sep 28, 2011 at 11:39 comment added Dror Speiser A book dedicated to this question is Cohen's Advanced Topics in Computational Number Theory.
Sep 28, 2011 at 11:12 history asked Sina CC BY-SA 3.0