Is there any method to find the Hilbert Class field of quadratic fields? Is there any bound for their dimensions? For example, if $4|d-1$ then $Q(\sqrt{d},i)$ is contained in the Hilbert class field of $Q(\sqrt{-d})$, therefore $Q(\sqrt{-d})$ isn't an UFD.
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6$\begingroup$ A book dedicated to this question is Cohen's Advanced Topics in Computational Number Theory. $\endgroup$– Dror SpeiserCommented Sep 28, 2011 at 11:39
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1$\begingroup$ 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. $\endgroup$– EmertonCommented Sep 28, 2011 at 12:18
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4$\begingroup$ 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) $\endgroup$– Tommaso CentelegheCommented Sep 28, 2011 at 12:37
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4$\begingroup$ For your second question, the answer is no by genus theory: The more primes that divide d, the bigger the class group. $\endgroup$– Cam McLemanCommented Sep 28, 2011 at 12:49
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1 Answer
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For the case of imaginary quadratic fields $K=\mathbb{Q}(\sqrt{d_K})$, the Hilbert class field can be given as $K\left(j(\frac{d_k+\sqrt{d_k}}{2})\right)$, where $j$ is the $j$-invariant. This is just a theoretical answer; and note that computing $j(\frac{d_k+\sqrt{d_k}}{2})$ in practice, is not easy.
answered Jan 23, 2021 at 14:19