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I want to construct the Hilbert class field of $K=\Bbb Q(\zeta_{23}).$ I have no clue how to construct it except that I know that $[H(K):K]=3$ from Sage. Any references or comments are appreciated.

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    $\begingroup$ Maybe find a cubic extension ramified only over 23? Then if you adjoin that to your Cyclotomic extension you will get something unramifed over 23 and hence unramified everywhere $\endgroup$
    – Asvin
    Dec 30, 2020 at 9:32
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    $\begingroup$ The claim in en.wikipedia.org/wiki/Hilbert_class_field is that you have to add a root of $x^3-x-1$ ; I didn't verify it. $\endgroup$ Dec 30, 2020 at 12:32

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The cyclotomic field $K=\mathbb{Q}(\zeta_{23})$ contains the quadratic field $F=\mathbb{Q}(\sqrt{-23})$, and $F$ has class number $3$ (it is the first quadratic field, when those are ordered by absolute value of the discriminant, that has class number divisible by $3$). Thus, $F$ has an unramified cubic extensions $H$, and since the degree of $K$ is coprime with $3$, the extensions $H$ and $K$ are disjoint, so that the compositum $HK$ is an unramified cubic extension of $K$. Thus, your problem is reduced to finding the Hilbert class field of $F$. Magma (and presumably also Sage?) will just give it to you. It is the splitting field over $\mathbb{Q}$ of the cubic polynomial $x^3-x+1$.

In summary, the Hilbert class field of $K$ is obtained by adjoining to $K$ a root of $x^3-x+1$.

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    $\begingroup$ small typo: "reduced to finding the Hilbert class field of F" instead of "reduced to finding the Hilbert class field of K" (but edit has to be at least 6 characters) $\endgroup$
    – Arno Fehm
    Dec 30, 2020 at 12:53
  • $\begingroup$ Thanks, @Arno, corrected. Hope you are well! $\endgroup$
    – Alex B.
    Dec 30, 2020 at 13:02
  • $\begingroup$ Thank you, @Alex, for this nice answer. Best wishes for the New Year! $\endgroup$
    – Arno Fehm
    Dec 30, 2020 at 13:55

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