TITLE: how to compute Hilbert class field of $\Bbb Q(\zeta_{23})$?
QUESTION [8 upvotes]: 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.

REPLY [18 votes]: 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 $x^3-x+1$.