Having computed the Hilbert class field of some 'smaller' examples like $\mathbb{Q}(\sqrt{-5})$, $\mathbb{Q}(\sqrt{-31})$ and $\mathbb{Q}(\sqrt{-17})$, I now want to compute the Hilbert class field of $K=\mathbb{Q}(\sqrt{d})$ for $d=730=2\cdot 5\cdot 73$.

The number field $K$ has class group $Cl_K\cong\mathbb{Z}/6\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$. More specifcally, we have $\mathcal{O}_K=\mathbb{Z}[\sqrt{730}]$ and $Cl_K$ is generated by the classes $[\mathfrak{p}_3]$ and $[\mathfrak{p}_5]$ of order $6$ and $2$ respectively, where $\mathfrak{p}_3=(3,1+\sqrt{730})$ is a prime of $\mathcal{O}_K$ of norm $3$ and $\mathfrak{p}_5=(5,\sqrt{730})$ the ramified prime of norm $5$.

It is now not to hard to find an unramified Abelian extension of $K$ with group $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$, by finding two distinct quadratic unramified extensions of $K$: the extension $K(\sqrt{2})$ can only be ramified at $2$, but $K(\sqrt{2})=K(\sqrt{5\cdot 73})$ can only be ramified at $5$ and $73$ (note that $5,73\equiv 1\bmod 4$), hence $K(\sqrt{2})$ is unramified at the finite primes. It is also totally real and hence it is contained in the Hilbert class field of $K$. The same reasoning works for the extensions $K(\sqrt{5})$ and $K(\sqrt{73})$, and this yields a non-cyclic unramified abelian extension $K(\sqrt{2},\sqrt{5})$ of degree $4$.

It now remains to find some unramified cubic extension $K(\alpha)$ of $K$, for then taking composita we would get that $K(\sqrt{2},\sqrt{5},\alpha)$ is the Hilbert class field of $K$, and this is were I'm stuck.

Here are my attempts:

Let $H$ be the Hilbert class field of $K$. I tried to compute the Galois group $G$ of $H/\mathbb{Q}$, which has order $24$. Note that $H$ contains at least $7$ quadratic subfields as $\mathbb{Q}(\sqrt{2},\sqrt{5},\sqrt{73})\subset H$, so $G$ has at least $7$ subgroups of index $2$. The classification of groups of order $24$ now implies that either $G\cong D_{6}\times\mathbb{Z}/2\mathbb{Z}$ or $G\cong \mathbb{Z}/6\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$. If the group would be abelian, then we have an intermediate cubic field $H/E/\mathbb{Q}$ with $E/\mathbb{Q}$ Galois of degree $3$. Now some rational prime $p$ must ramify in $E$, and since $E/\mathbb{Q}$ is Galois of prime degree, this would imply that $p$ is totally ramified in $E$. But then we have the Galois extension $EK/K$ of degree $3$, which must then also be totally ramified over a prime $\mathfrak{p}$ of $K$ over $p$: if it were unramified over $\mathfrak{p}$ then by multiplicativity of the ramificiation index in the tower $EK/K/\mathbb{Q}$, we see that $e(\mathfrak{B}/p)\leq 2$ for any prime $\mathfrak{B}$ of $EK$ over $\mathfrak{p}$, which contradicts that $E/\mathbb{Q}$ is totally ramified over $p$.

There has to be a simpler way to determine the group, but in any case the group is $D_6\times\mathbb{Z}/2\mathbb{Z}$. Now I tried to find a cubic extension of $\mathbb{Q}$ whose normal closure is obtained by taking the compositum of a $K$ to get an unramified Abelian extension of degree $3$ of $K$, as in my answer to this question:

Hilbert class field of a quadratic field whose class number is 3

however this produces subfields of $H$ of degree $6$ that are Galois over $\mathbb{Q}$, while there is only one normal subgroup of $D_6\times\mathbb{Z}/2\mathbb{Z}$ of index $6$, which is its center. Since $H/K$ has a unique intermediate field of degree $3$ over $K$, this will only work provided that somehow $K$ is already contained in the fixed field belonging to the center of $G$, and I have no clue how to proceed from here.

Any help would be highly appreciated!

  • 3
    $\begingroup$ Not that it's much (or any) help in this instance, but the Minkowski bound is roughly 27.0185. $\endgroup$ Commented Jun 6, 2017 at 19:56

1 Answer 1


For finding unramified cubic extensions of ${\mathbb Q}(\sqrt{m}\,)$, you have to find singular primary numbers (cubes of ideals) in ${\mathbb Q}(\sqrt{-3m})$. A quick search shows you that $\alpha = 197 + 3\sqrt{-3 \cdot 730}$ is such a number since its norm is $49^3$ and it is congruent to an integer modulo $3 \sqrt{-3}$ (read this with a grain of salt).

Now set $\theta = \sqrt[3]{\alpha} + \sqrt[3]{\alpha'}$; then \begin{align*} \theta^3 & = (\sqrt[3]{\alpha} + \sqrt[3]{\alpha'})^3 = \alpha + 3\sqrt[3]{\alpha}^2\sqrt[3]{\alpha'} + 3\sqrt[3]{\alpha}\sqrt[3]{\alpha'}^2 + \alpha' \\ & = (\alpha + \alpha') + 3\sqrt[3]{\alpha \alpha'}(\sqrt[3]{\alpha} + \sqrt[3]{\alpha'}) = 2 \cdot 197 + 3 \cdot 49 \cdot \theta, \end{align*} hence $\theta$ is a root of the polynomial $f(x) = x^3 - 147 x - 394$. pari confirms that adjoining a root of $f$ to ${\mathbb Q}(\sqrt{730}\,)$ gives you an unramified cyclic extension. More details concerning the construction can be found here.

  • $\begingroup$ Thank you for the answer! A minor generalisation of the hypothesis of my answer which I referenced to in the question (allowing an odd number of factors $2$ in the discriminant) shows that your polynomial $f$ indeed produces the desired cubic unramified extension of $\mathbb{Q}(\sqrt{730})$, which is very nice. However, I'm still confused a lot about how you came up with the polynomial. After reading $1.5.1$ and $1.5.2$ of the pdf you gave a link of, I still don't see how the construction could take place. For example, $m=730$ is not of the form $t^2-2$ so Proposition $1.5.2$ will not help. $\endgroup$
    – Tim.ev
    Commented Jun 13, 2017 at 12:15
  • $\begingroup$ I hope it's clear now. $\endgroup$
    – user23365
    Commented Jun 13, 2017 at 13:54
  • $\begingroup$ Well I can now see how $f$ is constructed from $\theta$, but I still don't see why I need to find such singular primary numbers in $\mathbb{Q}(\sqrt{-3m})$, and why you those to set $\theta$ the way you did. Is it possible to give a brief explanation of this? Thank you in any case. $\endgroup$
    – Tim.ev
    Commented Jun 13, 2017 at 15:43
  • 1
    $\begingroup$ Basically this is Galois theory. If you have a field $K$ not containing the cube roots of unity and if you set $L = K(\sqrt{-3})$, then any cyclic cubic extension of $L$ has the form $F = L(\sqrt[3]{\alpha})$. If $K/k$ is quadratic, then Galois theory tells you exactly when $L/k$ is normal, what its Galois group is, and which properties $\alpha$ needs to have for $L/k$ or $L/k(\sqrt{-3})$ etc. to be abelian. Essentially this is Kummer theory (see the section on eigenspaces in the manuscript). If you give me some time, I'll write down the cubic case explicitly. $\endgroup$
    – user23365
    Commented Jun 13, 2017 at 15:56
  • $\begingroup$ Ah that already explains a lot. an explicit description for the cubic case would be very nice! $\endgroup$
    – Tim.ev
    Commented Jun 13, 2017 at 16:02

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