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I tried constructing the Hilbert class field of $\Bbb Q(\zeta_{31})$ by imitating one of the problems from MO. I failed miserably as a quadratic field inside the cyclotomic field $\Bbb Q(\zeta_{31})$ has a class number different from $\Bbb Q(\zeta_{31})$. Thanks for your help in advance.

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This is a fun question, and I had already been thinking of making some comments on this in the question you link to. Apologies in advance for the long post.

Actually, the quadratic subfield of $\mathbb{Q}(\zeta_{31})$ is $\mathbb{Q}(\sqrt{-31})$, and has class number $3$. It is the second quadratic field, with respect to absolute value of discriminant, whose class number is divisible by $3$. However, that's not good enough this time for constructing the Hilbert class field of $\mathbb{Q}(\zeta_{31})$, since the latter has class group that is cyclic of order $9$. Moreover, the argument I gave in the question you link to for disjointness of the Hilbert class field of the quadratic and the cyclotomic field no longer works, because the degree of $\mathbb{Q}(\zeta_{31})$ is divisible by $3$.

First, let us get the second "obstacle" out of the way and give a better proof that the Hilbert class field $H$ of $F=\mathbb{Q}(\sqrt{-31})$ is disjoint from $K=\mathbb{Q}(\zeta_{31})$: just by thinking about how the Galois group of $\mathbb{Q}(\sqrt{-31})$ acts on elements, and hence on ideals, you will see that it acts by inversion on the class group of $\mathbb{Q}(\sqrt{-31})$. The way class field theory works, you can deduce that $H$ is Galois over $\mathbb{Q}$, the subgroup ${\rm Gal}(H/F)\cong \mathbb{Z}/3\mathbb{Z}$ is normal, and the quotient ${\rm Gal}(F/\mathbb{Q})$ acts on it by multiplication by $-1$, so that ${\rm Gal}(H/\mathbb{Q})$ is isomorphic to $S_3$. In particular, it is non-abelian, so $H$ cannot be contained in $K$, since its Galois group over $\mathbb{Q}$ is cyclic.

This tells you that the Hilbert class field of $F$, which one can compute to be obtainable by adjoining a root of $x^3+x+1$, gives you a piece of the Hilbert class field of $K$. But because this time the class group of $K$ is cyclic of order $9$, it does not give you everything.

Let $H_2$ denote the Hilbert class field of $K$. Before proceedings, let us think about the structure of $G={\rm Gal}(H_2/\mathbb{Q})$ (the fact that $H_2$ is Galois over $\mathbb{Q}$ again follows from general class field theory yoga). $G$ has a normal subgroup $N={\rm Gal}(H_2/K)$ that is cyclic of order $9$, and the quotient $G/N$ is cyclic of order $30$. First, I claim that this extension splits, so that $G$ is a semi-direct product $G\cong \mathbb{Z}/9\mathbb{Z}\rtimes \mathbb{Z}/30\mathbb{Z}$. In the case of $H/\mathbb{Q}$ we could see this simply by observing that the order of the normal subgroup was coprime to its index, and invoking Schur-Zassenhaus. In the current situation, that argument does not work, so instead we will exhibit a subgroup of $G$ that is cyclic of order $30$ and intersects $N$ trivially. There is a standard trick to this: take inertia at $31$. Let me call it $I$. It must be cyclic of order $30$, because $K$ is totally ramified at $31$ — we are using the fact that $\mathbb{Q}$ has no extensions that are unramified at all finite places — and it intersects $N$ trivially, since $H_2/K$ is everywhere unramified, while the extension cut out by $I$ is totally ramified at $31$.

Having established that $G = N\rtimes I$, we will have the complete structure of $G$ once we know how $I\cong G/N$ acts on $N$ by conjugation, or equivalently how ${\rm Gal}(K/\mathbb{Q})$ acts on the class group of $K$. The automorphism group of $N$ is cyclic of order $6$, and I claim that the image of $G/N$ in that automorphism group is everything. This will determine the whole group, since $G/N$, being cyclic of order $30$, has a unique quotient of order $6$. We already know that multiplication by $-1$ is in that image, because of what we said about the Galois group of $H$. Now, if $I\to {\rm Aut}N$ did factor through the quotient of order $2$, then the subgroup of $I$ that is cyclic of order $15$ would act trivially on $N$, and therefore would be normal in all of $G$. Moreover, it is contained (with index $2$) in the inertia subgroup $I$. It follows that its fixed field would be a Galois extension of $\mathbb{Q}$ that is unramified everywhere over $F$, and of degree $9$ over $F$. But wait, we said that $F$ only has class number $3$, not $9$, so this is impossible. It follows that $G/N\to {\rm Aut} N$ does not factor through a quotient of order $2$, but through a quotient of order $6$, i.e. is surjective.

This, finally, gives you a hint on how to find the Hilbert class field of $K$: applying all the same reasoning, we know in advance that the subgroup of $I$ that is cyclic of order $5$ (rather than $15$) will be normal in $G$, and that its fixed field will be an extension that is unramified of degree $9$ over the subfield of $F$ that is fixed by the subgroup of order $5$ inside ${\rm Gal}(F/\mathbb{Q})$. Now that you know in advance that this will succeed, you can fire up the computer and just wait for a few minutes: let $L$ be the subfield of $\mathbb{Q}(\zeta_{31})$ of degree $6$ over $\mathbb{Q}$. Magma will tell you that its class group is cyclic of order $9$ (we already knew this from our group theoretic considerations!) and will, after a few minutes, spit out a slightly horrendous looking polynomial of degree $9$ over $L$ whose root generates the Hilbert class field of $L$: $$ x^9 + \tfrac{1}{256}(351\alpha^4 + 22842\alpha^2 + 999)x^7 + \tfrac{1}{256}(-9585\alpha^4 + 33210\alpha^2 + 567)x^6 + \tfrac{1}{256}(56133\alpha^4 + 756702\alpha^2 + 26973)x^5 + \tfrac{1}{128}(-14096673\alpha^4 - 289073286\alpha^2 - 9985113)x^4 + \tfrac{1}{256}(837980397\alpha^4 + 2627921070\alpha^2 + 89938917)x^3 + \tfrac{1}{64}(-525358953\alpha^4 + 150497910738\alpha^2 + 5208850071)x^2 + \tfrac{1}{128}(500734949193\alpha^4 - 5434147475802\alpha^2 - 188657086959)x + \tfrac{1}{128}(329428602877167\alpha^4 + 3754393943660730\alpha^2 + 129532294910295), $$ where $\alpha\in L$ has minimal polynomial $$x^6 + 93x^4 + 899x^2 + 31$$ over $\mathbb{Q}$. Its compositum with $K$, obtained by adjoining to $K$ a root of the same polynomial, must then be the Hilbert class field of $K$.

Edit: Franz Lemmermeyer has found a much nicer polynomial that generates the same field over $L$, and therefore the same field over $K$: $$ x^9 - x^7 - 2x^6 + 3x^5 + x^4 + 2x^3 - x^2 + x - 3. $$

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  • $\begingroup$ That's impressive. Did you try to reduce the defining equation? One could use 'rnfpolredabs' or 'rnfpolredbest' in Pari/GP (there are probably similar commands in Sage or Magma). $\endgroup$ – François Brunault Jan 2 at 7:26
  • $\begingroup$ @FrançoisBrunault: thank you! I did not try to reduce the equation. If someone does, they should feel free to edit the answer. $\endgroup$ – Alex B. Jan 2 at 11:50
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    $\begingroup$ Can Magma verify that this is the same field as the one you get by adjoining a root of $x^9 − x^7 − 2x^6 + 3x^5 + x^4 + 2x^3 − x^2 + x − 3$? $\endgroup$ – Franz Lemmermeyer Jan 2 at 17:23
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    $\begingroup$ @AlexB. How we say the class group of K is cyclic? $\endgroup$ – SUNIL PASUPULATI Jan 3 at 5:21
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    $\begingroup$ @SunilPasupulati: Magma told me so (I did the computation conditional on GRH). However, the argument does not really depend on that. If the group was $C_3\times C_3$, the entire argument would still work: the extension would be split, and the subgroup of order $5$ in $I$ would still act trivially, so that you could find the Hilbert class field by looking at the sextic subfield of $\mathbb{Q}(\zeta_{31})$. $\endgroup$ – Alex B. Jan 3 at 13:06

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