# How to compute Hilbert class field of $\Bbb Q(\zeta_{31})$?

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.

## 1 Answer

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 $$K$$: $$x^9 - x^7 - 2x^6 + 3x^5 + x^4 + 2x^3 - x^2 + x - 3.$$

• 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). – François Brunault Jan 2 at 7:26
• @FrançoisBrunault: thank you! I did not try to reduce the equation. If someone does, they should feel free to edit the answer. – Alex B. Jan 2 at 11:50
• Can Magma verify that this is the same field as the one you get by adjoining a root of $x^9 − x^7 − 2x^6 + 3x^5 + x^4 + 2x^3 − x^2 + x − 3$? – Franz Lemmermeyer Jan 2 at 17:23
• @AlexB. How we say the class group of K is cyclic? – SUNIL PASUPULATI Jan 3 at 5:21
• @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 $\mathbb{Q}(\zeta_{31})$. – Alex B. Jan 3 at 13:06