# noncyclic class group example of a cubic number field

I'm trying to compute ideal class groups of various number fields, and now I'm a little familiar to a class group of a quadratic number field. However, I can't find any non-cyclic example of a class group of a cubic number field. In Marcus' "Number Fields" book, there are some exercises that deal with $$\mathbb{Q}(\sqrt[3]{m})$$ for integer $$m$$, but every such exercise has a cyclic class group. Is there any good example of a cubic number field that has a class group isomorphic to the Klein 4-group? How about biquadratic or quartic fields? (I think I can't do with quintic things...) Thanks in advance. Until now, I computed class groups of $$\mathbb{Q}(\sqrt{223}), \mathbb{Q}(\sqrt{226}), \mathbb{Q}(\sqrt{-30}), \mathbb{Q}(\sqrt{-89})$$.

• One side note, class groups which are elementary abelian group is a rare occurrence. For example, it is known that no imaginary quadratic field has class group $C_3\times C_3\times C_3$. This of course, does not say anything on cubic case, but at least give an idea on the situation. – pisco Oct 13 '18 at 10:18
• @pisco That's very interesting! Do you have any reference for it? Also, is there any single known finite abelian group that doesn't occur as a class group of a number field? – Seewoo Lee Oct 13 '18 at 18:44
• See here mathoverflow.net/questions/132838. Regarding your second question, I think it is still open. – pisco Oct 14 '18 at 6:15
• @pisco Oh I see... it is just brute force... – Seewoo Lee Oct 14 '18 at 18:03

This following example was found via a computer search for simple integral basis and small discriminant, so that the discriminant is easy to compute and the Minkowski bound is small.

I am not sure if this suffices as a good example though: The only method I know of computing class group is the basic one via Minkowski's bound and this example still involved quite a bit computation since the bound is fairly large at $$<38$$. I wasn't able to find a cubic extension, $$\mathbb Z_2\times \mathbb Z_2$$ class group with small bounds.

The stats was also checked on Sagemathcell to be sure.

Let $$f(x) = x^3 + 11x+21\in\mathbb Z[x]$$.

1. $$f(x)$$ is irreducible over $$\mathbb Z$$.
2. Let $$\alpha \in \mathbb C$$ be a root of $$f(x)$$ and consider the number field $$K=\mathbb Q(\alpha)$$.
We can show that $$\{1,\alpha,\alpha^2\}$$ is an integral basis and $$K$$ has prime discriminant $$-17231$$.
3. Hence Minkowski bound $$M_K=\frac{8}{9\pi} \sqrt{17231} < 38,$$ and we can find the list of non-principal ideals for each prime $$\leq 37$$.
4. Finally we can show that the class group $$H(K)$$ of $$K$$ is $$H(K)\cong \mathbb Z_2\times \mathbb Z_2$$, with generators $$<3,\alpha>,<3,\alpha-1>$$

Edit 1: We check that $$<3,\alpha>,<3,\alpha-1>$$ has order 2. Let \begin{align} I &:= <3,\alpha>^2 = <9,3\alpha,\alpha^2>\\ J &:= <3,\alpha-1>^2 = <9,3\alpha-3,\alpha^2-2\alpha+1> \end{align}

We claim that \begin{align} <9,3\alpha,\alpha^2> &= <2\alpha^2 - 3\alpha + 27>\\ <9,3\alpha-3,\alpha^2-\alpha+1> &= <\alpha+2> \end{align}

Clearly, we have $$2\alpha^2 - 3\alpha + 27 \in <9,3\alpha,\alpha^2> \implies <2\alpha^2-3\alpha+27> \subseteq <9,3\alpha,\alpha^2>$$ We obtain \begin{align} -(\alpha^2+3\alpha+2)(2\alpha^2 - 3\alpha + 27) &= 9\\ (\alpha^2 + 6\alpha + 7)(2\alpha^2 - 3\alpha + 27) &= \alpha^2 \end{align} Therefore $$9,\alpha^2\in <2\alpha^2 - 3\alpha+27>$$, which in turn gives $$3\alpha\in <2\alpha^2 - 3\alpha + 27>$$. Hence \begin{align} <9, 3\alpha,\alpha^2> &\subseteq <2\alpha^2 - 3\alpha + 27>\\ \implies <9, 3\alpha,\alpha^2> &= <2\alpha^2 - 3\alpha + 27> \end{align} This shows the first equivalence. On the other hand, \begin{align} (\alpha^2-2\alpha+15)(\alpha+2) &= 9\\ 3(\alpha+2)-9 &= 3\alpha-3\\ (\alpha+2)^2-2(3\alpha-3)-(9) &= \alpha^2-2\alpha+1 \end{align} Therefore $$9,3\alpha-3,\alpha^2-2\alpha+1\in <\alpha+2>$$. For the reverse containment, \begin{align} (\alpha + 2)(\alpha^2 - 2 \alpha + 1) + 5 (3 \alpha - 3) + 4 (9) &= \alpha+2 \end{align} shows that $$\alpha+2 \in <9,3\alpha-3,\alpha^2 - 2\alpha+1>$$. Therefore $$<9,3\alpha-3,\alpha^2-2\alpha+1> = <\alpha+2>$$

Another example might be $$f(x) = x^3+8x+60$$ with integral basis $$\{1,\alpha,\alpha^2/2\}$$.

• Can you add an explanation for the claim that the squares of these ideals are principal? I tried to work it out by hand, but had problems spotting a generator. The numbe $\alpha+1$ has norm nine, but didn't seem to fit. – Jyrki Lahtonen Oct 11 '18 at 5:14
• @JyrkiLahtonen Given the generators, derived from Sage, I have listed the computations to check that they equal the square of the ideals. However I have not found an easy way to find the generators starting from the square of ideals. – Yong Hao Ng Oct 11 '18 at 7:24
• Thank you! ${}{}$ – Jyrki Lahtonen Oct 11 '18 at 7:29
• Thank you for this example! I'll try to do it myself. – Seewoo Lee Oct 11 '18 at 20:36
• I just tried this myself and thank you for this example again. How did you show that $(3, \alpha)$ and $(3, \alpha-1)$ are not principal? I found a fundamental unit with SAGE (and use Artin's inequality to prove that a unit is indeed a fundamental unit) and proved that the above ideals are not principal using these. Is there any easier way to do this? – Seewoo Lee Oct 16 '18 at 2:10

Using a math engine, like PARI/GP one can check the class group of the polynomials of the form $$x^3-a$$. For $$a>0$$ we have that the first polynomial with root $$\alpha$$ for which $$\mathbb{Q}(\alpha)$$ has non-cyclic class group is $$65$$. Indeed we have that the class group of $$x^3-65$$ is $$\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/6\mathbb{Z}$$.

On the other side $$a=113$$ is the smallest $$a$$ s.t. $$\mathbb{Q}(a)$$ has the Klein 4-group as a class group, where $$\alpha=\sqrt[3]{113}$$. In between the $$a'$$s producing non-cyclic class groups are $$70, 86, 91, 110$$ and for all o them $$\mathbb{Q}(\sqrt[3]{a})$$ has $$\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$$ as a class group.

However considering the size of the discriminant and having to compute at least $$20$$ some ideals (let alone combine them later) by using Minkowski's Bound it will be too tedious to do it by hand.

• Thanks! Actually, I want this kind of example, but it is much bigger than I think... – Seewoo Lee Oct 11 '18 at 20:37

Here is a detailed computation of class group of $$\mathbb{Q}(\alpha)$$, with $$\alpha^3+11\alpha +21=0$$, the example given by @Yong Hao Ng. I want to illustrate the following two maxims of computational algebraic number theory:

• Computation of ideal class group is intractable
• Knowledge of units and class group often complement each other

Denote $$K=\mathbb{Q}(\alpha)$$, the discriminant is $$-17231$$, so an integral basis is $$\{1,\alpha,\alpha^2\}$$.

## 1.Finding generators and relations

This is usually the first step (for general number field) in computing class group in almost any algorithm: factor a large number of ideals with small norm, yielding a linear system, then pinpoint possible generators.

First we factors primes smaller than the Minkowski bound, primes not shown below are inert. \begin{aligned}(3)&=(3,\alpha)(3,\alpha-1)(3,\alpha+1) \\ (7)&=(7,\alpha)(7,\alpha^2-3)\\ (11)&=(11,\alpha-1)(11,\alpha^2+\alpha+1)\\ (13)&=(13,\alpha+3)(13,\alpha^2-3\alpha-6)\\ (17)&=(17,\alpha-2)(17,\alpha^2+2\alpha-2)\\ (19)&=(19,\alpha+7)(19,\alpha^2-7\alpha+3)\\ (23)&=(23,\alpha-8)(23,\alpha^2+8\alpha+6)\\ (29)&=(29,\alpha+4)(29,\alpha+6)(29,\alpha-10)\\ \end{aligned} Let $$\mathfrak{p}_1=(3,\alpha),\mathfrak{p}_2=(3,\alpha-1),\cdots,\mathfrak{p}_{17}=(29,\alpha+6),\mathfrak{p}_{18}=(29,\alpha-10)$$ which are named in their order of appearance. Now we find elements with small norm: $$N(\alpha) = -3\times 7 \qquad N(\alpha+1)=-3^2 \qquad N(\alpha-1)=-3\times 11 \qquad N(\alpha+3)=-3\times 13$$ $$N(\alpha+4)=3\times 29 \qquad N(\alpha^2+\alpha-2)=-3^3\times 11 \qquad N(\alpha^2-\alpha-2)=3^3\times 17$$ $$N(\alpha^2-\alpha+1)=3\times 13\times 19 \qquad N(\alpha^2-2\alpha-2)=3\times 13\times 23 \qquad N(3\alpha-1)=-23\times 29$$ $$N(4\alpha-1)=-3^2\times 13^2 \qquad N(4\alpha+7)=3\times 7\times 11 \qquad N(4\alpha+9)=3\times 17\times 19$$

Now we factor then into $$\mathfrak{p}_n$$. For example, to factor $$(\alpha)$$, we have three possibilities: $$(\alpha) = \mathfrak{p}_1\mathfrak{p}_4 \qquad \mathfrak{p}_2\mathfrak{p}_4 \qquad \mathfrak{p}_3\mathfrak{p}_4$$ I rule out the latter two. If $$(\alpha) = \mathfrak{p}_2\mathfrak{p}_4 = (3,\alpha+2)(7,\alpha-7)$$, then $$(\alpha+2)(\alpha-7)/\alpha \in \mathcal{O}_K$$, but this number is not integral. If $$(\alpha) = \mathfrak{p}_3 \mathfrak{p}_4 = (3,\alpha-2)(7,\alpha-7)$$, then $$(\alpha-2)(\alpha-7)/\alpha \in \mathcal{O}_K$$, this is neither integral as well. Hence $$(\alpha) = \mathfrak{p}_1\mathfrak{p}_4$$. Similarily, we can factor other ideals, giving $$(\alpha) = \mathfrak{p}_1\mathfrak{p}_4 \qquad (\alpha+1) = \mathfrak{p}_3^2 \qquad (\alpha-1) = \mathfrak{p}_2 \mathfrak{p}_6 \qquad (\alpha+3) = \mathfrak{p}_1 \mathfrak{p}_8$$ $$(\alpha+4) = \mathfrak{p}_3 \mathfrak{p}_{16} \qquad (\alpha^2+\alpha-2) = \mathfrak{p}_2^3 \mathfrak{p}_6 \qquad (\alpha^2-\alpha-2) = \mathfrak{p}_3^3 \mathfrak{p}_{10}$$ $$(\alpha^2-\alpha+1) = \mathfrak{p}_3 \mathfrak{p}_8 \mathfrak{p}_{12} \qquad (\alpha^2-2\alpha-2) = \mathfrak{p}_2 \mathfrak{p}_8 \mathfrak{p}_{14} \qquad (3\alpha-1) = \mathfrak{p}_{14} \mathfrak{p}_{18}$$ $$(4\alpha-1) = \mathfrak{p}_2^2 \mathfrak{p}_8^2 \qquad (4\alpha+7) = \mathfrak{p}_3 \mathfrak{p}_4 \mathfrak{p}_6 \qquad (4\alpha+9) = \mathfrak{p}_1 \mathfrak{p}_{10} \mathfrak{p}_{12}$$

We now have $$21$$ relations, and $$18$$ 'variables', hence forming the corresponding $$21\times 18$$ matrix $$A$$, the rank is $$18$$, apply Smith normal form on it, gives $$\begin{pmatrix} 1 & & & & & & & & & & & & & & & & 1 & 1 \\ & 1 & & & & & & & & & & & & & & & & \\ & & 1 & & & & & & & & & & & & & & 1 & 1 \\ & & & 1 & & & & & & & & & & & & & 1 & 1 \\ & & & & 1 & & & & & & & & & & & & 1 & 1 \\ & & & & & 1 & & & & & & & & & & & 1 & \\ & & & & & & 1 & & & & & & & & & & 1 & \\ & & & & & & & 1 & & & & & & & & & 1 & 1 \\ & & & & & & & & 1 & & & & & & & & 1 & 1 \\ & & & & & & & & & 1 & & & & & & & & 1 \\ & & & & & & & & & & 1 & & & & & & & 1 \\ & & & & & & & & & & & 1 & & & & & 1 & \\ & & & & & & & & & & & & 1 & & & & 1 & \\ & & & & & & & & & & & & & 1 & & & & 1 \\ & & & & & & & & & & & & & & 1 & & & 1 \\ & & & & & & & & & & & & & & & 1 & 1 & 1 \\ & & & & & & & & & & & & & & & & 2 & \\ & & & & & & & & & & & & & & & & & 2 \end{pmatrix} \vec{v} = 0$$ where $$\vec{v}$$ is the column matrix form by $$\mathfrak{p}_{18},\mathfrak{p}_{17},\cdots,\mathfrak{p}_1$$. Form this, we see that the ideal class group is generated by $$\mathfrak{p}_1, \mathfrak{p}_2$$, and $$\mathfrak{p}_1^2, \mathfrak{p}_2^2$$ are principal. Also note that $$\mathfrak{p}_{17}$$ is principal, a fact highly nonobvious. However, we cannot yet conclude the class group is $$C_2\times C_2$$. To show this, we have to check:

• $$\mathfrak{p}_1, \mathfrak{p}_2$$ are not principal.
• $$\mathfrak{p}_1, \mathfrak{p}_2$$ belongs to genuinely different ideal classes.

Both bullets are non-trivial to prove. Below I present a way which tell us with 'high confidence' the class number is $$4$$, hence $$C_2\times C_2$$ must be the class group. I do have an ad hoc (i.e. not the same as general class group algorithm) idea of rigorous establishing this, but the procedure takes much longer than the one presented below. Both the informal and rigorous method require knowledge of the fundamental unit.

## 2. Fundamental units

Now we find a fundamental unit of $$K$$. For general number field, a system of independent units can be found by computing the (left) nullspace of above coefficient matrix $$A$$. In our example, $$\vec{u}=\begin{pmatrix} 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 2 & -2 & 0 & -2 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ \end{pmatrix}$$ is a left null-vector of $$A$$ (i.e. $$\vec{u} A$$ is the zero vector), this says, in terms of ideals $$(3)^2 (\alpha+1)^{-1} (\alpha-1)^2 (\alpha+3)^{-2} (\alpha^2+\alpha-2)^{-2} (4\alpha-1) = (1)$$ Hence $$v = \frac{{9{{(\alpha - 1)}^2}(4\alpha - 1)}}{{(\alpha + 1){{(\alpha + 3)}^2}{{({\alpha ^2} + \alpha - 2)}^2}}} = 215-25\alpha +16\alpha^2$$ is a nontrivial unit. After embedding $$K$$ into $$\mathbb{R}$$ via $$\alpha \to -1.56238$$, we have $$v = 293.116$$. We show $$v$$ is fundamental, for cubic field with only one real embedding, this can be done via

Let $$u>1$$ be the fundamental unit of a real cubic field with one real embedding, then $$u^3 > \frac{|\delta|-27}{4}$$ where $$\delta$$ is the discriminant of the field.

Applying this we have $$u>16.2626$$, thus we have to show $$\sqrt{v} = 17.1206$$ is not in $$K$$. There is cheap way to check this: note that $$101\mid (6^3+11\times 6+21)$$, we have a homomorphism (such map always exists for monogenic number field): $$\mathcal{O}_K \to \mathbb{F}_{101}: \alpha \mapsto 6$$ then $$v$$ sends to $$215-25\times 6+16\times 6^2$$, which can be checked is not a quadratic residue modulo $$101$$. Now we can easily compute the regulator $$R$$, which is $$5.68057$$.

## 3. The class number

The analytic class number formula says $$\frac{hR}{w} = 2^{-r_1} (2\pi)^{-r_2} \sqrt{|\delta|} \prod_p \frac{1-1/p}{\prod_{\mathfrak{p}|p} (1-1/N(\mathfrak{p}))}$$

Now we approximate the product by taking finitely many terms: $$h \approx \frac{R}{2\pi} \sqrt{17231} \prod_{p\leq 29} \frac{1-1/p}{\prod_{\mathfrak{p}|p} (1-1/N(\mathfrak{p}))} = 4.27026$$ If we count first $$100$$ primes, then $$h\approx 4.10765$$. So we have sufficiently confidence that $$h=4$$. This shows the class group of $$K$$ is $$C_2 \times C_2$$.