noncyclic class group example of a cubic number field

Question

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 $\mathbb{Q}(\sqrt{223}), \mathbb{Q}(\sqrt{226}), \mathbb{Q}(\sqrt{-30}), \mathbb{Q}(\sqrt{-89})$.

Answer 1

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.

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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 $$ <9,3\alpha-3,\alpha^2-2\alpha+1> = <\alpha+2> $$

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

Answer 2

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.