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 $\mathbb{Q}(\sqrt{223}), \mathbb{Q}(\sqrt{226}), \mathbb{Q}(\sqrt{-30}), \mathbb{Q}(\sqrt{-89})$.
Noncyclic class group example of a cubic number field
algebraic-number-theoryideal-class-group
Related Solutions
I don't know if this is helpful, but for $E = \mathbb{Q}(\sqrt{-6})$ case....
You know that $H/E$ is quadratic, and thus $H = E(\sqrt{D})$ for some $D \in E$. $H/E$ is an unramified extension, however all number fields are ramified over $\mathbb{Q}$, so clearly $D$ must factor into ramified primes of $E$.
I looked over $2$ first. I don't remember why, but I convinced myself at one time that for a quadratic extension to be unramified, you need to find a $D$ whose prime factorization is square. (hopefully this is fresh in your mind so you'll be able to work out why!) So I observe
$$(2, \sqrt{-6})^2 = (2)$$
However, $2$ is clearly (?) not a square. Thus $E(\sqrt{2})$ is an unramified extension.
I also vaguely recall a general theorem that the Hilbert class field of $\mathbb{Q}(\sqrt{m})$ consists of adjoining square roots of the prime factors of $m$, with appropriate signs.
To complement mathmos's nice answer (Ariel is that you?):
It turns out several examples of such fields have been constructed:
In Yamamura, Ken. “On Unramified Galois Extensions of Real Quadratic Number Fields.” Osaka Journal of Mathematics 23, no. 2 (1986): 471–478 the author constructs infinitely many examples of real quadratic fields with (strictly) unramified extensions with Galois group $A_5$. He also finds examples where the real quadratic has class number one.
Yamamura gives the example $\mathbf Q(\sqrt{36497})$ which we see has class number one in http://www.lmfdb.org/NumberField/2.2.36497.1, moreover $f(x) = x^{5} \mathstrut -\mathstrut 2 x^{4} \mathstrut -\mathstrut 3 x^{3} \mathstrut +\mathstrut 5 x^{2} \mathstrut +\mathstrut x \mathstrut -\mathstrut 1 $ has Galois group $S_5$, all roots real and discriminant $36497$. So if $K$ is the splitting field of $f$ we have
$$K/\mathbf Q(\sqrt{36497})$$
and the author shows that this extension is unramified at all finite primes (but we have real roots so this is everywhere unramified) so this is the sort of extension you were looking for, with non-abelian Galois group $A_5$.
In Brink, David. “Remark on Infinite Unramified Extensions of Number Fields with Class Number One.” Journal of Number Theory 130, no. 2 (February 1, 2010): 304–6. https://doi.org/10.1016/j.jnt.2009.08.013. You can find a $\operatorname{PSL}_2(\mathbf F_7)$ example over a real biquadratic field.
He also expands on the work of Yamamura and shows $\mathbf{Q}(\sqrt{36497}, \sqrt{2819 \cdot 103})$ is class number one and has an infinite unramified extension!
Best Answer
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 $$ \begin{align} I &:= <3,\alpha>^2 = <9,3\alpha,\alpha^2>\\ J &:= <3,\alpha-1>^2 = <9,3\alpha-3,\alpha^2-2\alpha+1> \end{align} $$
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 $\{1,\alpha,\alpha^2/2\}$.