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I know that $\mathbb{Q}(\zeta_{23})$ has class number 3, and I am wondering how I can find ideal representatives of the two nonprincipal classes in the class group. I have tried looking at examples but it just seems like the answer is pulled out of a hat. Help would be appreciated.

user165580
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  • In fact Kummer was the first proving that $\mathbb{Z}[\zeta_{23}]$ is not an UFD, and his proof was based exactly on finding some strange elements with a lot of trial and error. – Crostul Jul 21 '14 at 18:20

2 Answers2

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I don't know what the best algorithm is, but for sure in principle you can use the following theorem (whose proof you can find in Marcus' book "Number Fields"):

Let $K$ be a number field with number ring $\mathcal O_K$. Let $\{\alpha_1,\dots,\alpha_n\}$ be an integral basis of $\mathcal O_K$ and set $\lambda= \prod_{i=1}^n\sum_{j=1}^n|\sigma_i(\alpha_j)|$ where the $\sigma_i$ are the embeddings of $K$ into $\mathbb C$. Then every ideal class of the ideal class group has a representative with norm $\leq \lambda$.

This reduces your search to decomposing the ideals $p\mathcal O_K$ for primes $p\leq \lambda$.

Ferra
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Let $K = \mathbb Q(\sqrt{-23})$ and $L = \mathbb Q(\zeta_{23}).$ Then taking the norm from $L$ to $K$ induces an isomorphism from the class group of $L$ to the class group of $K$.

The class group of $K$ is generated by the ideal $\mathfrak p := (2,(1+\sqrt{-23})/2)$. Now you can easily check that $\mathfrak p$ is inert in $L$, and if we let $\mathfrak P$ denote the prime lying over it (which is simply the ideal $(2,(1 +\sqrt{-23})/2),$ but now understood in $\mathcal O_L$) then $N_{L/K}(\mathfrak P) = \mathfrak p^{11},$ which is equivalent to $\overline{\mathfrak p}$ in $Cl_K$ (since this group has order $p$) (and in particular is again a generator).

So $(2,(1+\sqrt{-23})/2)$ generates the ideal class group of $L$.

user160609
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