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Let $ \omega=e^{2 \pi/3} $ be a primitive third root of unity. And let $ \phi= \frac{1+\sqrt{5}}{2} $ be the golden ratio.

The ring of integers $ \mathbb{Z}[\omega] $, called the Eisenstein integers, has class number 1 (it is a Euclidean domain and thus a PID). The ring of integers $ \mathbb{Z}[\phi] $ also has class number $ 1 $ (again is a Euclidean domain and thus a PID).

Does the ring $ \mathbb{Z}[\omega,\phi] $ have class number $ 1 $? I'm guessing not but I'm not sure.

Context and motivation: I'm interested in the subgroup of $ SU(3) $ isomorphic to $ 3.A_6 $ https://math.stackexchange.com/a/4535587/724711 its character values generate the ring $ \mathbb{Z}[\omega,\phi] $. Given this result https://groupprops.subwiki.org/wiki/Linear_representation_is_realizable_over_principal_ideal_domain_iff_it_is_realizable_over_field_of_fractions it would be interesting to know if this degree $ 3 $ character of $ 3.A_6 $ is defined over a PID or not, equivalently to know if $ \mathbb{Z}[\omega,\phi] $ has class number $ 1 $.

Ian Gershon Teixeira
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1 Answers1

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The LMFDB says yes: https://www.lmfdb.org/NumberField/4.0.225.1 .

If you wanted to prove it yourself, the strategy (as always) is to:

  1. compute the discriminant and the Minkowski bound,

  2. check whether primes with norm lower than the Minkowski bound are principal.

Edit: as pointed out by Jyrki Lahtonen, we do need to check that $\mathbb{Z}[\omega,\phi]$ is the ring of integers of its fraction field!

This can likely be done “by hand”, but here is a direct way.

We need to show that $\mathbb{Z}[\omega] \otimes \mathbb{Z}[\phi]$ is normal. Now, $i: \mathbb{Z}[\omega] \rightarrow \mathbb{Z}[\omega] \otimes \mathbb{Z}[\phi]$ is a base change of $\mathbb{Z} \rightarrow \mathbb{Z}[\phi]$, which is étale after inverting $5$, so that $i$ is étale after inverting $5$. Since its source is normal, $\mathbb{Z}[\omega] \otimes \mathbb{Z}[\phi][1/5]$ is normal.

Similarly, $\mathbb{Z}[\phi] \rightarrow \mathbb{Z}[\omega,1/3] \otimes \mathbb{Z}[\phi]$ is étale and its domain is normal, so $\mathbb{Z}[\omega] \otimes \mathbb{Z}[\phi][1/3]$ is normal.

Therefore $\mathbb{Z}[\omega] \otimes \mathbb{Z}[\phi]$ is normal.

Aphelli
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  • It may be worth adding that the ring OP described is the full ring of integers of $\Bbb{Q}(\omega,\sqrt5)$. So figuring out the class number is all we need :-) – Jyrki Lahtonen Aug 25 '24 at 07:29
  • @JyrkiLahtonen: indeed (although I don’t think that’s entirely trivial to check and we need to spell it out)! – Aphelli Aug 25 '24 at 11:39
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    When $K$ and $L$ are number fields with relatively prime discriminants, $\mathcal O_{KL} = \mathcal O_K\mathcal O_L$. The example $K = \mathbf Q(\omega)$ and $L = \mathbf Q(\sqrt{5})$ fits this, with discriminants $-3$ and $5$. – KCd Aug 27 '24 at 13:27
  • @KCd: indeed, the argument I give extends to this general situation! – Aphelli Aug 27 '24 at 14:50