# Number fields with same zeta function?

Given a zeta function $\zeta_K$ of some number field $K$ how much information will this give us about $K$? Specifically, if two number fields have the same zeta function, what shared properties are they known to have? Is there a way to construct distinct number fields that have the same zeta function?

All constructions of pairs of arithmetically equivalent number fields arise in the following way: start with a Galois extension $F/M$ with Galois group $G$, let $H$, $H'$ be two subgroups that give rise to isomorphic permutation representations $\mathbb{C}[G/H]\cong \mathbb{C}[G/H']$. Then, the fixed fields $K=F^H$ and $K'=F^{H'}$ will have the same zeta function. If $H$ and $H'$ are conjugate, then the fields $K$ and $K'$ are in fact isomorphic, so they share almost all the interesting properties. But otherwise you get non-isomorphic fields.

They will always have the same number of real and complex embeddings, the same discriminant and the same number of roots of unity. Also, the product $h(K)R(K)$ will be the same, where $h$ is the class number and $R$ is the regulator. However each of the terms by itself need not be the same, as shown in numerous examples by Bart de Smit. As far as I know, it is still an open problem whether the $p$-part of the class numbers can differ for arbitrary $p$. There is no reason whatsoever to doubt that it can, and de Smit has proposed a general construction (i.e. suitable $G$, $H$, $H'$) that should work for any $p$, and that is in fact the smallest group that has any hope of producing arithmetically equivalent fields with different $h_p$, but it has not been proven that it always does. For small $p$ the proof goes by producing lots of Galois extensions with a suitable $G$, using a computer algebra package, until one finds one that happens to give $K$ and $K'$ with different $p$-parts of class numbers.

In a similar direction as above, the torsion of the odd-numbered $K$-groups of the rings of integers is always the same for arithmetically equivalent fields. Also, the quotient $$\frac{|K_{2n}(\mathcal{O}_K)|\cdot R_n(\mathcal{O}_K)}{|K_{2n}(\mathcal{O}_{K'})|\cdot R_n(\mathcal{O}_{K'})}$$ is a power of 2 (probably trivial, but this is not known), where $R_n$ is the higher Borel regulator. Again, there is no reason to expect the single terms to be equal, but to give concrete examples is probably computationally out of reach at the moment.

• To make the first paragraph even more concrete, the condition imposed on $H$ and $H'$ is equivalent to saying for every conjugacy class $C$ in $G$ that $H \cap C$ and $H' \cap C$ have the same size. – KConrad Nov 9 '11 at 1:16
• I understand the idea in the first paragraph can also be used to write down examples of non-isometric compact Riemannian manifolds which are isospectral (which I think is equivalent to having the same Selberg zeta function...?). – Qiaochu Yuan Nov 9 '11 at 14:04
• Dear Qiaochu, yes, I think that's correct, although I don't know the details. – Alex B. Nov 9 '11 at 15:01
• Qiaochu: Yes, the construction is exactly the same. Look in this paper of Sunada: jstor.org/stable/1971195?seq=1 – Ben Webster Nov 9 '11 at 20:58

Coming very late to the party, here is a small complement to Alex's excellent answer. There is a recent paper of Marcolli and Cornelissen (arXiv link) which among other things discusses this question. The following two points give partial answers to the question posed here:

1. If two number fields are Galois over $\mathbb{Q}$ and have the same zeta function, then they are isomorphic.

2. In general, one can say something similar if one is willing to consider all twists of the zeta function by Dirichlet characters. More precisely, assume that there is an isomorphism between the Pontryagin duals of the abelianized Galois groups of the two number fields. Assume also that whenever two characters are identified under this isomorphism, the corresponding twisted zeta functions agree. Then the two number fields are isomorphic. See Theorem 2 in the paper linked above for more details.

The introduction of the paper actually gives a rather good overview over criteria which can or cannot determine whether two number fields are isomorphic.

• Hi Andreas, greetings from Korea! Just a small note: your point 1. follows immediately from my first paragraph together with Keith's characterisation in his comment to my answer. Indeed, let $H$ and $H'$ correspond to the two fields. Since the fields are Galois over $\mathbb{Q}$, $H$ and $H'$ are normal in $G$. But then they are both unions of conjugacy classes of $G$. If in addition you know that the intersections with all conjugacy classes have the same size, then it immediately follows that $H=H'$. – Alex B. Dec 30 '11 at 16:55