Two algebraic number fields are said to be arithmetically equivalent if they share the same Dedekind zeta function. If this is the case, they must have certain invariants in common among which is the product $\,$ $hR$ $\,$ where $h$ is the class number and $\,$$R$$\,$ is the regulator.

Questions: 1) Let K and L be non-isomorphic but arithmetically equivalent number fields. On the face of it, the equality $\,$ $h(K)R(K)$ = $h(L)R(L)$ $\,$seems surprising. Does this really give an unexpected rational relation of dependence among the various logarithms involved or is there a simpler explanation if one passes to a suitable common extension field of K and L and expresses everything in terms of the information there?

2) Are there any examples of near misses where one zeta function is almost but not quite equal to another? $\,$ That is, if two Dedekind zeta functions (expanded as Dirichlet series) agree for all but finitely many terms, must they be identical?

Thanks very much.

  • 2
    $\begingroup$ For (2), the multiplicity one theorem for automorphic representations states that if two automorphic representations are isomorphic at almost every place, then they are globally isomorphic. In particular, this holds for the zeta function. $\endgroup$ Jul 16, 2016 at 1:57
  • 2
    $\begingroup$ In fact, (2) was used in class field theory back when the complete definition of Artin's L-functions at the ramified primes was not yet known. $\endgroup$ Jul 16, 2016 at 14:29

2 Answers 2


(1) As you suspected, there is an algebraic explanation after passing to a suitable extension, at least to the fact that $R(K)/R(L)\in\mathbb{Q}^\times$.

By Chebotarev, $K$ and $L$ have the same Galois closure, say $M$. Let $G$ be the Galois group of $M/\mathbb{Q}$. Then we have $K = M^H$ and $L = M^J$ for some subgroups $H,J\subset G$. The Dedekind zeta functions can be interpreted as Artin L-functions of the corresponding permutation representations: $$ \zeta_K(s) = L(\mathbb{C}[G/H],s)\text{ and }\zeta_L(s) = L(\mathbb{C}[G/J],s). $$ Since these zeta functions are equal, the corresponding Artin representations are isomorphic: $\mathbb{C}[G/H]\cong\mathbb{C}[G/J]$, and since permutation representations are defined over $\mathbb{Q}$, we even have $\mathbb{Q}[G/H]\cong\mathbb{Q}[G/J]$ (but by your assumption that $K$ and $L$ are not isomorphic, i.e $H$ and $L$ are not conjugate, i.e. $G/H$ and $G/J$ are not isomorphic as $G$-sets).

Now the unit group $U(M)$ is acted upon by $G$, and by Galois descent we have $U(K) = U(M)^H$ and $U(L) = U(M)^J$. The group $U(M)^H$ is the union of the images of the homomorphisms in $Hom_H(\mathbb{Z},U(M))$, which is isomorphic to $Hom_G(\mathbb{Z}[G/H],U(M))$, and similarly for $L$ and $J$. Since $\mathbb{Q}[G/H]\cong\mathbb{Q}[G/J]$, we get that there exists a $\mathbb{Q}$-vector space isomorphism $\mathbb{Q}\otimes_\mathbb{Z}U(K) \to \mathbb{Q}\otimes_\mathbb{Z}U(L)$, in other words a group homomorphism $U(K)\to U(L)$ with finite kernel and cokernel. Checking that this is compatible with the complex embeddings, you obtain that the quotient of their regulators is rational.

You can find more details on this type of relations in On Brauer-Kuroda type relations of S-class numbers in dihedral extensions by A. Bartel.

(2) By Chebotarev, if the Dedekind zeta functions agree on a density $1$ subset of primes, then they are equal.


(2) it follows from Haseo Ki's proof of the strong multiplicity one theorem for the Selberg class, whose Dedekind Zeta functions are elements.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.