Arithmetic Equivalence

We discuss some basic facts about arithmetic equivalence of number fields.

Our exposition largely follows Andrew Sutherland. We take the following (albeit nonstandard) definition.

Definition 1. Fix number fields \(K\) and \(K’\). Then \(K\) and \(K’\) are arithmetically equivalent if and only if, for all but finitely many rational primes \(p\), the factorization types of \(p\mathcal O_K\) and \(p\mathcal O_{K’}\) are the same. We may write this as \(K\sim K’\).

Our main goal will be to present the following permanence properties.

Theorem 2. The following quantities are preserved by arithmetic equivalence.

1. The Dedekind \(\zeta\)-function.
2. The set of unramified primes and the factorization types for all unramified primes.
3. The number of real embeddings and the number of complex embeddings.
4. The degree.
5. The discriminant.
6. The normal closure.
7. The normal core, which is the largest subfield Galois over \(\mathbb Q\).
8. The isomorphism class of the group of units.

The Dedekind \(\zeta\)-Function

The most difficult of the equivalences in Theorem 2 will be the Dedekind \(\zeta\)-function.

Remark 3. For a sense of scale, we note that \(K\sim K’\) already implies that \(\zeta_K\) and \(\zeta_{K’}\) are the same up to finitely many Euler factors. Indeed, if \(p\) is a rational prime, then the Euler factor of \(\zeta_K\) above \(p\) is given by \[\prod_{\mathfrak p\mid p\mathcal O_K}\frac1{1-\operatorname N(\mathfrak p)^{-s}},\] so this Euler factor is entirely determined by the number of primes \(p\) splits into and the inertia degrees of those primes.

The main input to Theorem 2 will be the functional equation.

Definition 4. Fix a number field \(K\) of signature \((n_1,n_2)\). Then the completed Dedekind \(\zeta\)-function is \[\Xi_K(s)=\left|\operatorname{disc}\mathcal O_K\right|^{s/2}\Gamma_{\mathbb R}(s)^{n_1}\Gamma_{\mathbb C}(s)^{n_2}\zeta_K(s),\] where \(\Gamma_{\mathbb R}(s)=\pi^{-s/2}\Gamma(s/2)\) and \(\Gamma_{\mathbb C}(s)=2(2\pi)^{-s}\Gamma(s)\).

Theorem 5. Fix a number field \(K\). Then \(\zeta_K(s)\) admits a meromorphic continuation to \(\mathbb C\), which is holomorphic everywhere except for a simple pole at \(s=1\). Furthermore, there is the functional equation \[\Xi_K(s)=\Xi_K(1-s).\]

Proof. This is a difficult theorem and thus omitted. \(\blacksquare\)

Proposition 6. Fix number fields \(K\) and \(K’\). Then the following are equivalent.

• \(K\) and \(K’\) are arithmetically equivalent.
• \(\zeta_K=\zeta_{K’}\).

Proof. As discussed in Remark 3, arithmetic equivalence is equivalent to having the equality of \(\zeta_K\) and \(\zeta_{K’}\) up to finitely many Euler factors. Namely, we are using the fact that the coefficients of a Dirichlet series can be recovered from the holomorphic function by integrating in vertical strips. Once the coefficients of the Dirichlet series are recovered, one can use the \(p\)-power coefficients to recover the Euler factor at \(p\) and hence the factorization type for unramified primes.

Thus, it only remains to show that having an equality up to finitely many Euler factors implies full equality. To this end, let \(S\) be a finite set of rational primes containing all rational primes where the Euler factors of \(\zeta_K\) and \(\zeta_{K’}\) disagree. Now, let \(D_K\) and \(D_{K’}\) be the absolute discriminants, and let \((n_1,n_2)\) and \((n_1’,n_2’)\) be the signatures of \(K\) and \(K’\) respectively. Then \[\frac{\Xi_K(s)}{\Xi_{K’}(s)}=\left|\frac{D_K}{D_{K’}}\right|^{s/2}\Gamma_{\mathbb R}(s)^{n_1-n_1’}\Gamma_{\mathbb C}(s)^{n_2-n_2’}\prod_{p\in S}\frac{E_{K’,p}(p^{-s})}{E_{K,p}(p^{-s})},\] where \(E_{K,p}(p^{-s})\) and \(E_{K’,p}(p^{-s})\) are the Euler factors. We now hit this equality with the functional equation and compare zeroes and poles.

• The terms \(\left|D_K/D_{K’}\right|^{s/2}\) and \(\left|D_K/D_{K’}\right|^{(1-s)/2}\) will admit no zeroes or poles.
• The function \(\Gamma_{\mathbb R}(s)\) only admits a zero or pole along \(2\mathbb Z_{\le0}\), and the function \(\Gamma_{\mathbb R}(1-s)\) only admits a zero or pole along \(1+2\mathbb Z_{\ge0}\).
• Similarly, the function \(\Gamma_{\mathbb C}(s)\) only admits a zero or pole along \(\mathbb Z_{\le0}\), and the function \(\Gamma_{\mathbb C}(1-s)\) only admits a zero or pole along \(\mathbb Z_{\ge1}\).
• Lastly, recall that the Euler factor \(E_{K,p}(p^{-s})\) looks like some product of terms of the form \(\left(1-p^{-sf}\right)\), so we can only admit a zero or pole when \(p^{-s}\) is a root of unity, which amounts to requiring \(\operatorname{Re}s=0\) and \(\operatorname{Im}s\in\frac{2\pi}{\log p}\mathbb Q\).

The above factor-by-factor analysis reveals that \(\Xi_K(s)/\Xi_{K’}(s)\) and \(\Xi_K(1-s)/\Xi_{K’}(1-s)\) have disjoint sets of zeroes and poles. But these are of course the same function by Theorem 5, so we are forced to conclude that these quotients have no zeroes or poles or at all!

We will be able to convert the previous paragraph into the result after a little work. Note that \(\frac{2\pi i}{\log p}\cdot\frac ab\) will be a root of \(E_{K,p}(p^{-s})\) exactly with multiplicity equal to the number of inertial degrees \(f(\mathfrak p/(p))\) dividing \(b\) (where \(a/b\) is assumed to be in reduced form). Thus, any nontrivial factor in the product \(\prod_{p\in S}E_{K’,p}(p^{-s})/E_{K,p}(p^{-s})\) will contribute roots or poles with nonzero imaginary part. This is illegal, so we conclude that all terms must be \(1\). It follows that \(\zeta_K=\zeta_{K’}\). \(\blacksquare\)

The proof above actually equips us to show much more of Theorem 2.

Corollary 7. Fix number fields \(K\) and \(K’\). If \(K\sim K’\), then \(K\) and \(K’\) have the same signature and degree.

Proof. Because the degree is twice the number of total number of complex embeddings, it is enough to show the equality of signatures, which we do in two steps.

1. We claim that \(n_2=n_2’\). Indeed, otherwise, \(\Xi_K(s)/\Xi_{K’}(s)\) will admit a zero or pole at \(s=-1\), which is impossible.
2. We claim that \(n_1=n_1’\). As in the previous step, we simply note that \(n_1\ne n_1’\) would then imply that \(\Xi_K(s)/\Xi_{K’}(s)\) will admit a zero or pole at \(s=-2\), which is impossible. \(\blacksquare\)

Remark 8. Here is a more direct proof of the degree equality: choose any rational prime \(p\) which is unramified in both \(K\) and \(K’\) and admits the same factorization type in both \(K\) and \(K’\). Then \(p\) splits into \(r\) primes, which we may say have inertia degrees \((f_1,\ldots,f_r)\). But the fundamental identity shows us that \[[K:\mathbb Q]=\sum_{i=1}^rf_i=[K’:\mathbb Q],\] as desired.

Corollary 9. Fix number fields \(K\) and \(K’\). If \(K\sim K’\), then \(K\) and \(K’\) have the same discriminant.

Proof. With an equality of signatures and Dedekind \(\zeta\)-functions, we see that \[\frac{\Xi_K(s)}{\Xi_{K’}(s)}=\left|\frac{\operatorname{disc}\mathcal O_K}{\operatorname{disc}\mathcal O_{K’}}\right|^s.\] The functional equation of Theorem 5 thus forces \(\left|\operatorname{disc}\mathcal O_K\right|=\left|\operatorname{disc}\mathcal O_{K’}\right|\).

It remains to pin down the equality of signs. We will show that the sign of the discriminant depends only on the number of complex embeddings, which will complete the proof by Corollary 7. Because the discriminant of an element of \(K\) will differ from \(\operatorname{disc}\mathcal O_K\) only up to a square, it is enough to compute the sign of \(\operatorname{disc}\alpha\) where \(K=\mathbb Q(\alpha)\). For this, let \(\{\sigma_1,\ldots,\sigma_n\}\) be the set of embeddings, so \[\operatorname{disc}\alpha=\det\begin{bmatrix}\sigma_1(1) & \cdots & \sigma_1(\alpha^{n-1}) \\\vdots & \ddots & \vdots \\\sigma_n(1) & \cdots & \sigma_n(\alpha^{n-1})\end{bmatrix}^2,\] where \(n\) is the degree of \(K\).

Now, consider the action by complex conjugation on the determinant \[\det\begin{bmatrix}\sigma_1(1) & \cdots & \sigma_1(\alpha^{n-1}) \\\vdots & \ddots & \vdots \\\sigma_n(1) & \cdots & \sigma_n(\alpha^{n-1})\end{bmatrix},\] which is the square root of the discrimimant. On one hand, complex conjugation will fix the square root if and only if \(\operatorname{disc}\alpha>0\). On the other hand, compex conjugation will swap conjugate rows of complex embeddings while fixing rows of real embeddings, in effect multiplying the entire determinant by \((-1)^{n_2}\), where \(n_2\) is the number of pairs of complex embeddings. We conclude that \[\operatorname{sgn}\operatorname{disc}\mathcal O_K=(-1)^{n_2},\] as required. \(\blacksquare\)

Corollary 10. Let \(K\) and \(K’\) be arithmetically equivalent number fields. Then the normal closures of \(K\) and \(K’\) are isomorphic.

Proof. Let \(L\) and \(L’\) be the normal closures of \(K\) and \(K’\), respectively.

We will use the fact that totally split primes recognize Galois extensions. To be explicit, let \(\operatorname{Spl}(M)\) denote the set of rational primes which are totally split in a number field \(M\). Then being totally split is preserved by taking the normal closure, so \(\operatorname{Spl}(K)=\operatorname{Spl}(L)\) and \(\operatorname{Spl}(K’)=\operatorname{Spl}(L’)\). But now recall from this post that having the equality \[\operatorname{Spl}(L)=\operatorname{Spl}(L’)\] shows that \(L\cong L’\), so we are done because checking Euler factors of Proposition 6 yields this equality. \(\blacksquare\)

Remark 11. The above proof is dishonest because one does not actually need to appeal to Proposition 6. Indeed, to conclude that \(L=L’\), one technically only needs the equality \[\operatorname{Spl}(L)=\operatorname{Spl}(L’)\] up to a finite number of primes.

Prime-Splitting With Group Theory

In this subsection, we will provide a group-theoretic characterization of arithmetic equivalence, and it will help us prove the rest of Theorem 2.

To start, we provide a group-theoretic way to compute prime-splitting in non-Galois extensions using Frobenius elements.

Proposition 12. Let \(K\) be a number field, and let \(L\) be an extension of \(K\) which is Galois over \(\mathbb Q\). Set \(G=\operatorname{Gal}(L/\mathbb Q)\) and \(H=\operatorname{Gal}(L/K)\) for brevity. Fix a rational prime \(p\) unramified in \(L\), and choose a prime \(\mathfrak q\) of \(L\) above it.

1. The map \(G\to\{\mathfrak p\subseteq\mathcal O_K:\mathfrak p\mid p\mathcal O_K\}\) given by \(\sigma\mapsto(\sigma\mathfrak q\cap K)\) descends to a bijection from the double coset space \(H\backslash G/D_{\mathfrak q}\).
2. The inertia degree of the prime \(\sigma\mathfrak q\cap K\) over \(p\) is the size of \(H\backslash H\sigma D_{\mathfrak q}\).

Proof. We show the two parts separately.

1. For the surjectivity of the map, we note that any prime \(\mathfrak p\) of \(K\) above \(p\) lies under some prime \(\mathfrak q’\) of \(L\). But \(G\) acts transitively on the primes of \(L\) above \(p\), so we may find \(\sigma\in G\) with \(\mathfrak q’=\sigma\mathfrak q\), whereupon \(\mathfrak p=\sigma\mathfrak q\cap K\) follows.

   It remains to show injectivity. Well, suppose that \(\sigma\mathfrak q\cap K=\tau\mathfrak q\cap K\) for \(\sigma,\tau\in G\). Then there are two primes \(\sigma\mathfrak q\) and \(\tau\mathfrak q\) of \(L\) which lie over the same prime of \(K\), so there is an element \(h\in H\) for which \(h\sigma\mathfrak q=\tau\mathfrak q\). It follows that \(\tau^{-1}h\sigma\in D_{\mathfrak q}\), so \(H\sigma D_{\mathfrak q}=H\tau D_{\mathfrak q}\).

2. Everything is unramified, so \(D_{\mathfrak q}\) is cyclic (generated by the Frobenius), so the size \(f\) of \(H\backslash H\sigma D_{\mathfrak q}\) is exactly the least positive integer for which \[H\sigma\mathrm{Frob}_{\mathfrak q}^f=H\sigma.\] Rearranging, this is the least positive integer for which \(\mathrm{Frob}_{\sigma\mathrm q}^f\in H\). By reducing (which will not affect nontriviality because everything is unramified), we see that this is the least power of the Frobenius which acts trivial on the residue field \(\sigma\mathrm q\cap L^H\). This is exactly the inertia degree of \(\sigma\mathrm q\cap K\). \(\blacksquare\)

Here is our group-theoretic characterization.

Definition 13. Two subgroups \(H\) and \(H’\) of \(G\) are Gassmann equivalent if and only if one has an equality \[\{\left|H\backslash H\sigma C\right|:H\sigma C\in H\backslash G/C\}=\{\left|H’\backslash H’\sigma C\right|:H’\sigma C\in H’\backslash G/C\}\] of multisets for any cyclic subgroup \(C\subseteq G\).

Corollary 14. Fix number fields \(K\) and \(K’\). Let \(L\) be a Galois extension of \(\mathbb Q\) containing \(KK’\) with Galois group \(G\), and set \(H=\operatorname{Gal}(L/K)\) and \(H’=\operatorname{Gal}(L/K’)\). The following are equivalent.

• \(K\) and \(K’\) are arithmetically equivalent.
• \(H\) and \(H’\) are Gassmann equivalent.

Proof. By Proposition 12, the factorization type in \(K\) of any unramified rational prime \(p\) is given exactly by the multiset \[\{\left|H\backslash H\sigma\langle\mathrm{Frob}_{\mathfrak q}\rangle\right|:H\sigma C\in H\backslash G/\langle\mathrm{Frob}_{\mathfrak q}\rangle,\}\] where \(\mathfrak q\) is some chosen prime in \(L\) above \(p\). Thus, Gassmann equivalence implies having the same factorization type for all unramified primes, which implies arithmetic equivalence.

Conversely, arithmetic equivalence implies having equality \[\{\left|H\backslash H\sigma C\right|:H\sigma C\in H\backslash G/C\}=\{\left|H’\backslash H’\sigma C\right|:H’\sigma C\in H’\backslash G/C\}\] whenever \(C=\langle\mathrm{Frob}_{\mathfrak q}\rangle\). By the Chebotarev density theorem, any cyclic subgroup takes this form, so Gassmann equivalence follows. \(\blacksquare\)

Here is a more useful way to test Gassmann equivalence.

Proposition 15. Fix subgroups \(H\) and \(H’\) of a group \(G\). Then the following are equivalent.

• \(H\) and \(H’\) are Gassmann equivalent.
• For each \(\sigma\in G\), we have \[\left|\{H\tau:H\tau=H\tau\sigma\}\right|=\left|\{H’\tau:H’\tau=H’\tau\sigma\}\right|.\]
• There is an isomorphism \(\mathbb C[H\backslash G]\cong\mathbb C[H’\backslash G]\) of (permutation) representations of \(G\).
• For any conjugacy class \(c\), we have \(\left|c\cap H\right|=\left|c\cap H’\right|\).

Proof. We start with the equivalence of the first three points.

• Note that the multiset \[\{\left|H\backslash H\tau\langle\sigma\rangle\right|:H\tau C\in H\backslash G/\langle\sigma\rangle\}\] is exactly the cycle type of \(\sigma\) acting on \(H\backslash G\). Thus, Gassmann equivalence is equivalent to the cycle types of any \(\sigma\) acting on \(H\backslash G\) or \(H’\backslash G\) being the same. In particular, this implies that the number of fixed points is the same, meaning \[\left|\{H\tau:H\tau=H\tau\sigma\}\right|=\left|\{H’\tau:H’\tau=H’\tau\sigma\}\right|.\]
• The number of fixed points of \(\sigma\) acting on \(G\backslash H\) is exactly the trace of \(\sigma\) acting on the permutation representation \(\mathbb C[H\backslash G]\). Thus, the second point implies that the characters of \(\mathbb C[H\backslash G]\) and \(\mathbb C[H’\backslash G]\) are the same, so isomorphism follows.
• Having an isomorphism \(\mathbb C[H\backslash G]\cong\mathbb C[H’\backslash G]\) certainly implies that \(\sigma\) acting on either \(H\backslash G\) or \(H’\backslash G\) will admit the same eigenvalues and therefore have the same cycle type. Gassmann equivalence follows.

It remains to show equivalence with the fourth point. We will show that the second and fourth points are equivalent. Quickly, note that we always have \(\left|H\right|=\left|H’\right|\): the fourth point implies this by summing over conjugacy classes, and the third point implies this by taking dimensions.

Now, for any \(\sigma\in G\), let \(G\cdot\sigma\) denote the conjugacy class. To start, note \[\left|H\cap G\cdot\sigma\right|\cdot\left|C(\sigma)\right|=\left|\{\tau\in G:\tau\sigma\tau^{-1}\in H\}\right|\] by the Orbit–stabilizer theorem. The right-hand side can be rearranged into \[\left|\{\tau\in G:H\tau\sigma=H\tau\}\right|,\] which of course is \[\left|H\cdot\right|\cdot\left|\{H\tau\in H\backslash G:H\tau\sigma=H\tau\}\right|.\] We conclude that \(\left|H\cap G\cdot\sigma\right|\) and \(\left|\{H\tau\in H\backslash G:H\tau\sigma=H\tau\}\right|\) disagree only by some factor depending on \(\left|H\right|\) and \(\sigma\), so the equivalence of the second and fourth points follows. \(\blacksquare\)

Corollary 16. Let \(K\) and \(K’\) be arithmetically equivalent number fields. Then the normal cores of \(K\) and \(K’\) are isomorphic.

Proof. Let \(L\) be the Galois closure of \(KK’\). Then the normal sub-extensions of \(L/\mathbb Q\) are in bijection with the normal subgroups of \(G=\operatorname{Gal}(L/\mathbb Q)\), so the normal core of \(K\) is given by the fixed field of the smallest normal subgroup of \(G\) containing \(H=\operatorname{Gal}(L/K)\). An analogous statement holds for the normal core of \(K’\) with the subgroup \(H’=\operatorname{Gal}(L/K’)\). We are thus left to show that \[N_G(H)\stackrel?=N_G(H’).\] To see this, note the normalizer of a subgroup \(H\) is formed by (iteratively) taking the group generated by the conjugacy classes intersecting \(H\). Because the conjugacy classes intersecting \(H\) and \(H’\) are the same by combining Corollary 14 and Proposition 15, we conclude that the normalizers are the same, so we are done. \(\blacksquare\)

Corollary 17. Let \(K\) and \(K’\) be arithmetically equivalent number fields. Then \(\mathcal O_K^\times\cong\mathcal O_{K’}^\times\).

Proof. Note that the unit groups are finitely generated abelian groups, so their isomorphism type is given by their torsion groups and their ranks. The rank is given by Dirichlet’s unit theorem: it is determined by the signature, which is the same for \(K\) and \(K’\) by Corollary 7.

It remains to show that \(\mathcal O_{K,\mathrm{tors}}^\times\cong\mathcal O_{K’,\mathrm{tors}}^\times\). Well, the torsion subgroup of a unit group is exactly the roots of unity, so it is enough to check that \[K\cap\mathbb Q(\mu_\infty)\stackrel?=K’\cap\mathbb Q(\mu_\infty),\] where \(\mathbb Q(\mu_\infty)\) is the cyclotomic closure of \(\mathbb Q\). However, \(K\cap\mathbb Q(\mu_\infty)\) is abelian over \(\mathbb Q\) and will therefore be contained in the normal core of \(K\), so Corollary 16 tells us that \(K\cap\mathbb Q(\mu_\infty)\subseteq K’\). Thus, \[K\cap\mathbb Q(\mu_\infty)\subseteq K’\cap\mathbb Q(\mu_\infty),\] and the other inclusion follows by symmetry. \(\blacksquare\)

Remark 18. Suppose \(K\sim K’\) so that \(\zeta_K=\zeta_{K’}\) by Proposition 6. Then the residues at \(s=1\) are the same, allowing us to compare class number formulae of \(K\) and \(K’\). Theorem 2 shows that many of the quantities present in the class number formula are already the same, leaving us only with \[\left|\mathrm{Cl}_K\right|\cdot\left|\mathrm{Reg}_K\right|=\left|\mathrm{Cl}_{K’}\right|\cdot\left|\mathrm{Reg}_{K’}\right|.\] However, it turns out that the class number alone is not determined by arithmetic equivalence class.