4 added 474 characters in body

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.

3 added 1 characters in body

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 $K'F^{H'}$ 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$. There is no reason whatsoever to doubt that it can, but it has not been proven. In a similar direction as above, the torsion of the odd-numbered$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. 2 added 4 characters in body 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$F^{H'}$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 where whether the $p$-part of the class numbers can differ for arbitrary $p$. There is no reason whatsoever to doubt that it can, but it has not been proven.

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.

1