##
**On the arithmetic determination of the trace.**
*(English)*
Zbl 1326.11066

Let \(K\) be a number field with ring of integers \(\mathcal{O}_{K}\). We write \(t_{K} : \mathcal{O}_{K} \times \mathcal{O}_{K} \rightarrow \mathbb{Z}\) for the symmetric \(\mathbb{Z}\)-bilinear form given by \((x,y) \mapsto \mathrm{Tr}_{K/\mathbb{Q}}(xy)\), where \(\mathrm{Tr}_{K/\mathbb{Q}}\) is the usual trace map. The integral trace form \(q_{K} : \mathcal{O}_{K} \rightarrow \mathbb{Z}\) given by \(x \mapsto t_{K}((x,x))=\mathrm{Tr}_{K/\mathbb{Q}}(x^{2})\) is the integral quadratic form associated to the pairing \(t_{K}\). Two number fields \(K\) and \(L\) of equal degree \(n\) over \(\mathbb{Q}\) are said to have isometric integral trace forms whenever the integral quadratic forms \(q_{K}\) and \(q_{L}\) are equivalent, that is, if \(t_{K}\) and \(t_{L}\) are represented by symmetric integral \(n \times n\) matrices \(A\) and \(B\), respectively, then there exists some \(C \in \mathrm{GL}_{n}(\mathbb{Z})\) such that \(A=CBC^{t}\). A necessary condition for \(K\) and \(L\) to have isometric integral trace forms is that they have the same signature and discriminant. Thus, since a quadratic field is completely determined by its discriminant, two quadratic fields have isometric integral trace forms if and only if they have the same discriminant. It is of interest to find necessary and sufficient conditions for two number fields to have isometric integral trace forms in other cases, and the article under review gives several such results for number fields satisfying certain hypotheses.

One of the main results is that for non-totally real number fields that are only tamely ramified over \(\mathbb{Q}\), the integral trace pairing is completely determined by the discriminant, signature, and a finite set of positive integers that only depend on the factorization of ramified primes. As a consequence, the author deduces the following result: two non-totally real cubic number fields have isometric integral trace forms if and only if they have the same discriminant (note that here the fields in question are not required to be tamely ramified over \(\mathbb{Q}\)). Another result is as follows. Let \(L,K\) be non-totally real number fields with equal signature and fundamental discriminant (that is, their discriminants are equal to each other and to the discriminant of a quadratic field), and assume further that \(2\) is at most tamely ramified in both \(L/\mathbb{Q}\) and \(K/\mathbb{Q}\). Then the integral trace forms of \(L\) and \(K\) are isometric if and only if for every odd prime \(p\) that ramifies in \(K/\mathbb{Q}\), the number of primes of \(\mathcal{O}_{K}\) above \(p\) has the same parity as the number of primes of \(\mathcal{O}_{L}\) above \(p\).

One of the main results is that for non-totally real number fields that are only tamely ramified over \(\mathbb{Q}\), the integral trace pairing is completely determined by the discriminant, signature, and a finite set of positive integers that only depend on the factorization of ramified primes. As a consequence, the author deduces the following result: two non-totally real cubic number fields have isometric integral trace forms if and only if they have the same discriminant (note that here the fields in question are not required to be tamely ramified over \(\mathbb{Q}\)). Another result is as follows. Let \(L,K\) be non-totally real number fields with equal signature and fundamental discriminant (that is, their discriminants are equal to each other and to the discriminant of a quadratic field), and assume further that \(2\) is at most tamely ramified in both \(L/\mathbb{Q}\) and \(K/\mathbb{Q}\). Then the integral trace forms of \(L\) and \(K\) are isometric if and only if for every odd prime \(p\) that ramifies in \(K/\mathbb{Q}\), the number of primes of \(\mathcal{O}_{K}\) above \(p\) has the same parity as the number of primes of \(\mathcal{O}_{L}\) above \(p\).

Reviewer: Henri Johnston (Exeter)

### MSC:

11R21 | Other number fields |

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