Trace of a product of two elements of integral basis I am struggling with the idea of the trace of $b_ib_j$ where $b_1, \dots, b_n$ form an integral basis of some algebraic number field $K$.
I know the trace is the trace of the linear combination of $b_ib_jb_1, \dots, b_ib_jb_n$ thing in terms of $b_1, \dots, b_n$. So is there anything (a formula or something) that makes life easier when computing the trace? Or does it always depend on what the basis is?
 A: Since you don’t mention the base field, I suppose it’s $\mathbf Q$. What you seek is then the theory of the discriminant of $K/\mathbf Q$, which is explained in any textbook on ANT, e.g. in Marcus’ « Number Fields », chapter 2. Given $K$ of absolute degree $n$, one begins to define the disc. of an $n$-tuple $a_1,…, a_n$ of elements of $K$ as the square of the determinant of the matix $\sigma_i (a_j)$, where the $\sigma_i$ are the $n$ embeddings of $\mathbf Q$ in an algebraic closure, and one shows that disc$ (a_1,…, a_n)$ equals the determinant of the matrix $T(a_i.a_j)$ , where $T$ is the trace from $K$ to $\mathbf Q$.  It is easy to see that for 2 integral bases $(a_i)$ and $(b_j)$, the discriminants are equal : this defines the (absolute) discriminant disc ($K$) over $\mathbf Q$.
Given now an extension $K/k$ of number fields, of degree $n$, one can repeat the definition of the disc. of an $n$-tuple of elements of $K$ over $k$, but things get more complicated when dealing with the rings of integers. It is classical that the ring of integers $A_K$ is an $A_k$-submodule of a free $A_k$-module of rank $n$ (and is such a free module if  $A_k$ is principal). For an $n$-tuple $(a_i)$ of elements of $A_K$, it is clear that $T(a_i.a_j) \in A_k$, where $T$ is this time the trace from $K$ to $k$. The (relative) discriminant disc ($K/k$) is then defined as the ideal of $A_k$ generated by the bases of $K/k$ which are contained in $A_K$. 
