Trace pairing

For a finite field extension $L/K$, the trace pairing is the bilinear formIt is nondegenerate exactly when $L/K$ is separable.

For a monic degree-$n$ polynomial $g$ over a commutative ring $R$, let $A=R[X]/(g)$ and let $\lambda$ take the coefficient of $X^{n-1}$ in the reduced representative. The pairing $(a,b)\mapsto \lambda (ab)$ is unimodular on $1,X,\dots ,X^{n-1}$: reversing one basis order gives a triangular matrix with diagonal ones. When $A$ embeds in a separable field extension, Lagrange interpolation identifies $\lambda (a)$ with $\mathrm{Tr}(a/g^{\prime }(x))$. This yields the derivative formula for the inverse different.

For an $R$-lattice $M$ in $L$, its trace-dual lattice isFor the integral closure $S$ of $R$, the trace dual $S^{\vee }$ is the inverse different.

For a nonzero fractional ideal $\mathfrak{b}$ of a number field, its trace-dual lattice is the fractional idealThe equality follows from the definition of the inverse different and invertibility of fractional ideals. Under the positive metric on the Minkowski embedding of a number field, the Euclidean dual lattice of $j(\mathfrak{b})$ is $\overline{j({\mathfrak{b}}^{\vee })}$: coordinatewise complex conjugation converts the positive inner product into the trace pairing. This assertion does not require that conjugation preserve the number field itself.

For the integral closure $S$ of a Dedekind domain $R$ in a finite separable extension, the discriminant ideal is locally generated bywhere $({\alpha }_i)$ is a local $R$-basis of $S$. Intrinsically it is the image of $({\det }_{R}S{)}^{\otimes 2}$ under the determinant of the trace pairing.

The relative discriminant ${\mathfrak{d}}_{L/K}$ of a finite extension of number fields is the discriminant ideal of ${\mathcal{O}}_L$ over ${\mathcal{O}}_K$. It satisfiesand its prime divisors are exactly the ramified finite primes.

The relative discriminant of a finite extension of number fields is the relative norm of a fractional ideal applied to its different ideal. Consequently, at $\mathfrak{p}$,where $f_{\mathfrak{P}/\mathfrak{p}}$ is the residue-field degree and $d_{\mathfrak{P}}$ is the different exponent. Over $\mathbb{Q}$, $N({\mathfrak{D}}_{E/\mathbb{Q}})=|D_E|$.