discriminant

Last updated January 27, 2022

Dedekind domains

Let $\mathcal{O}$ be a dedekind domain and $k$ its field of fractions, $K$ a finite separable extension. Suppose $N$ is an $\mathcal{O}$- lattice in $K$.

The discriminant of $N$ is the $\mathcal{O}$- fractional ideal in $k$ $$\mathfrak{d}\mathcal{O}(N)\coloneqq (D\mathcal{O}(N):N)\mathcal{O}$$ We may just write $\mathfrak{d}{K/k}$ for $\mathfrak{d}_\mathcal{O}(\mathcal{O}_K)$. The latter is an ideal in $\mathcal{O}$.

Properties

Suppose $N,M\subset K$ are $\mathcal{O}$-lattices. Then

1. If $N=\langle{e_i}\rangle$ is a free $\mathcal{O}$-module, then $$\mathfrak{d}\mathcal{O}(N)=(\text{det }Tr(e_ie_j))$$ Notation: The generating element on the right is denoted $\Delta\mathcal{O}(N)\in k^\ast/(\mathcal{O}^\ast)^2$. If $\mathcal{O}=\mathbb{Z}$, then we may just write $\Delta(N)$, and $\Delta_K$ for $N=\mathcal{O}_K$.
2. $$\mathfrak{d}\mathcal{O}(N)=\mathfrak{d}\mathcal{O}(M)(M:N)^2_\mathcal{O}$$
3. Suppose $N\subset M$. Then $\mathfrak{O}(N)\subset\mathcal{O}(M)$, with equality iff $M=N$.
4. $$\mathfrak{d}\mathcal{O}(N)=\prod\mathfrak{p}\mathfrak{p}^{v_\mathfrak{p}(\mathfrak{d}_{\mathcal{O}_\mathfrak{p}}(N_\mathfrak{p}))}$$