fractional ideal of $\mathcal O_K$. Note that there exists an invertible $K$-linear transformation of $V$ such that $\varphi(X)=Y$, so that $$\label{eq:ind-det}
[X:Y]_{\mathcal O_K}=(\det \varphi)\mathcal O_K$$ (see [@CF] for details and [@DCDJA] for an overview).

\[remark:nodvr\] Let $L/K$ be an extension of number fields and let $X, Y$ be full $\mathcal{O}_K$-lattices in $L$. In this setting, the index $[X:Y]_{\mathcal{O}_K}$ is defined to be the only ideal of $\mathcal{O}_K$ with the following property: for every maximal ideal $\mathfrak{p}$ of $\mathcal{O}_K$, the $\mathfrak{p}$-adic valuation of $[X:Y]_{\mathcal{O}_K}$ is equal to that of $[X\otimes_{\mathcal{O}_K}\mathcal{O}_{K_\mathfrak p}:Y\otimes_{\mathcal{O}_K}\mathcal{O}_{K_\mathfrak p}]_{\mathcal{O}_{K_\mathfrak p}}$, where $K_\mathfrak p$ is the completion of $K$ at $\mathfrak{p}$. Note that a formula analogous to holds if $X$ and $Y$ are free over $\mathcal{O}_K$, which happens for instance whenever $K$ has class number $1$.

Suppose now that $X$ is an $\mathcal O_K$-lattice and let $V := X \otimes_{\mathcal{O}_K} K$ be the corresponding finite-dimensional vector space. Let $B$ be a non-degenerate symmetric $K$-bilinear form on $V$.

\[def:Discriminant\] We define the *discriminant* of $X$ by means of the module index: $$\operatorname{disc}_{B}(X):=[X':X]_{\mathcal O_K},$$ where $X'$ is the dual module of $X$ with respect to $B$, namely $$X'=\{v\in V\mid B(v,X)\subseteq \mathcal O_K\}.$$

With these definitions, for $Y\subseteq X$ we have the 