have $L=K[G]\alpha$). In this case the element $\alpha$ is called a normal basis generator. Clearly, $\alpha$ can be chosen to lie in $\mathcal{O}_L$, and it is easy to see that $\alpha \in \mathcal{O}_L$ is a normal basis generator if and only if $\mathcal{O}_K[G]\alpha$ is a *full* $\mathcal{O}_K$-lattice in $L$; this happens precisely when $[\mathcal O_L:\mathcal{O}_K[G]\alpha]$ is finite. Therefore, in order to compute $m(L/K)$, it suffices to consider elements $\alpha \in \mathcal{O}_L$ that are normal basis generators.

Let $X\supseteq Y$ be $\mathcal O_K$-lattices of the same rank ($\mathcal O_K$ is a principal ideal domain, so the rank of an $\mathcal O_K$-lattice, which is free, is well defined). Choosing two bases for $X$ and $Y$, we easily see that the quotient $X/Y$ is of the form $\oplus_{i=1}^r \mathcal O_K/(\pi_K^{a_i})$, hence in particular finite, and we define the *module index* $[X:Y]_{\mathcal O_K}$ as the ideal of $\mathcal O_K$ generated by $\pi_K^{\sum_{i=1}^r a_i}$. The *subgroup index* is instead defined as $[X:Y]:=|X/Y|$, and we have $$[X:Y]=N([X:Y]_{\mathcal O_K}),$$ where $N$ is the ideal norm. More generally, if $V$ is a finite-dimensional $K$ vector space and $X$ and $Y$ are full $\mathcal{O}_K$-lattices in $V$, then we set $[X:Y]_{\mathcal O_K}=[X:X\cap Y]_{\mathcal O_K}[Y:X\cap Y]^{-1}_{\mathcal O_K}$, and this is a 