free $\Lambda$-module. Then $\alpha$ is also such that $K  N=A \alpha$ is free over $A$. The inclusion $\Lambda\subseteq \mathfrak{A}(A, N)$ is clear. Conversely, let $a\in \mathfrak{A}(A, N)$; then $a\alpha\in N=\Lambda \alpha$, so there exists $\lambda\in\Lambda$ such that $a\alpha =\lambda\alpha$ in $A$. Since $K  N$ is freely generated by $\alpha$, this means that $a=\lambda\in \Lambda$.

\[productindex\] Let $L/K$ be a Galois extension of $p$-adic fields with Galois group $G$. Then $$m(L/K) = [\mathfrak{A}_{L/K}:\mathcal{O}_K[G]] \cdot \min_{\alpha \in \mathcal{O}_L} [\mathcal{O}_L : \mathfrak{A}_{L/K}   \alpha].$$ In particular, $\mathcal{O}_L$ is free over $\mathfrak{A}_{L/K}$ if and only if $m(L/K) = [\mathfrak{A}_{L/K}:\mathcal{O}_K[G]]$. In this case, given an element $\beta$ that realises the minimal index $m(L/K)$, the elements realising $m(L/K)$ are precisely those of the form $\lambda \beta$ for $\lambda \in \mathfrak{A}_{L/K}^*$. Equivalently, an element $\beta \in \mathcal{O}_L$ realises the minimum if and only if it generates $\mathcal{O}_L$ over $\mathfrak{A}_{L/K}$.

Let $\beta\in \mathcal{O}_L$ be an element such that $$m(L/K) = [\mathcal{O}_L : \mathcal{O}_K[G]   \beta].$$ By the definition of the associated order, we have that $\mathfrak{A}_{L/K}   \beta\subseteq \mathcal {O}_L$, so $$m(L/K) = [\mathcal{O}_L : \mathfrak{A}_{L/K}   \beta]\cdot [\mathfrak{A}_{L/K}   \beta : \mathcal{O}_K[G]   \beta].$$ Since $\beta$ generates a normal basis for $L/K$, we have that the $K$-linear map $\varphi_\beta \colon K[G]\to K[G] \beta$ 