Determinants of free modules This should be easy, but I'm stuck. 

Let $A$ be a DVR and $B$ a finite algebra over $A$ that is free as an $A$-module. For $b \in B$ one can define the norm of $b$ as the determinant of multiplication by $b$ on $B$. Suppose that there is a basis $f_1, \ldots, f_n$ of $B$ over $A$ and $n$ elements $a_i \in A$ such that $a_1f_1, \ldots, a_n f_n$ is a basis of the free $A$-module $bB$. Then, one should show that $N(b) = a_1 \cdots a_n$. How can this be done?

Regards,
Heidar

Multiplication by $b$ has image $bB$ and sends the basis $\{f_1,\dots,f_n\}$ to $\{bf_1,\dots,bf_n\}$. This is also an $A$-basis of $bB$ (why?). Now write $bf_i=\sum_{j=1}^na_{ij}a_jf_j$, where $A=(a_{ij})$ is an invertible matrix. Then $N(b)=a_1\cdots a_n\det A$. (Note that since $A$ is invertible, then $\det A$ is an invertible element of $A$.) 
Edit. As noted @Jiangwei Xue in his comment to the question this is the maximum we can get.
