Norm for a finite free ring extension ($N_{B/A}(b)=\det (m_b)$)

If $B$ is a commutative $A$-algebra which is a finite free module over $A$, its norm sends $b$ to the determinant of multiplication $m_b:B\to B$. The determinant is independent of the chosen basis and is multiplicative because $m_{bc}=m_b{m}_c$. It commutes with scalar extension. On a separable field extension it becomes the product over embeddings, recovering the field norm. An invertible $b$ has invertible norm.