Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 27 / 1 / a / Solution

Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 30 / 2 / Solution

{title2=$N_{B/A}(b)=\det(m_b)$}

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 $m_{bc}=m_bm_c$. 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.

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 Incoming links: Norm for a finite free ring extension