Reduced norm

Remarks

The value $\mathrm{Nrd}(a)$ lies in $F$ and does not depend on the choice of $L$ and $\varphi$; independence of $\varphi$ follows from the Skolem–Noether theorem. The reduced norm is multiplicative, and $a$ is invertible in $A$ if and only if $\mathrm{Nrd}(a) \ne 0$. It relates to the norm of left multiplication by $N(a) = \det(x \mapsto ax) = \mathrm{Nrd}(a)^n$, the determinant being taken over $F$. Example: for $A = M_n(F)$, take $L = F$ and $\varphi$ the identity, so $\mathrm{Nrd}$ is the usual determinant. Example: for the Hamilton quaternions $\mathbb{H}$ over $\R$, with $n = 2$ and $L = \C$, one finds $\mathrm{Nrd}(a + bi + cj + dk) = a^2 + b^2 + c^2 + d^2$. Non-example: the map $a \mapsto \det(x \mapsto ax)$ on $\mathbb{H}$ is not the reduced norm, since it equals $(a^2 + b^2 + c^2 + d^2)^2$. The reduced norm differs from the field norm in general. But when $A$ contains a subfield $K$ of degree $n$ over $F$, the reduced norm of $a \in K$ is the field norm $N_{K/F}(a)$.