What is the reduced norm map?

Question

This is a basic question about the reduced norm homomorphism.

Let $A$ be a central simple $K$-algebra and $P$ a f.g. projective $A$-module. I know that $\operatorname{End}_A(P)$ is also a central simple $K$-algebra.

Then how is the reduced norm map $\operatorname{Nrd}:\operatorname{End}_A(P)^\times\to K^\times$ defined? What is its explicit formula?

Many thanks.

Answer 1

There are several possible definitions. One of them is to choose an embedding $K\to \overline{K}$ into an algebraic closure : then $A\otimes_K \overline{K}$ is isomorphic to $M_r(\overline{K})$ and $\operatorname{End}_A(P)\otimes_K \overline{K}\simeq M_n(\overline{K})$.

Then the reduced norm is the composition of $\operatorname{End}_A(P)\to \operatorname{End}_A(P)\otimes_K \overline{K}$ and the determinant map $M_n(\overline{K})\to \overline{K}$.

You may check that this composition has images in $K$, and does not depend on the isomorphism $\operatorname{End}_A(P)\otimes_K \overline{K}\simeq M_n(\overline{K})$