Given a map $f:M_n(k)\to k$ (with $k$ some field) such that $f(AB)=f(A)f(B)$ for all matrices $A$ and $B$, is it necessarily the case that $f$ factors through the determinant, i.e. does there exist a multiplicative map $g:k\to k$ such that $f=g\circ\det\,$? Are constraints on $k$ necessary?
A simple corollary would be that nonzero multiplicative maps on subgroups of the general linear group $GL_n(k)$ factor through multiplicative maps on the units $k^\times\to k^\times$.
Two definitions of $\det$ I'm aware of (written with our setting in mind):
- The unique alternating mulilinear map (of column vectors in a matrix) sending $I$ to $1_k$.
- The trace of the map induced by $A$ on the $n$th exterior power $\mathrm{Alt}^nk^n$.
By Gaussian elemination, any multiplicative map $f$ from matrices to the base field is determined by its values on the matrices representing elementary row operations and upper triangular matrices.
One stumbling block is that it seems hard, in general, to fully characterize the multiplicative maps on the base field $k$. With a finite field it would just be integer powers and the zero map, but with the reals you get all (positive) real powers too, and some funky stuff may occur with other fields.