Multiplicative functions $\phi : M_n(F) \longrightarrow F$ with $\phi(I) = 1$

Question

Let $F$ be an infinite field and let $f \in F[x_{11},x_{12},...,x_{nn}]$ be an arbitrary polynomial in $n^2$ variables. Consider the function $\phi : M_n(F)\longrightarrow F$ defined by $\phi((a_{ij})) = f(a_{11},a_{12}, ..., a_{nn})$ and suppose that $\phi(I_n) = 1$ and $\phi(AB)= \phi(A)\phi(B)$, for any $A,B \in M_n(F)$. Is it true that $f$ is equal to some power of the determinant (considered as a polynomial of $n^2$ valiables $x_{ij}$).

Comment: When $f$ is a homogeneous polynomial then the problem is known to be true but I have no idea for the general case.

Answer 1

The answer is yes. Classically, there is a group endomorphism $g$ of $F^*$ such that $\phi(M)=g(\det M)$ for all non-singular $M$ (this is obvious if $|F|=2$, otherwise one uses the fact that $[GL_n(F),GL_n(F)]=SL_n(F)$). Then, $g$ is a polynomial map from $F$ to itself that satisfies $g(XY)=g(X)g(Y)$. From there, it is easy to see that $g : x \mapsto x^k$ for some non-negative integer $k$. Thus, $f(M)=(\det M)^k$ for all non-singular $M$; thus $(f-\det^k)\det$ vanishes everywhere on $M_n(F)$, and hence it is zero as $F$ is infinite. Therefore, $f=\det^k$.

Answer 2

Your condition implies that $\phi$ defines a homomorphism $GL_n(F)\rightarrow F^*$. Any such homomorphism annihilates the commutator subgroup $SL_n(F)$, hence factors as $GL_n(F) \,{\buildrel {\det}\over {\longrightarrow}}\,F^*\,{\buildrel {\lambda }\over {\longrightarrow}}\, F^*$, where $\lambda $ is a homomorphism. Since $\lambda \circ \det $ is a polynomial in $(a_{ij})$, $\lambda $ must be given by a polynomial in one variable, and this implies $\lambda (x)=x^n$ for some integer $n$.

Answer 3

It seems to be a theorem of Schur that every polynomial representation of $GL(n)$ is a direct sum of homogeneous representations, so the general result follows. For a lot of information on this, check out J. A. Green's Polynomial Representations of GL_n (Spring LNM 830).