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. 



Multiplicative functions $\phi : M_n(F) \longrightarrow F$ with $\phi(I) = 1$
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, $f=\det^k$.


