Let $F$ be a field and $D$ a function on $n\times n$ matrices over $F$ (with values in $F$). Suppose $D(A\cdot B)=D(A)\cdot D(B)$ forall $A,B$. Show that either $D(A)=0$ for all $A$, or $D(I_n)=1$. In the latter case show that $D(A)\neq 0$ whenever $A$ is invertible.
In either $P$ or $Q$ kind of problems. One can show if $P$ don’t hold, then $Q$ hold, or if $Q$ don’t hold, then $P$ hold.
Approach(1): If $D(I_n)\neq 1$. Put $B=I_n$, we have $D(A\cdot I_n)$ $=D(A)$ $=D(A)\cdot D(I_n)$, $\forall A\in M_n(F)$. By elementary properties of field, $D(A)-D(A)\cdot D(I_n)$ $=D(A)\cdot (1-D(I_n))$ $=0$. Which implies $D(A)=0$ or $1-D(I_n)=0$. If $1-D(I_n)=0$, then $D(I_n)=1$. Which contradicts our initial assumption of $D(I_n)\neq 1$. So $D(A)=0$, $\forall A\in M_n(F)$.
Suppose $D(I_n)=1$. Let $A$ be invertible matrix. Then $\exists A^{-1}\in M_n(F)$ such that $A\cdot A^{-1}$ $=A^{-1}\cdot A$ $=I_n$. Put $B=A^{-1}$, we have $D(A\cdot A^{-1})$ $=D(A)\cdot D(A^{-1})$ $=D(I_n)$ $=1$. Assume towards contradiction, $D(A)=0$. Then $D(A)\cdot D(A^{-1})$ $=1$ $=0$. Thus we reach contradiction. Hence $D(A)\neq 0$.
Approach(2): If $D(B)\neq 0$, for some $B\in M_n(F)$. By hypothesis, $D(B\cdot I_n)$ $=D(B)\cdot D(I_n)$ $=D(B)$. By elementary properties of field, $D(B)\cdot D(I_n)-D(B)$ $=D(B)\cdot (D(I_n)-1)$ $=0$. Which implies $D(B)=0$ or $D(I_n)-1=0$. Since $D(B)\neq 0$, we have $D(I_n)-1=0$. Hence $D(I_n)=1$. Is my proof correct?
Que: This exercise is not valid for commutative ring with identity. Because we don’t have condition $1\neq 0$. Am I right?