A special definition of determinant

Question

Let $R$ be a commutative ring. Let $D:M_n(R)\to R$ be a function. We say that $D$ is a determinant if

• $D(B)=aD(A)$, whenever $B$ is obtained by multiplying a row of $A$ by $a$.
• If $A,B,C$ are identical except the $i$-th row, and $C_i=A_i+B_i$, where subscripts denote the $i$-th row vector, then $D(C)=D(A)+D(B)$.
• $D(A)=0$ if two rows of $A$ are identical.

Now, one can prove that if $D$ is a determinant and $D(I)=1$, then $D(A)D(B)=D(AB)$.

The question is, though, can I reverse this, and deduce that if $D$ satisfy $D(A)D(B)=D(AB)$, then $D$ is a determinant? This might not be true, so perhaps we need more assumptions.

PS: $D(I)=1$ can actually be proved from $D(A)D(B)=D(AB)$ if there exists $M$ such that $D(M)\neq 0$. Then $D(M)=D(MI)=D(M)D(I)$, so $D(I)=1$.

Answer 1

No. Let $\sigma:R\to R$ be any function that is multiplicative but not additive. Then $D=\sigma\circ\det$ is multiplicative but it does not satisfy the requirement in your second bullet point.

E.g. let $R=\mathbb R$ and $\sigma(x)=x^2$, so that $D(A)=\det(A)^2$. Then $D$ is multiplicative, but $$ 0=D\left(\begin{bmatrix}0&0\\ 0&0\end{bmatrix}\right) \ne D\left(\begin{bmatrix}1&0\\ 0&1\end{bmatrix}\right) +D\left(\begin{bmatrix}-1&0\\ 0&1\end{bmatrix}\right) =1+1. $$

Answer 2

No, take the function $D:M_n(R)\to R$ with $D(A)=1$ for all $A \in M_n(R).$ Clearly $D(AB)=1=1*1=D(A)D(B)$ but $D$ is not a determinant.