abstract algebra - Is every multiplicative function from the matrix ring $text{Mat}_n(R)$ to $R$ a function of $text{det}$?

Suppose that $R$ is a commutative ring with $1$, and for $n \in \mathbb{N}$, let $\text{Mat}_n(R)$ be the set of $n \times n$ matrices with entries in $R$.

It is well known that the determinant function $\text{det} : \text{Mat}_n(R) \rightarrow R$ is multiplicative, i.e.

$$
\text{det}(AB) = \text{det}(A) \text{det}(B)
$$

$\text{det}$ is certainly not unique in this respect; there are lots of functions $g : \text{Mat}_n(R) \rightarrow R$ which are multiplicative. For a start, there are the constant $1$ and constant $0$ functions, as well as the indicator function of "is invertible". More strangely, for $R = \mathbb{R}$, are the functions

$$
g(A) =
\begin{cases}
e^{f(\log(\left|\det(A)\right|))} &\text{if} \det(A) \neq 0 \\
0 &\text{if} \det(A) = 0
\end{cases}
$$

where $f : \mathbb{R} \rightarrow \mathbb{R}$ is any solution of the Cauchy functional equation $f(x+y) = f(x)+f(y)$ : non-continuous solutions to this equation are really badly behaved.

However, I have not found any examples of multiplicative functions which are not themselves a function of $\det$.

Is there any such function which is not a function of det?

That is, is there a ring $R$, an $n \in \mathbb{N}$ and $g : \text{Mat}_n(R) \rightarrow R$ which is multiplicative, and not a function of det, i.e. there exist $A,B \in \text{Mat}_n(R)$ such that

$$
\begin{align}

\det(A) &= \det(B) \\
g(A) &\neq g(B)
\end{align}
$$

Answer

Suppose $R$ satisfies the following conditions.

• $R$ has an element $x\neq 1$ with $x^2=1$.


• There is a surjective ring homomorphism $q:R\to\mathbb{F}_2$.

For example, $R=\mathbb{Z}$ and $x=-1$, or $R=S\times\mathbb{F}_2$ where $S$ has an element $s$ of multiplicative order $2$, and $x=(s,1)$.

Then $q$ induces a ring homomorphism $\text{Mat}_2(R)\to\text{Mat}_2(\mathbb{F}_2)$,
which I'll also denote by $q$.

An element $X$ of $\text{GL}_2(\mathbb{F}_2)$ acts on the three non-zero vectors of $\mathbb{F}_2^2$, and I'll say $X$ is even or odd depending on whether it acts by an even or odd permutation.

There is a multiplicative function $g:\text{Mat}_2(R)\to R$ with

$$g(A)=
\begin{cases}
0&\mbox{if $q(A)$ is not invertible}\\
1&\mbox{if $q(A)$ is invertible and even}\\
x&\mbox{if $q(A)$ is invertible and odd.}
\end{cases}$$

But $g(A)$ is not a function of $\det(A)$, since for $A_1=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ and $A_2=\begin{pmatrix}1&1\\0&1\end{pmatrix}$, $\det(A_1)=\det(A_2)$ but $g(A_1)=1\neq x=g(A_2)$.