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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

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

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    $\begingroup$ Here's one observation. If you restrict your multiplicative function to $\mathrm{GL}_n(R)$ you get a group map $\mathrm{GL}_n(R)\to R^\times$. If you assume further that $R$ is a field with at least three element then this follows from the fact that the derived subgroup of $\mathrm{GL}_n(R)$ is $\mathrm{SL}_n(R)$. $\endgroup$ Commented Sep 14, 2015 at 13:17
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    $\begingroup$ @AlexYoucis, good point, however, I don't think the OP asks for $g(Id)$ to be $1$ so we might not get a group map... The only thing we know is that $g(Id)^2=g(Id)$ (it could be $4$ if we are working modulo $6$ for instance). $\endgroup$ Commented Sep 14, 2015 at 13:23
  • $\begingroup$ It's still fine with the assumption that $R$ is a field, though. $\endgroup$ Commented Sep 14, 2015 at 13:26

1 Answer 1

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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 $g(A_1)=1\neq x=g(A_2)$.

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  • $\begingroup$ Wonderful, intuitive example! $\endgroup$ Commented Sep 15, 2015 at 11:22
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    $\begingroup$ @SamCappleman-Lynes By the way, I think the example I've given is about the simplest, but it's very specific to $n=2$. There are more interesting and sophisticated examples based on calculations in algebraic $K$-theory (about which I'm far from an expert). For example, a famous example is $R=\mathbb{R}[x,y]/(x^2+y^2-1)$, whose "reduced Whitehead group" $SK_1(R)$ is known to be cyclic of order $2$. It turns out that this implies that for any $n>1$ there's a multiplicative function of the kind you're asking about, taking values of $0$ (for non-invertible matrices), $1$ and $-1$. $\endgroup$ Commented Sep 15, 2015 at 15:03

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