Is there a commutative ring with a "generalized determinant"?

Question

Does there exist a

commutative ring(-with-a-1) $R$
and
positive integer $n$
and
function $\hspace{.04 in}f$ from [the set of $n$-by-$n$ matrices over $R$] to $R$

such that

$f$ is linear in each row and each column separately
and
$f$ of the $n$-by-$n$ identity matrix is $1_R$
and
for all $n$-by-$n$ matrices $M\hspace{-0.03 in}$, if $M$ is invertible then $\hspace{.04 in}f(M)$ is a unit
and
$f$ is not the restriction of determinant to $\hspace{.04 in}f\hspace{.02 in}$'s domain

?


I'm inspired by this answer.

Answer 1

For $R=\mathbb Z/4\mathbb Z$ the permanent has all the properties you list, and it differs from the determinant when $n\ge 2$.

(For any matrix, the difference between its permanent and its determinant is a multiple of $2$, so in $\mathbb Z/4\mathbb Z$ the permanent is a unit iff the determinant is).

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In general, we can set $R=\mathbb Z/m\mathbb Z$ whenever $m$ is not square-free, and then let $\bar m$ be the product of $m$'s prime factors (without multiplicity) and consider $$ f((a_{ij})) = \sum_{\sigma \in S_n} (h(\sigma) + \operatorname{sgn}(\sigma))\cdot \prod_i a_{i,\sigma(i)} $$ for any function $h: S_n \to \bar mR $ such that $h({\rm id})=0$. If $h$ is not identically zero, then $f$ will differ from the determinant, but satisfy all your conditions.