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

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

?

• Do you also require $f$ to preserve multiplication? Jun 17, 2016 at 19:34
• Not here, although if the answer is yes then that will probably be a follow-up question. ​ ​
– user57159
Jun 17, 2016 at 19:38
• Is it on purpose that you left out the property of singular matrices having determinant $0$? Jun 17, 2016 at 19:47
• @HenningMakholm : ​ Yes - I do not want to exclude rings with zero-divisors. ​ ​ ​ ​
– user57159
Jun 17, 2016 at 19:51
• It's even possible to define an analogue (with various degrees of usefulness) of the determinant for matrices over a noncommutative ring $A$; see the first chapter or Serre's Trees for a sample application. Jun 17, 2016 at 21:13

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).
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.
• "The permanent always differs from the determinant by a multiple of 2, so in Z/4Z the permanent is a unit iff the determinant is." This confused me the first few times I read it until I realized that when you wrote "differs" you meant that literally: i.e., subtraction. Perhaps saying that the determinant and the permanent are congruent mod $2$ would be more clear to the me's of the world? (Anyway: nice answer. +1) Jun 18, 2016 at 17:14
• @PeteL.Clark: I fear it might invite confusion if I start talking about "congruent mod 2" in a ring that isn't $\mathbb Z$ itself, but I've tried to reword it, attempting to be more explicit. Is this clearer? Jun 18, 2016 at 17:21
• Well, $2$ is a canonical element of an arbitrary ring...anyway, your edit should be easily understandable to all. Jun 18, 2016 at 18:27