Does there exist a

commutative ring(-with-a-1) $R$
positive integer $n$
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
$f$ of the $n$-by-$n$ identity matrix is $1_R$
for all $n$-by-$n$ matrices $M\hspace{-0.03 in}$, if $M$ is invertible then $\hspace{.04 in}f(M)$ is a unit
$f$ is not the restriction of determinant to $\hspace{.04 in}f\hspace{.02 in}$'s domain


I'm inspired by this answer.

  • $\begingroup$ Do you also require $f$ to preserve multiplication? $\endgroup$
    – Berci
    Jun 17, 2016 at 19:34
  • $\begingroup$ Not here, although if the answer is yes then that will probably be a follow-up question. ​ ​ $\endgroup$
    – user57159
    Jun 17, 2016 at 19:38
  • $\begingroup$ Is it on purpose that you left out the property of singular matrices having determinant $0$? $\endgroup$ Jun 17, 2016 at 19:47
  • $\begingroup$ @HenningMakholm : ​ Yes - I do not want to exclude rings with zero-divisors. ​ ​ ​ ​ $\endgroup$
    – user57159
    Jun 17, 2016 at 19:51
  • $\begingroup$ 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. $\endgroup$
    – anomaly
    Jun 17, 2016 at 21:13

1 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).

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.

  • 1
    $\begingroup$ "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) $\endgroup$ Jun 18, 2016 at 17:14
  • $\begingroup$ @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? $\endgroup$ Jun 18, 2016 at 17:21
  • $\begingroup$ Well, $2$ is a canonical element of an arbitrary ring...anyway, your edit should be easily understandable to all. $\endgroup$ Jun 18, 2016 at 18:27

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