Let be a prime. Compute the Jordan canonical form over of the matrix whose diagonal entries are 0 and whose off-diagonal entries are 1. ()
Let . The computation we carried out here shows that over (indeed, over any ring with 1). So the minimal polynomial of divides . Note that and are nonzero (any off-diagonal entry is 1), so the minimal polynomial of is precisely .
Let be an -module via as usual. Now the invariant factors of divide , and we have .
We claim that has dimension 1 over . Indeed, if , then .
Using Lemmas 2 and 3 from this previous exercise, we have . Note that any mapping which realizes an isomorphism as -modules is necessarily also an -vector space isomorphism; since has dimension 1, the right hand side of this isomorphism also has dimension 1 over . Now one of the left summands is , corresponding to the minimal polynomial. Thus the remaining summands must be trivial; that is, there are no invariant factors which are not divisible by , and every invariant factor other than the minimal polynomial is exactly .
So the elementary divisors of are ( times) and . The corresponding Jordan canonical form is where if , if , and otherwise.