INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 4 (2004), #A17

A FINITE RING POLYNOMIAL

B. Sury

Stat-Math Unit, Indian Statistical Institute, Bangalore, 560 059, India

                           sury@isibang.ac.in

Received: 1/1/04, Revised: 9/28/04, Accepted: 10/22/04, Published: 10/25/04

Abstract

For any finite commutative ring A with unity, the polynomial

S_{A}(x) = ∏_{a∈A}(x − a)

was considered in [1] as an analogue of the sine function and studied with reference to Kronecker’s Jugendtraum. A question posed there was whether, for A = ZZ/nZZ, the additivity

S_{A}(x + y) = S_{A}(x) + S_{A}(y)

holds if, and only if, n is a prime. We prove this very easily and show, more generally, that if additivity holds for A, then A has characteristic a prime and, further, for the ring A which is the direct sum of r copies of ZZ/pZZ for a prime p, we have

S_{A}(x) = (x^{p} − x)^{p^{r−1}}.

For any finite commutative ring A with unity, consider the polynomial

S_{A}(x) = ∏_{a∈A}(x − a).

This finite analogue of the sine function was studied in [1] where the question was posed for A = ZZ/nZZ as to whether the additivity

S_{A}(x + y) = S_{A}(x) + S_{A}(y)

holds if, and only if, n is a prime. This is extremely easy to prove and here we observe, more generally, the following :

Theorem. Let A be any finite commutative ring with unity having cardinality n and let S_{A}(x) ∈ A[x] be defined to be ∏_{a∈A}(x − a). If

S_{A}(x + y) = S_{A}(x) + S_{A}(y)

holds, then A has characteristic a prime. In particular, for ZZ/nZZ, additivity holds if, and only if, n is prime. Further, for the ring A which is the direct sum of r copies of ZZ/pZZ for a prime p, we have

S_{A}(x) = (x^{p} − x)^{p^{r−1}}.

In particular, additivity holds in all these cases.

Proof. Let us start by observing that S_{A}(x) = x^{n} + f(x) where f is a polynomial consisting of terms in x, of degree smaller than n. Thus S_{A}(x + y) contains the terms \binom{n}{r}x^{r}y^{n−r} for 1 ≤ r < n which, if nonzero, contribute to S_{A}(x + y) − S_{A}(x) − S_{A}(y). Thus, additivity forces all the binomial coefficients \binom{n}{r} to be zero in A for all 1 ≤ r < n.

If n = ∏_{i=1}^{k} p_{i}^{l_{i}} for different primes p_{i}, then since ( \begin{matrix}n \\ p_{i}^{l_{i}} \end{matrix} ) is coprime to p_{i} and since

n = \binom{n}{1} is itself zero, we obtain that _{\frac{n}{p_{i}^{l_{i}}}} is zero in A. But these k numbers do not have a common factor, which gives a contradiction unless there is only one prime.

Let us write n = p^{l}. Then the binomial coefficient ( \begin{matrix}p^{l} \\ p^{l−1} \end{matrix} ) is zero in A. This number is

of the form pd for some (p, d) = 1; hence we once again obtain, by using p^{l} = 0 = pd, that p = 0 in A. This proves the first assertion and answers question 1 of [1] affirmatively.

Now, we proceed further and consider rings which are direct sums of a finite number of copies of ZZ/pZZ and show that in these cases, the function S_{A}(x) = (x^{p} − x)^{p^{r−1}}. Let us consider A = ZZ/pZZ ⊕ ZZ/pZZ first so as to make the pattern clear. Now S_{A}(x) = ∏_{0≤i,j<p}(x − (i, j)). Further, it is trivial to see that the corresponding function of ZZ/pZZ is

x(x − 1) ⋯ (x − p + 1) = x^{p} − x.

In other words, the k-th symmetric polynomials σ_{k} in the natural numbers 1, 2, ⋯, p − 1 are multiples of p for 1 ≤ k < p − 1. Note that, for this ring A,

(x − (0, 0))(x − (1, 1)) ⋯ (x − (p − 1, p − 1)) = x(x − 1_{A})(x − 2(1_{A})) ⋯ (x − (p − 1)1_{A})

where 1_{A} is the unity of A. Hence

x(x−1_{A})(x−2(1_{A})) ⋯ (x−(p−1)1_{A}) = x^{p} −(σ_{1})1_{A}x^{p−1} +⋯−(σ_{p−2})1_{A}x^{2} −x = x^{p} −x.

Thus, we have the subproduct

S_{0}(x) := (x − (0, 0))(x − (1, 1)) ⋯ (x − (p − 1, p − 1)) = x^{p} − x.

Now, we claim that the product S_{A}(x) can be broken up into p products each of which equals the above subproduct S_{0}(x). It is clear that the product of the p − 1 factors

S_{i}(x) = (x − (0, i))(x − (1, i + 1)) ⋯ (x − (p − 1, i − 1)),

for i = 1, 2, ⋯, p − 1, when multiplied by S_{0}(x), give S_{A}(x). We claim that S_{i} = S_{0} for all i. Clearly, S_{i}(x) = S_{0}(y) where y = x − (0, i). Therefore,

S_{i}(x) = y^{p} − y = (x − (0, i))^{p} − (x − (0, i)) = x^{p} − x.

Hence, we have proved, for A = ZZ/pZZ ⊕ ZZ/pZZ, that S_{A}(x) = (x^{p} − x)^{p}.

For general r, the proof is essentially the same as that given for r = 2, taking

S_{0}(x) = (x − (0, ⋯, 0))(x − (1, ⋯, 1)) ⋯ (x − (p − 1, ⋯, p − 1))

except that there are now p^{r−1} subproducts in the obvious manner. In fact, ZZ/pZZ acts on A by

j.(i_{1}, ⋯, i_{r}) = (i_{1} − j, ⋯, i_{r} − j)

and S_{0} is an orbit of the action. Since each orbit clearly has p elements, there are p^{r−1} orbits. Note that for any orbit S_{i}, the subproduct equals x^{p} − x as seen before. Hence, it follows that for the direct sum A of r copies of ZZ/pZZ, we have S_{A}(x) = (x^{p} − x)^{p^{r−1}}. This proves the theorem.

It may also be interesting to study this function for finite noncommutative rings, where we fix any order of the elements for defining the product.

References

[1] N. Kurokawa, Eva-Marie Muller-Stuler, H. Ochiai, and M. Wakayama, Kronecker’s Jugendtraum and ring sine functions, J. Ramanujan Math. Soc., Vol. 17 (2002) 211-220.

Acknowledgments

It is a pleasure to thank Professor H. N. Ramaswamy who drew my attention to this question and gave me a copy of the paper of Kurokawa et. al.