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\centerline{\smalltt INTEGERS:
 \smallrm ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 4
(2004), \#A17}
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{\uppercase{{\bf A Finite Ring Polynomial}}}
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{\bf B. Sury}\\
{\smallit Stat-Math Unit, Indian Statistical Institute, Bangalore,  
560 059, India}\\
{\tt sury@isibang.ac.in}
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\centerline{\smallit Received: 1/1/04, Revised: 9/28/04, 
Accepted: 10/22/04, Published: 10/25/04}
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\centerline{\bf Abstract}

\noindent
For any finite commutative ring $A$ with unity, the polynomial
$$S_A(x) = \prod_{a \in A} (x - a)$$
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 $A= \Z/n\Z$, 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 $\Z/p\Z$ for a prime $p$, we have
$$S_A(x) = (x^p-x)^{p^{r-1}}.$$

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\markright{\smalltt INTEGERS: \smallrm ELECTRONIC 
JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 4 (2004), \#A17\hfill}
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\noindent
For any finite commutative ring $A$ with unity, consider the polynomial
$$S_A(x) = \prod_{a \in A} (x - a).$$
This finite analogue of the sine function was studied in $[1]$
where the question was posed for $A= \Z/n\Z$ 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 :
\newpage

\noindent
{\bf Theorem.}
{\it Let $A$ be any finite commutative ring with unity having 
cardinality $n$ and let $S_A(x) \in A[x]$ be defined to be
$\prod_{a \in A} (x-a)$. If
$$S_A(x+y) = S_A(x)+S_A(y)$$
holds, then $A$ has characteristic a prime.
In particular, for $\Z/n\Z$, additivity holds if, and only if,
$n$ is prime.\\
Further, for the ring $A$ which is the direct sum of $r$ copies
of $\Z/p\Z$ for a prime $p$, we have
$$S_A(x) = (x^p-x)^{p^{r-1}}.$$
In particular, additivity holds in all these cases.}

\noindent
{\it 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 ${n \choose r} x^r y^{n-r}$
for $1 \leq r < n$ which, if nonzero, contribute to
$S_A(x+y)-S_A(x)-S_A(y)$.
Thus, additivity forces all the binomial coefficients
${n \choose r}$ to be zero in $A$ for all $1 \leq r < n$.\\
If $n = \prod_{i=1}^k p_i^{l_i}$ for different primes $p_i$, then
since $\left( \begin{array}{c}n \\ p_i^{l_i} \end{array}\right)$
is coprime to $p_i$
and since
$n = {n \choose 1}$ 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
$\left( \begin{array}{c} p^l \\ p^{l-1} \end{array}\right)$
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.
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\noindent
Now, we proceed further and consider rings which are direct sums
of a finite number of copies of $\Z/p\Z$ and show that in these
cases, the function $S_A(x) = (x^p-x)^{p^{r-1}}$.\\
Let us consider $A = \Z/p\Z \oplus \Z/p\Z$ first so as to make the
pattern clear.\\
Now $S_A(x) = \prod_{0 \leq i,j<p} (x- (i,j))$.
Further, it is trivial to see that the corresponding function of $\Z/p\Z$ is
$$x(x-1) \cdots (x-p+1) = x^p-x.$$
In other words, the $k$-th symmetric polynomials $\sigma_k$ in the natural 
numbers
$1,2, \cdots, p-1$ are multiples of $p$ for $1 \leq k < p-1$.\\
Note that, for this ring $A$,
$$(x-(0,0))(x-(1,1)) \cdots (x-(p-1,p-1))
= x(x-1_A)(x-2(1_A)) \cdots (x-(p-1)1_A)$$
where $1_A$ is the unity of $A$.\\
Hence
$$x(x-1_A)(x-2(1_A)) \cdots (x-(p-1)1_A) = 
x^p - (\sigma_1)1_A x^{p-1} + \cdots - (\sigma_{p-2})1_A x^2 - x
= x^p-x.$$
Thus, we have the subproduct
$$S_0(x) := (x-(0,0))(x-(1,1)) \cdots (x-(p-1,p-1)) = x^p-x.$$
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
$$S_i(x) = (x-(0,i))(x-(1,i+1)) \cdots (x-(p-1,i-1)),$$
for $i = 1,2, \cdots, p-1$, 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 $A = \Z/p\Z \oplus \Z/p\Z$,
that $S_A(x) = (x^p-x)^p$.
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\noindent
For general $r$, the proof is essentially the same as that given for
$r=2$, taking
$$S_0(x) = (x-(0,\cdots, 0))(x-(1,\cdots, 1)) \cdots (x-(p-1,\cdots, p-1))$$
except that there are now $p^{r-1}$ subproducts in the obvious manner.
In fact, $\Z/p\Z$ acts on $A$ by 
$$j. (i_1, \cdots, i_r) = (i_1-j, \cdots, i_r-j)$$
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 $\Z/p\Z$,
we have $S_A(x) = (x^p-x)^{p^{r-1}}$.
This proves the theorem.

\noindent
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.
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\section*{\normalsize References}
$[1]$  {\bf N. Kurokawa, Eva-Marie Muller-Stuler, H. Ochiai, and
M. Wakayama}, {\it Kronecker's Jugendtraum and ring sine functions},
J. Ramanujan Math. Soc., Vol. 17 (2002) 211-220.
 
\section*{\normalsize 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.

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