Product of elements in Finite ring Let $R$ be a finite ring with or without unity. Can one give a short description of the product of all elements of $R^{\times}$?
Here, we denote by $R^{\times}$ the set of all nonzero elements of $R$, not (as usual) the set of all invertible elements of $R$.
 A: To summarize some of the comments, thanks to peoplepower for completing the solution.
We have three situations:
Case 1: The ring is an integral domain. Then it is a field. Let $n=p^k$ be the number of elements.
Then $R^*$ is a cyclic group. Let $a$ be a generator. Then you product is
$$P=a\cdot a^2 \cdot...\cdot a^{n-1}=a^{\frac{n(n-1)}{2}}$$
If $p=2$ then since $a^{n-1}=1$ the product is one. Otherwise, $n-1 \nmid \frac{n(n-1)}{2}$ thus $P\neq 1$ and $P^2=1$. This means $P=-1$.
For domains, the product is always $-1$.
Case 2: The ring contains at least two zero divisors. If any of the zero divisors has the property that $x^2 \neq 0$, we are done, because the definition of zero divisors guarantees then that the product is zero.
Otherwise, let $x,y$ be two distinct zero divisors. Then we know that $xx=yy=0$.
This shows that $x(xy)=0$ and $(xy)y=0$.
If $xy=0$, we are done, if $xy=x\neq y$ then $(xy)y=0$; otherwise $x(xy)=0$ is the product of non-zero distinct elements.  
Case 3: The ring contains exactly one zero divisor $x$. Let $R^X=\{ y_1,..,y_k,x \}$. Then since $y_1,..,y_k$ are not zero divisors we have 
$y_1..y_kx \neq 0$. But this is a zero divisor, thus 
$$y_1..y_kx=x \,.$$
In this case, the product is the unique zero divisor.
A: $0$. Since $0$ is an element, it is in the product of all elements (this is true since a ring $R$ is an abelian group under $+$, so has an additive identity).
EDIT: I agree with DoctorBatmanGod that rings are generally very complicated objects. If we're talking about a ring that isn't an integral domain, then the result may or may not be $0$, for example, $\Bbb Z_4$ and $\Bbb Z_6$, respectively. Even if we restrict ourselves to the units in an arbitrary ring, then we might end up with an empty product since we can choose a ring that is comprised of zero divisors. On the other hand, if we are talking about integral domains, then it is possibly easier as then it is a field. However, that need not get us out of the woods, so to speak. In a boolean ring, every element is its own multiplicative inverse and additive inverse, hence the product of all of the nonzero elements will precisely be $x_1\cdots x_n$. Thus, there isn't any reduction we can do aside from calculate the entire product.
End result: I'd say that there isn't a general method that applies to all rings, even when they're finite. There are too many additional structures on the algebra portion that affect what you want.
A: For commutative rings:
1) If $\,R\,$ has non-zero divisors of zero then the product's obviously zero unless all of the zero divisors are nilpotent, which can occur only with $\,\Bbb Z/4\Bbb Z\,$ (why?) , and in this case the product equals $\,2\,$=the non-zero nilpotent element.
2) If $\,R\,$ is an integral domain then it is a field (why? This is a nice exercise), and thus the set of all its non-zero elements is a multiplicative group. 
i) First subcase: $\,|R|=p^n\,\,,\,p\,$ an odd prime. Since there exists only one element of order 2 here (why?), namely $\,-1\,$ , the wanted product is precisely $\,-1\,$ (why?)
ii) Second subcase: $\,|R|=2^n\,$ . Since here all the non-zero elements have odd multiplicative order, the result is $\,1\,$
A: An arbitrary finite ring in an incredibly complicated object. If the ring isn't an integral domain then the answer is $0$. Otherwise the simplest representation would probably be exactly as you put the question, $\prod_{r \in R}r$. Even with groups, which are "easier" to manipulate, finding what a product is given elements and rules governing products of those elements can get really tricky. 
