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 $R$.
On
$0$. 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.
On
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.
On
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 $\,1\,$
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.