product of rings I feel a need to apologies for this question, since it seems to be to basic to be asked. 
in this question I am primarily concerned with commutative rings and therefore all rings here are assumed to be commutative (and unital of course), though the non-commutative analogue is also interesting.
a ring $R$ is called connected if there are no two non-zero rings $R_1$ and $R_2$ such that $R\cong R_1\times R_2$ (the terminology comes from the fact that $R$ is connected precisely when $Spec R$ is connected as a topological space). the question is about the possibility to represent a general ring $R$ as a (not necessarily finite) product of (non-zero) connected rings. two questions are in place
1) given a ring $R$, does such representation always exist?
2) if it exists, is it unique (in an appropriate sense)?
I think I was able to prove it for noetherian rings (for which such a product will always be finite) by showing the existence of "minimal" non-zero idempotents and that there are finitely many of them, in a rather technical but very routine and straightforward way . this already makes me feel a bit uncomfortable, since in atiyah-macdonald in the proof of the structure theorem for artin rings (p.90), the uniqueness part is proved by a very specialized argument and using "heavy guns" like primary decomposition. since artin rings are notherian and local rings are connected (and this was established in the book before p.90) it would seem more natural to deduce it from the more general and interesting by itself statement with an easy and elementary proof.
so, to conclude, I would like know if I am right that the answer to both questions is positive for noetherian rings and whether it can fail for general rings.
 A: The positive result in this direction is that every ring is a subdirect product of subdirectly irreducible rings.
Counterexamples to your version are plenty among Boolean rings (essentially any Boolean ring which is not a product of copies of $Z_2$ is a counterexample). Google has found me a pretty good reference on all this stuff.
A: I think a non-Noetherian counterexample comes from choosing an algebraic closure $K = \overline{\mathbb{F}_p}$ of the finite field $F = \mathbb{F}_p$, and setting $R = K \otimes_F K$.  The spectrum of $R$ is the absolute Galois group scheme of $F$ over $\operatorname{Spec} K$, and is isomorphic to $\widehat{\mathbb{Z}}$ as a topological group.  The underlying topological space has continuum cardinality, and is totally disconnected with no isolated points.  In particular, it does not have a topological disjoint union decomposition into connected components.  The ring $R$ is not a continuum product of nonzero subrings, since (among other reasons) it is countable.
Edit: For any fixed positive integer $n$, one may choose $n+1$ nonempty closed and open subsets of $\operatorname{Spec} R$ to decompose $R$ as a product of $n+1$ nonzero rings.  Therefore, there is no product decomposition of $R$ into finitely many connected factors.  $R$ cannot be expressed as an infinite product of nonzero rings (connected or not), since it is a countable ring.
A: For a locally noetherian topological space $X$, the connected components of $X$ are open (EGA I, 6.1.9), in particular $X$ is a disjoint union of connected spaces (namely its connected components). Every locally noetherian scheme $X$ is locally noetherian as a topological space, and thus it is a disjoint union of connected locally noetherian schemes. If $X$ is also quasicompact, this has to be a finite union. In particular, every noetherian ring $R$ can be written as a finite product of connected noetherian rings.
The uniqueness is rather clear in the topological picture, because the summands / factors are just the connected components.
A: Consider
$$ R = \{a\in\prod_{k=1}^\infty\mathbb{Z} :
    \exists n \;\forall j,k>n \quad a_j=a_k\} = 
     c(\mathbb{Z}) + \bigoplus_{k=1}^\infty \mathbb{Z}
$$
(where $c:\mathbb{Z}\to R$ is the inclusion of constant sequences). 
This is countable, and so cannot split as an infinite product of nontrivial rings.  Moreover,
for each $a\in R$ there exist integers $n,m>0$ such that $\prod_{k=-n}^n(ma-k)=0$.  This is another reason why $R$ cannot split as an infinite product of copies of $\mathbb{Z}$, 
(This is more or less the same as Qiaochu Yuan's comment to Mark Meilstrup's answer.)
