Directly indecomposable rings Is every ring the (possibly infinite) direct product of directly indecomposable rings?
I believe the answer is no, but I'm not positive and don't know any explicit examples.

A reduction: If $R$ is a unital, associative ring, then define $B(R)$ to be set of all central idempotents of $R$, $B(R) = \{ e \in R: e^2 = e, er=re ~(\forall r \in R) \}$. $B(R)$ is a ring under the operations $e \oplus f = e+f -ef$ and the standard multiplication from $R$. 
If $R$ was a direct product of directly indecomposable rings $R_i =  Re_i$ for $i \in I$, then the elements of $B(R)$ are exactly the $e_J$ for $J \subseteq I$; $e_J \oplus e_K = e_{J \oplus K}$ where $J \oplus K$ is symmetric-difference; and $e_J \cdot e_K = e_{J \cap K}$.
In particular, $B(R) \cong \mathbb{Z}/2\mathbb{Z}^I$ is what is called a complete boolean algebra. Hence any ring in which $B(R)$ is not a complete boolean algebra is an example.
If $B(B(R)) = B(R)$, then I think this will basically work, though I still wouldn't mind the details being stated clearly (especially in algebraic language).
Anderson–Fuller page 102 provides an example to those who understand the topology of $\mathbb{Q}$, but I'm not such a person. I think the finite-cofinite boolean algebra is likely more my speed, but I'm not sure it is not isomorphic to a complete boolean algebra.
(I think the struck out portions might be wrong; I'd appreciate corrections.)
 A: Let $R$ be the subring of $\prod_{i\in\omega}\Bbb F_2$ generated by $\oplus_{i\in\omega}\Bbb F_2$ and the identity. (This is just the unitization of the ideal $\oplus_{i\in\omega}\Bbb F_2$, and it basically looks like $\{x+kI\mid x\in \oplus_{i\in\omega}\Bbb F_2, k\in \Bbb F_2\}$.)
Part 1: 
We claim that any directly irreducible factor of $R$ would have to be $\Bbb F_2$. 
Suppose that $I\lhd R$ is such a factor, and that $x\neq y$ are two nonzero elements in $I$. Since $x\neq y$, you can pick an idempotent $e$ in $R$ such that $ex\neq 0$ and $ey=0$. (Just use the projection on a coordinate where they differ.) But then $eI\oplus(1-e)I$ is a nontrivial decomposition of $I$, as $0\neq x\in eI$ and $0\neq (1-e)y\in (1-e)I$. Since this is impossible, $I$ is just a copy of $\Bbb F_2$.
Part 2:
$R$ isn't a direct product of copies of $\Bbb F_2$. 
Note first that $R$ is an essential $R$ submodule of $\prod_{i\in\omega}\Bbb F_2$. If $R$ were a direct product of fields, it would be a self-injective ring, but it can't be self-injective and have the proper $R$ module extension $R\subsetneq \prod\Bbb F_2$. (Alternatively, one can say that the injective submodule $R$ should split out of $\prod\Bbb F_2$, but it can't because a nontrivial essential submodule can't split the big module.)
So, $R$ is another example of a ring which can't be decomposed into directly irreducible rings.

While we're near the topic, I think I'm remembering correctly that every ring does decompose into a subdirect product of subdirectly irreducible rings.
A: Let $R \subseteq \mathbb{Z}/2\mathbb{Z}^\omega$ consist of all sequences $(a_n)$ such that there is some $k$ with $a_n = a_{n+2^k}$ for all $n=0,1,\dots \in \omega$. The operations (addition and multiplication) are coordinate-wise. The zero is the constant 0 sequence with $k=0$, and the one is the constant 1 sequence with $k=0$. Every element is a central idempotent.
If $R$ is a direct product of directly indecomposable rings, then it has at least one directly indecomposable direct factor, $S=  R e_S$, with identity $e_S$ and period $k$. If $e_S = (a_n)$ and $a_i \neq 0$ with $i < 2^k$, then consider the element $e_T = (b_n) \in R$ with $b_n = 0$ unless $n \equiv i \mod 2^{k+1}$. Notice that $e_S \neq e_T$ as they differ in the $(i+2^k)$th coordinate, but that $e_S \cdot e_T = e_T$ since the only nonzero entries of $e_T$ occur when the corresponding entry in $e_S$ is 1. Hence we get a non-trivial direct product decomposition of $S = S e_T \times S (1-e_T)$ since $e_T = e_T e_s \in R e_S = S$ is a nontrivial central idempotent of $S$.
Hence, not only does $R$ not have a (possibly infinite) direct product decomposition into directly indecomposable rings, it has no directly indecomposable direct factors.

Example:
Take $e_S = (1,0,0,0,1,0,0,0,1,0,0,0,1,0,\dots)$ with period $2^2$. Then $S=Re_S$ consists of those (periodic) sequences which are nonzero at most in every fourth coordinate. In particular, $e_T = (1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,\dots)$ with period $2^3$ is inside $S$ and $S(1-e_T) = R e_s(1-e_T) = R(e_S - e_T)$ so $e_S - e_T = (0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,\dots)$ with period $2^3$ is also in $S$.
This allows us to write $S$ as the direct product of two of its subrings, those whose nonzero entries are at $0 \equiv n \mod 8$ and those at $4 \equiv n \mod 8$.
Of course each of these subrings is a direct product of rings defined by similar conditions mod 16, and those are direct products mod 32, etc.
