TITLE: Directly indecomposable rings
QUESTION [11 upvotes]: 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.)

REPLY [5 votes]: 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.