Directly indecomposable rings

Solution 1:

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.