Can one decompose any ring into a (possibly infinite) product of indecomposable rings?

Question

Let $R$ be a ring. Do there exist indecomposable rings $R_i$, $i\in I$, such that $R={\prod}_{i\in I} R_i$?

Answer 1

HINT:

Try $R$ the ring of continuous functions from $C= \{0,1\}^{\mathbb{N}}$ to $\mathbb{R}$.

Answer 2

Let $R\subset\mathbb Z_2^{\mathbb N}$ be the set of sequences with elements in $\mathbb Z_2$ which are constant from a rank on. Then $R$ is a boolean ring. If $R\simeq\prod_{i\in I}R_i$ with $R_i$ indecomposable, then $R_i$ is isomorphic to a subring of $R$, hence $R_i$ is boolean. Moreover, since $R_i$ is indecomposable we must have $R_i\simeq\mathbb Z_2$. This shows that $R\simeq\mathbb Z_2^{I}$, but for cardinality reasons this is false.