# Is every commutative ring isomorphic to a product of directly irreducible rings?

In the following, all rings are assumed to be commutative, with multiplicative identity. A ring $$R$$ is said to be directly irreducible if it is not isomorphic to the direct product of two non trivial rings. An equivalent condition is that $$R$$ does not contain any idempotent element other than 0 and 1.

It is not hard to prove that any noetherian ring $$R$$ is isomorphic to a finite direct product of directly irreducible rings (if this wasn't the case, then you could "split" $$R$$ indefinitely and produce an infinite chain of ideals). Moreover, the factors are isomorphic up to reordering. Is it true that every ring $$R$$ is isomorphic to a (possibly infinite) direct product of directly irreducible rings?

• What's true is that there is a notion of "subdirect product" and "subdirectly irreducible" such that every commutative ring is a subdirect product of subdirectly irreducible rings. See this for example. Jul 23, 2020 at 17:53

No. For instance, let $$R$$ be the Boolean ring of subsets of $$\mathbb{N}$$ that are either finite or cofinite. Any quotient of $$R$$ is also a Boolean ring, and the only directly irreducible Boolean ring is $$\mathbb{F}_2$$. But $$R$$ is not a product of copies of $$\mathbb{F}_2$$, for instance because it is countably infinite.
More generally, in any ring $$R$$, the set of idempotent elements form a Boolean algebra $$B$$. If $$R\cong \prod_{i\in I}R_i$$ is a product of directly irreducible rings, then $$B$$ would be isomorphic to the power set algebra $$\mathcal{P}(I)$$. So, if $$B$$ is not a power set algebra, then $$R$$ cannot be a product of directly irreducible rings.
Note moreover that if $$R\cong \prod R_i$$ is a product of directly irreducible rings, then the projections $$R\to R_i$$ are exactly the quotient maps $$R\to R/(1-e)$$ where $$e$$ ranges over the atoms of the Boolean algebra $$B$$ (i.e., the minimal nonzero idempotents of $$R$$). So, a ring $$R$$ is isomorphic to a product of directly irreducible rings iff the canonical map $$R\to\prod_{e}R/(1-e)$$ is an isomorphism, where $$e$$ ranges over the atoms of $$B$$ (note that such a quotient $$R/(1-e)$$ always is directly irreducible).
Using this criterion, here is an example of a ring that is not a product of directly irreducible rings even though its Boolean algebra of idempotents is a power set algebra. Let $$k$$ be an infinite field, let $$I$$ be an infinite set, and let $$R$$ be the ring of functions $$I\to k$$ that take only finitely many values. Then the Boolean algebra of idempotents in $$R$$ is $$\mathcal{P}(I)$$, since the characteristic function of every subset of $$I$$ is in $$R$$. However, the quotient maps $$R\to R/(1-e)$$ for atoms $$e$$ are exactly the evaluation maps $$R\to k$$ at elements of $$I$$, so the canonical map $$R\to\prod R/(1-e)$$ is just the inclusion $$R\to k^I$$. Since $$R$$ is not all of $$k^I$$, it cannot be a product of directly irreducible rings.