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.)

  • $\begingroup$ I believe, I am currently down to proving (1) $B(B(R))=B(R)$ and (2) the finite-cofinite boolean algebra is not isomorphic to a direct product of fields. $\endgroup$ – Jack Schmidt Jun 2 '13 at 20:32
  • $\begingroup$ In case both (1) and (2) are actually false, another possible ring is the subring of $\mathbb{Z}/2\mathbb{Z}^\omega$ consisting only of those sequences that are periodic with a period a power of 2. $\endgroup$ – Jack Schmidt Jun 2 '13 at 20:44
  • $\begingroup$ I haven't figured out if (1) and (2) work. But the periodic thing actually worked just fine. $\endgroup$ – Jack Schmidt Jun 2 '13 at 21:04

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.

  • $\begingroup$ For the bottom, yes, that is Birkhoff's theorem, page 205 of Lam's First Course. Original paper is maybe Birkhoff (1944) ams.org/mathscinet-getitem?mr=10542 -- appears to be generally true, somewhat by definition, but is a good replacement for the direct product idea that fails. $\endgroup$ – Jack Schmidt Jun 3 '13 at 20:21
  • $\begingroup$ @JackSchmidt It's a shame subdirect products aren't exercised more often. I remember only a few results about them, but it seems like we should use them more often. $\endgroup$ – rschwieb Jun 3 '13 at 20:25
  • $\begingroup$ I wonder if "every abelian group is a subdirect product of cocyclic groups" is just an expression for the inclusion of an abelian group in its divisible hull = injective envelope. Maybe we do use equivalent ideas, but without realizing it. $\endgroup$ – Jack Schmidt Jun 3 '13 at 20:34
  • $\begingroup$ @JackSchmidt I feel like this is every application I know: semiprimitive rings are subdirect products of primitive rings; reduced rings are subdirect products of domains; strongly regular rings are subdirect products of division rings. I guess it is a boon to those who like structure. $\endgroup$ – rschwieb Jun 3 '13 at 20:38

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.


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.

  • $\begingroup$ I wanted to bounce something off you which I had trouble analyzing: suppose in $\prod_{i\in\omega} F_2$ you consider the subring $R$ generated by $\oplus_{i\in\omega}F_2$ and the identity. It surely has indecomposable direct factors (maybe they are all isomorphic to $F_2$?), but I can't see how it would be written as a product of directly indecomposable rings. Did you happen to process this example as you thought about this problem? $\endgroup$ – rschwieb Jun 3 '13 at 13:58
  • $\begingroup$ I believe that is another way of describing the (2) ring of finite-cofinite sets. I was unable to convince myself either way, mostly because of the existence of some but probably not enough indecomposable direct factors. In other words, the associated lattice has atoms (but not enough atoms?). I realized that I personally found a lack of any atoms much more convincing than a lack of suprema of certain infinite sets. You probably know: can you prove your ring $R$ is not a direct product of fields? (That would answer (2) for me, and be very helpful.) $\endgroup$ – Jack Schmidt Jun 3 '13 at 14:07
  • $\begingroup$ Isn't self-injectivity of rings stable under arbitrary products? If so, and if it were a product of fields, then it'd be a self-injective ring. But since it is dense in $\prod_{i\in\omega} F_2$ as an $R$ submodule, it would have to be equal to $\prod_{i\in\omega} F_2$, which is absurd. $\endgroup$ – rschwieb Jun 3 '13 at 14:50
  • $\begingroup$ Thanks! Self-injective is a nice way to look at it. Can “dense” be replaced with “essential”? Alternatively, R is not a direct summand of that direct product, so R is not an injective R-module. $\endgroup$ – Jack Schmidt Jun 3 '13 at 15:16
  • $\begingroup$ Yeah, I meant to say essential, but dense was what came out. This is a von Neumann regular ring, which is nonsingular, and if my memory isn't failing, density and essentiality coincide for nonsingular rings. Regardless of that, essential was what I was getting at :) $\endgroup$ – rschwieb Jun 3 '13 at 15:44

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