# Are all Boolean rings semi-artinian?

I'm basically only interested in the above question. I have played with this a little bit and have not found an obvious way to prove it. However, I am not familiar with ways for proving something is semi-artinian in general (i.e. what seem to be set-theoretic methods, or discussions of the so-called socles).

If this is not generally true, is it likely to be true for the power set of the natural numbers?

Thanks for any help on this.

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Perhaps I should clarify that a semi-artinian $R$ ring is one such that every homomorphic image of $R$ in an $R$-module has a simple submodule. –  Jon Beardsley Oct 14 '12 at 16:39
The socle method is to prove that every nonzero $R$-module $M$ has a nonzero socle. A socle is the sum of the minimal nonzero submodules of $M$. –  Alexander Gruber Oct 14 '12 at 17:00

I don't believe it's true for general Boolean rings. However, for any set $S$, $\mathscr{P}(S)$, the power-set of $S$ (as a Boolean ring) is definitely semi-artinian. Take some element $s\in S$, and look at the image of the submodule $\{\{s\},\emptyset\}$ under any $\mathscr{P}(S)$-module homomorphism. Its image is clearly a minimal submodule.
If you're looking for a specific counterexample, I would guess you may want to start by looking at $R = S/I$ where $S = \prod_J \mathbb{Z}/2$ and $I = \bigoplus_J \mathbb{Z}/2$ for some infinite set $J$. This is Boolean because $S$ is Boolean. –  Manny Reyes Oct 18 '12 at 20:30