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I am reading Atiyah, Macdonald's book "Commutative algebra".

There is an exercise, which states that that if a ring with identity has idempotent element $\ne 0,1$, then the ring is a direct product of 2 other rings (for the proof if $x$ is this idempotent, consider $A = xA \oplus (1-x)A$). In this paper it is proved that a boolean ring with identity is isomorphic to a subdirect product of the copies of $\mathbb{Z}/2\mathbb{Z}$.

My question is: am I right that the finite boolean ring is in fact isomorphic to direct product of the copies of $\mathbb{Z}/2\mathbb{Z}$?

Here is my proof: consider a boolean algebra $A$, all it's elements are idempotents, so if it is not $\mathbb{Z}/2\mathbb{Z}$, there is decomposition of this ring into $2$ rings (they are also boolean, since homomorphic image of boolean ring is obviously boolean), we proceed with them similarly, until we obtain a decomposition of $A$ into a product of copies of $\mathbb{Z}/2\mathbb{Z}$.

If the statement I made is false, please, say where did I make a mistake.

Thank you very much!

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Sorry guys, I forgot to say that I am only interested in finite case. I will edit it now. –  Sergey Finsky Feb 16 '13 at 15:26

1 Answer 1

up vote 2 down vote accepted

It is correct that any finite boolean ring is a finite product of copies of $\mathbb{F}_2$, and in that case your proof works. However, the infinite boolean rings are far more interesting! For example, inside $\mathbb{F}_2^{\mathbb{N}}$, you can consider the subring generated by $e_1,e_2,\dotsc$, i.e. the vector space generated by $1,e_1,e_2,\dotsc$. It consists of those sequences whose support is finite or cofinite.

More generally, for every totally disconnected compact hausdorff space $X$, aka Stone space, the ring $C(X,\mathbb{F}_2)$ of continuous functions $X \to \mathbb{F}_2$ is a boolean ring, and the famous Stone duality asserts that $X \mapsto C(X,\mathbb{F}_2)$ establishes an anti-equivalence between the category of Stone spaces and the category of boolean rings. The Stone-Čech compactifications of discrete spaces correspond to products of $\mathbb{F}_2$ (in particular, the finite discrete spaces correspond to finite products of $\mathbb{F}_2$), but there are of course more Stone spaces resp. boolean rings. The ring mentioned above corresponds to $\mathbb{N} \cup \{\infty\}$, the one-point compactification of $\mathbb{N}$.

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Thank you! Yes, the reason I tangled up is that I haven't notice that the proof of the theorem in the paper, I provided, was for general(infinite) case. –  Sergey Finsky Feb 16 '13 at 15:29
    
The content of the paper is trivial with some basics of commutative algebra. A boolean ring is reduced, and the only boolean integral domain is $\mathbb{F}_2$. Hence we get an embedding $R \to \prod_{\mathfrak{p}} R/\mathfrak{p} = \prod_{\mathfrak{p}} \mathbb{F}_2$. But this is not a classification, in my opinion, it is only the beginning. One has to notice that the image consists precisely of the continuous functions $\mathrm{Spec}(R) \to \mathbb{F}_2$, so that $R \cong C(\mathrm{Spec}(R),\mathbb{F}_2)$. –  Martin Brandenburg Feb 16 '13 at 15:34

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