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I'm trying to prove that every finite Boolean ring, $R$, is isomorphic to a finite number of copies of $\mathbb{Z} / 2 \mathbb{Z}$:

\begin{align} R \cong \mathbb{Z} / 2 \mathbb{Z} \; \times \cdots \; \times \; \mathbb{Z} / 2 \mathbb{Z} \end{align}

I'm aware that there are answers to this question on this website. However, almost all of the answers seem to use facts about modules and vector spaces defined over fields. This question is posed in section 7.6 in Dummit and Foote's Abstract Algebra. This section is part of the chapters which introduces the basics of ring theory (rings, homomorphisms, ideals, Chinese Remainder Theorem etc.).

I'm not even sure how to being answering this question. I know that if $R$ were an integral domain, then we would have that $R \cong \mathbb{Z} / 2 \mathbb{Z} $.

Please only give hints, considering that that the statement has to be proved using the material only included up till chapters 9 in the textbook.

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    $\begingroup$ This might be helpful, see specifically the links in the comment of Bill Dubuque. $\endgroup$ – Krish Sep 13 '17 at 15:37
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This answer assumes that you already believe Boolean rings are commutative. That proof appears elsewhere on the site and is well-known.

I'm also assuming that you believe it has an identity, which can also be proven in various ways, but it is usually assumed.

I'll also try to write it so that you can try to quit reading at any point to pursue the idea I'm describing, if you've decided you've read enough for a hint. Good luck.

I think the easiest angle to pursue is to use idempotents. In a Boolean ring, every element is idempotent, that is, $x^2=x$.

The thing to notice is that if $e$ is idempotent,

  1. $eR$ is a commutative ring with identity $e$;
  2. $1-e$ is also an idempotent; and
  3. $R=eR\oplus (1-e)R$

None of these involves anything more than basics of ring theory that you mentioned.

Now, of course you see that the idempotents $\{0,1\}$ give trivial splittings of $R$ into $R\oplus\{0\}$ or $\{0\}\oplus R$, so the interesting cases are when you have an idempotent $e\notin\{0,1\}$.

Take your finite Boolean ring and start splitting into smaller pieces this way. Obviously if $e\notin\{0,1\}$, the pieces $eR$ and $(1-e)R$ have strictly fewer elements than $R$. Each piece that you get is another finite Boolean ring (obvious, right?)

This splitting can't go on forever. The question becomes: when do I hit bottom? Obviously, if your ring had two elements (just the additive and multiplicative identities) you would be down to $\mathbb Z/2\mathbb Z$ and done.

So what if your ring has more than two elements? Then it has an element $x$ which is not the additive or multiplicative identity, and $x^2=x$, so you can again split using the idempotents $x$ and $1-x$ into a strictly smaller ring. This establishes that you can't split anymore precisely when the piece is a copy of $\mathbb Z/2\mathbb Z$.

So there you have it: you can refine $R$ over and over again into smaller pieces until it is a direct sum of copies of $\mathbb Z/2\mathbb Z$.

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  • $\begingroup$ Yes, I was able to follow through with the proof now. But I'm having trouble motivating that at each step the ideal $Re$ must have $1$. If that isn't true, the construction breaks down. $\endgroup$ – Junaid Aftab Sep 16 '17 at 16:28
  • $\begingroup$ @JunaidAftab I told you what the identity is: $e$. $\endgroup$ – rschwieb Sep 16 '17 at 16:44
  • $\begingroup$ Oh, my bad. I was trying to force that $1$, the identity of the original ring, $R$, must be in $Re$. That forces $Re$ to generate the whole of $R$, meaning no progress is made at factorizing $R$ at each step. $\endgroup$ – Junaid Aftab Sep 16 '17 at 16:48
  • $\begingroup$ Thanks for the response. $\endgroup$ – Junaid Aftab Sep 16 '17 at 16:48

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