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This question wants me to use a previous result, that if $e$ is a central idempotent in $R$ then $R \cong Re \oplus R(1-e)$. I was thinking about maybe some sort of induction would help here but I can't see exactly how I am going to get to the result I want.

I would like to know how to do the problem this way, because this is the way the book recommends. However, I also have heard that this problem can be done another way, by realizing that $R$ is a vector space over $Z_2$ and there has a finite basis, and i am under the impression that the result is not far away knowing this, but I can't quite connect the dots.

If somebody could guide me along these two different routes of solving this problem I would really appreciate it! Thanks!

Math is hard
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  • Also consider looking at https://math.stackexchange.com/questions/305556/boolean-ring-representation-as-direct-product or https://math.stackexchange.com/q/1398323/29335 – rschwieb Sep 25 '18 at 18:53
  • And to argue along the lines you want: Just note that if $e\notin{0,1}$, $Re$ and $R(1-e)$ are also boolean rings with strictly smaller cardinality than $R$, with identities $e$ and $1-e$ respectively. You can decompose such a ring repeatedly until literally the thing you have is ${0,1}$, and no further decomposition can occur. Since $R$ is finite, you can clearly do this for all the pieces of $R$, so you eventually wind up with $\prod F_2$ for some finite number of copies of $F_2$ – rschwieb Sep 25 '18 at 18:56
  • Or perhaps https://math.stackexchange.com/q/2428051/29335 is an even better duplicate. Please, please, please look for duplicates before you post. In this case searching "finite boolean" seemed to suffice. – rschwieb Sep 25 '18 at 18:59
  • @rschwieb You say in your second comment that you can decompose such a ring repeatedly. Forgive me if I'm missing something, but can you explain exactly how you do that? Thank you. – MathsIsFun Mar 15 '20 at 11:21
  • @Emily that was explained in the sentence before the one you are quoting, and the first sentence of the post . $R=eR\oplus (1-e)R$ – rschwieb Mar 15 '20 at 11:23
  • @rschwieb Apologies, I'm still not seeing what to do. If using your sentence before, surely you just end up going in a circle over and over, surely it doesn't reduce down? – MathsIsFun Mar 15 '20 at 11:26
  • @Emily Do you understand that $|eR|$ and $|(1-e)R|$ are strictly smaller than $|R|$ when $e$ is nontrivial? – rschwieb Mar 15 '20 at 11:28
  • @rschwieb I should, but I'm not confident. Would you mind explaining please? – MathsIsFun Mar 15 '20 at 11:31
  • I've also realised I'm working with $R \cong Re \otimes R(1-e)$ rather than $R \cong Re \oplus R(1-e)$ in the question. – MathsIsFun Mar 15 '20 at 11:37
  • @Emily The elements of the decomposition $eR\oplus(1-e)R$ are in one to one correspondence with the Cartesian product $eR\times (1-e)R$, and it is elementary to prove for finite sets $|A\times B|=|A||B|$. I can’t make it any more elementary. You’ll actually have to engage your mind from here on. – rschwieb Mar 15 '20 at 12:30
  • @Emily The ring is commutative so $eR=Re$ and it makes no difference. – rschwieb Mar 15 '20 at 12:32
  • Let us continue this discussion in chat. – MathsIsFun Mar 15 '20 at 19:30

1 Answers1

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  1. A boolean ring is always commutative: $$a+b=(a+b)^2 =a^2 +b^2 + ab + ba = a+b+ab+ba$$ Hence $ab+ba=0$. Applying the same with $a=b=x$ yields $x+x=0$, so $x=-x$ for all $x$, and we have $ba=-ab=ab$.
    And, by definition, also every element is idempotent, so we can arbitrarily pick a nontrivial element $e$, until there is such, and use induction and the previously proved claim.

  2. By the above result ($x+x=0$), we indeed get a $\Bbb Z_2$ vector space structure in $R$, and since it's finite, it has a finite basis. All you have to verify is that multiplication of $R$ coincides with the coordinatewise multiplication once a basis is fixed.

Berci
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