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Recall: a Boolean ring is a (commutative) ring $R$ where $\forall x \in R: x^2=x$.

I don't really know how to proceed. I have tried some things like

  • If $x,y \ne 0$ in $R$ such that $x^2=x$ and $y^2=y$, then by the fact that $R$ is local, we must have $x=y$, meaning $R$ would be a ring with 2 elements, or $\Bbb{F}_2$.
  • If $x \ne 1$ with $x^2=x$ then by some steps we should conclude $x=1$, by which we conclude the only non-zero element in $R$ is 1, so $R = \Bbb{F}_2$.

I haven't managed to actually prove either of those. Could someone help me out?

rschwieb
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soggycornflakes
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2 Answers2

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Let $R$ be a local ring and $M=R\setminus R^{*}$ its maximal ideal.

  • If $R$ is a boolean (local) ring, then it is a field
    It is sufficient to show $M=0$.
    Let $x\in M$. Note that this implies $x-1\notin M$, otherwise you get $1\in M$, which is clearly absurd. Thus you have that $x-1\in R^{*}$.
    But $x^2=x$ implies $x(x-1)=0$. Since $x-1$ is a unit, it can't be a zero divisor, so necessarily $x=0$. This shows that any boolean local ring is a field.
  • Any boolean domain is isomorphic to $\mathbb{F}_2$
    Let $R$ be a boolean integral domain. We know that $R\neq 0$.
    Let $0\neq x\in R$. We want to show that $x=1$.
    As above we have $x(x-1)=0$ and, since $x\neq 0$ and $R$ is a domain, we get $x-1=0$, which proves the assertion.
Temoi
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  • How can you know that $M = R\backslash R^\ast$? Why is the maximal field necessarily the ring without units? Couldn't it be smaller than that? – soggycornflakes Jan 12 '24 at 15:10
  • Let $x\in R\setminus R^{*}$ such that $x\notin M$. Thus $x$ is contained in a maximal ideal of $R$ and clearly this maximal ideal can't be $M$. This contradicts the fact that $R$ is a local ring. – Temoi Jan 12 '24 at 15:14
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    So in a local ring, if $M$ is the maximal ideal, one always has that $M = R\backslash R^\ast$? – soggycornflakes Jan 12 '24 at 15:15
  • Yes, that's true! Actually it is quite straightforward to see that if $R$ is a ring where $R\setminus R^{}$ is an ideal, then $R$ is a local ring with $R\setminus R^{}$ as maximal ideal. – Temoi Jan 12 '24 at 15:16
  • Thanks for your help! – soggycornflakes Jan 12 '24 at 15:17
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In a local ring the only idempotents are $0$ and $1$. (This is easy: $eR+(1-e)R=R$ but $eR$ and $(1-e)R$ can't be proper at the same time.)

Everything in a boolean ring $R$ is idempotent.

So for every $x\in R$, $x=0$ or $x=1$.

That's $F_2$.

rschwieb
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  • Well, the statement that a local ring only admits $0$ and $1$ as idempotents is true, but I am sorry to say your argument seems not very well-written. A better argument: Assume $e$ is an idempotent. That is $0=e-e^2=e(1-e)$. Since $e+(1-e)=0$, one of them must be a unit. (Otherwise they are both contained in the unique maximal ideal, a contradiction.) On the other hand, if $e\notin{0,1}$, then both $e$ and $1-e$ are non-zero, and both of them are zero-divisors other than zero, and a fortiori are non-units, which is absurd. All in all your answer is very good. – Asigan Jan 13 '24 at 11:30
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    What? $e+(1-e)= 0$? Then $1=0$? This must be a mistake of some sorts but I don't know what you mean. – soggycornflakes Jan 13 '24 at 16:37
  • @Asigan the way I’m suggesting is simpler than that. To fill in another detail, if $eR=R$ then $e$ is a unit, hence 1. For the same reason if $(1-e)R=R$, we have $1-e=1$ so that $e=0$. Sorry you don’t like it but it seems like a good easy exercise to suggest. I like to leave easy matters as exercises sometimes. – rschwieb Jan 13 '24 at 21:58
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    @rschwieb Well, it is my fault that I did not catch your method. So you mean that $e+(1-e)=1$ implies that they can not both be units, while the only idempotent being a unit is $1$ (since, if $e$ is a unit idempotent, then $e=e^{-1}(ee)=e^{-1}e=1$), and this implies $e=1$ or $1-e=1$. This method is also very good. – Asigan Jan 14 '24 at 05:28
  • @soggycornflakes Sorry, I meant $e+(1-e)={\color{SkyBlue}1}$. – Asigan Jan 14 '24 at 05:30