Quotient ring and Boolean algebra correspondence

While reading a probability paper titled Gröbner bases and factorisation in discrete probability and Bayes (can't find free version, sorry), I came across the explanation:

The set of indicator functions $I_i$ form a Boolean algebra that is in one-to-one correspondence with the quotient of the ring $\mathbb{R}[x_1, \ldots, x_n]$ with respect to the relationships $x_i(x_i - 1)=0$ where $x_i$ represents $I_i$.

$I_i$ is defined to be the indicator function $I_i(\omega) = 1 \;\textrm{if}\; \omega \in A_i \;\textrm{else}\; 0$, $i = 1, \ldots, n$.

I have a rudimentary understanding of rings, but can you please explain the correspondence shown here?

Cheers

Edit: In the rest of the paper, the authors don't seem to limit themselves to $\mathbb{F}_2$. In fact, I can't see a need for the use of the minus sign in $x_i(x_i-1)$ if the field is $\mathbb{F}_2$. I can't verify if the coefficients of the polynomials are disregarded yet.

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2 Answers

A nice exposition of Stone's cryptomorphism between Boolean algebras and Boolean rings is in Burris and Sankappanavar's textbook A Course in Universal Algebra. There you'll also find a presentation of Boolean products - a powerful tool that facilitates the transfer of Boolean algebra results to other algebraic structures - leading to very general representation theories.

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I assume $i$ is an element of $\{ 1, 2, ... n \}$. The correspondence comes from identifying $x_i$ with the indicator function of $\{ i \}$.

But the sentence you quote is imprecise. If you wanted a Boolean algebra (more precisely, a Boolean ring), you would take the base field to be $\mathbb{F}_2$, not $\mathbb{R}$. If you wanted real multiples of indicator functions, you wouldn't say "the set of indicator functions."

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