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Suppose we have a finite, boolean ring $A$ induced by a finite, commutative, boolean monoid $X$ containg $0$ as in:
this question. You only need the first few paragraphs of that long post.

Suppose I have a principal ideal $(i) \subset A$ and I want to test whether $u \in (i)$ efficiently. That is whether or not you can write $u = ri$ for some $r \in A$. But I can't very well test all posible elements in $r \in A$, so I have to come up with a short cut.

Does math have an answer?

Note that every ideal in a boolean ring is principal. See this answer for proof of it.

Daniel Donnelly
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    “Note that at every ideal in a Boolean ring is principal” That is not true and is not what your reference says. You omitted the finite generation condition. I know in your case you are only interested in finite rings but it’s still not good to omit it. – rschwieb Oct 23 '19 at 03:05
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    What is the stuff after “induced by...” supposed to contribute? What special subclass of the already very simple class of finite Boolean rings does it single out? – rschwieb Oct 23 '19 at 03:18

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In any Boolean ring at all, $x\in (y)$ if and only if $xy=x$. That seems to be amply efficient in your context.

rschwieb
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