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I have some questions about finite rings of sets and I'll be very grateful for any help.

Let E be some fixed non-empty set. Suppose we are given some finite ring of subsets of set E, i.e. some non-empty system $S \subset E$ such that

$\forall A,B \in 2^E ~~ A \vartriangle B \in S$

$\forall A,B \in 2^E ~~ A \cap B \in S$

1) Does it necessarily have a unity? (or in more abstract form: is there necessarily a unity in the finite commutative ring in which multiplication is idempotent?)

2) Suppose we're given some finite system of subsets of E. Is there any algorithm for building the minimal ring of sets which contains this system? I know that from n sets using union, intersection and set difference we mat build at most $~2^{2^n}$ different sets so this ring must be finite.

3) If we know that S is a ring with unity (and so it's a boolean ring) how can we build an isomorphism from our ring of sets to some ring $ B^n = (\{0,1\}^n, +, \cdot)$ where

$0 + 0 = 1 + 1 = 0$

$0 + 1 = 1 + 0 = 1$

$0 \cdot 1 = 0 \cdot 0 = 1 \cdot 0 = 0$

$1 \cdot 1 = 1$

Thanks in advance!

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1)… – user26857 May 14 '13 at 17:24
@YACP I was just going to post the same reference :-) And I believe that my answer there covers both 1) and 3) – Andreas Caranti May 14 '13 at 17:25
up vote 1 down vote accepted

Some elementary help seeing why it has unity:

Notice that $(A\vartriangle B)\vartriangle(A\cap B)=A\cup B$, so the ring actually contains finite unions of its members. Since the ring is finite, $E:=\cup S\in S$. Therefore, $E$ contains all members of $S$, and so $E\cap A=A$ for all $A$.

This also gives a partial start on item 2. Suppose that $G$ is a collection of sets that you want to generate a ring of sets with. Notice that both operations only produce sets which have elements in the two sets you started with. So, any set you generate will not escape $\cup G$. Then the generated ring is contained inside the powerset $\mathcal{P}(\cup G)$.

Ring theoretically, the collection of finite sums of finite products of generators produces the ring.

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