Symmetric Difference with Intersection forms Ring

Theorem
Let $$S$$ be a set.

Then $$\left({\mathcal P \left({S}\right), *, \cap}\right)$$ is a commutative ring with unity, in which the unity is $$S$$.

This ring is not an integral domain.

Proof

 * It has been established that $$\left({\mathcal P \left({S}\right), *}\right)$$ is an abelian group, where $$\varnothing$$ is the identity and each element is self-inverse.


 * From Power Set with Intersection is a Monoid, we know that $$\left({\mathcal P \left({S}\right), \cap}\right)$$ is a commutative monoid whose identity is $$S$$.


 * We have that Intersection Distributes over Symmetric Difference.


 * Thus $$\left({\mathcal P \left({S}\right), *, \cap}\right)$$ is a commutative ring with a unity which is $$S$$.


 * Next we find that $$\forall A \in \mathcal P \left({S}\right): A \cap \varnothing = \varnothing = \varnothing \cap A$$. Thus $$\varnothing$$ is indeed the zero.

As set intersection is not cancellable, it follows that $$\left({\mathcal P \left({S}\right), *, \cap}\right)$$ is not an integral domain.

Alternative Proof
From Power Set Closed under Symmetric Difference and Power Set Closed under Intersection, we have that both $$\left({\mathcal P \left({S}\right), *}\right)$$ and $$\left({\mathcal P \left({S}\right), \cap}\right)$$ are closed.

Hence $$\mathcal P \left({S}\right)$$ is a ring of sets, and hence a commutative ring.

From Intersection with Subset is Subset‎, we have $$A \subseteq S \iff A \cap S = A$$. Thus we see that $$S$$ is the unity.

Also during the proof of Power Set with Intersection is a Monoid, it was established that $$S$$ is the identity of $$\left({\mathcal P \left({S}\right), \cap}\right)$$.

We also note that set intersection is not cancellable, so $$\left({\mathcal P \left({S}\right), *, \cap}\right)$$ is not an integral domain.

The result follows.

Comment
The same does not apply to symmetric difference and union unless $$S = \varnothing$$.

For a start, the identity for union and symmetric difference is $$\varnothing$$ for both, and the only way the identity of both operations in a ring can be the same is if the ring is null.

Unless $$S \ne \varnothing$$, then this can not be the case.

Also note that union is not distributive over symmetric difference. From Symmetric Difference of Unions, $$\left({R \cup T}\right) * \left({S \cup T}\right) = \left({R * S}\right) \setminus T$$.