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, $\left({R \cup T}\right) * \left({S \cup T}\right) = \left({R * S}\right) \setminus T$.