Symmetric Difference with Intersection forms Ring

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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 1

From Symmetric Difference on Power Set forms Abelian Group, $\struct {\powerset S, *}$ is an abelian group, where $\O$ is the identity and each element is self-inverse.

From Power Set with Intersection is Monoid, $\struct {\powerset S, \cap}$ is a commutative monoid whose identity is $S$.

Also Intersection Distributes over Symmetric Difference.

Thus $\struct {\powerset S, \cap}$ is a commutative ring with a unity which is $S$.

From Intersection with Empty Set:

$\forall A \in \powerset S: A \cap \O = \O = \O \cap A$

Thus $\O$ is indeed the zero.


However, from Set Intersection Not Cancellable, it follows that $\struct {\powerset S, *, \cap}$ is not an integral domain.

$\blacksquare$


Proof 2

From Power Set is Closed under Symmetric Difference and Power Set is 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 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.

$\blacksquare$


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


Sources