Symmetric Difference with Intersection forms Ring

Theorem

Let $S$ be a set.

Let:

$\symdif$ denote the symmetric difference operation

$\cap$ denote the set intersection operation

$\powerset S$ denote the power set of $S$.

Then $\struct {\powerset S, \symdif, \cap}$ is a commutative ring with unity, in which the unity is $S$.

This ring is not an integral domain.

From Symmetric Difference on Power Set forms Abelian Group, $\struct {\powerset S, \symdif}$ 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 $S$.

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

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

Thus $\O$ is indeed the zero.

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

$\blacksquare$

From Power Set is Closed under Symmetric Difference and Power Set is Closed under Intersection, we have that both $\struct {\powerset S, \symdif}$ and $\struct {\powerset S, \cap}$ are closed.

Hence $\powerset S$ 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 $\struct {\powerset S, \cap}$.

We also note that set intersection is not cancellable, so $\struct {\powerset S, \symdif, \cap}$ is not an integral domain.

The result follows.

$\blacksquare$

1964: Steven A. Gaal: Point Set Topology ... (previous) ... (next): Introduction to Set Theory: $1$. Elementary Operations on Sets
1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $21$. Rings and Integral Domains: Example $21.3$
2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $\S 2$: Problem $6 \ \text{(iii)}$