Symmetric Difference with Intersection forms Ring/Proof 1

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.

Proof

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 $S$.

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 $\struct {\powerset S, \symdif, \cap}$ is not an integral domain.

$\blacksquare$

Sources

• 1970: B. Hartley and T.O. Hawkes: Rings, Modules and Linear Algebra ... (previous) ... (next): Chapter $1$: Rings - Definitions and Examples: $2$: Some examples of rings: Ring Example $6$