A strange zero divisor

Let S=\{a,b,c\} be a three element set. Equip its power set \mathcal{P}(S) with two operations + and . defined as: A+B=(A-B)\cup(B-A) and A.B=A\cap B, for A,B\in\mathcal{P}(S). Then \left(\mathcal{P}(S),+,.\right) is a ring. The zero-divisors in this ring are the singleton subsets and the two element subsets.

Recall that an element a in a ring is a zero-divisor if a\neq 0 and there exists another b\neq 0 for which a.b=0 (technically, a left zero-divisor).

The zero element in \left(\mathcal{P}(S),+,.\right) is obviously the empty set \varnothing. In addition, any singleton subset has an empty intersection with another singleton subset: \{a\}\cap\{b\}=\varnothing (we’ve assumed implicitly that a\neq b).

Consider a two element subset, e.g. \{a,b\}. We have \{a,b\}\cap\{c\}=\varnothing. Similarly, \{b,c\}\cap\{a\}=\varnothing and \{a,c\}\cap\{b\}=\varnothing. So the two element subsets are also zero-divisors.

The entire set itself, namely \{a,b,c\} is not a zero-divisor.

What is strange about these zero divisors as the title suggests? Nothing.

This concludes today’s snippet. Thanks for visiting!