Post #1186301

Ganesan's Theorem: If R is a commutative ring with exactly n > 0 zero divisors, then |R| ≤ (n+1)^2. (Conventions: R does not necessarily have a unity; 0 itself is not a zero divisor.) Proof: Let a_0 = 0, let a_1,...,a_n be the n zero divisors, and set a := a_1. Let b ≠ 0 be such that ab = 0. For each x in R, (xa)b = 0, and thus xa = a_i for some i = 0,...,n. For each i, let A_i = {x | xa=a_i }. Now suppose |R| ≥ (n+1)^2+1 = n(n+1)+(n+2). By the pigeonhole principle, some A_i has at least n+2 elements, say, r_1,...,r_{n+2}. Then the n+1 elements r_1-r_2,...,r_1-r_{n+1} are nonzero and distinct, and satisfy a(r_1-r_i)=0 for each i. This contradicts the assumption that there are exactly n zero divisors. Therefore |R|≤(n+1)^2. QED This is not Ganesan's proof, which, although easy, is not as elementary. Commutativity isn't important; the proof actually shows that a not necessarily commutative ring R with exactly n>0 *left* zero divisors has order no more than (n+1)^2. I can't take 100% credit for the proof. The basic idea for n=1 and n=2 appeared in a Quora answer by computer scientist David Ash in response to a question asking if there are noncommutative rings with unity with exactly two zero divisors. (Answer: no, because there are no noncommutative unital rings of order less than 8.) I noticed the connection of Ash's argument to Ganesan's Theorem and went from there.