Hahaha! I will!

Solid torus - Wikipedia

As unimaginative as it is, "solid torus" seems to be common:
en.wikipedia.org/wiki/Solid_t...

That still doesn't seem right though. Ugh, I think I'm going to have to give up

Warning Forever - Wikipedia

Ah. In that case I wonder if you're talking about Warning Forever:

en.wikipedia.org/wiki/Warning...

How far back in time are we talking here?

Ah, I see. Hrm, that's a tough one.

Was it possibly Bosses Forever?

My younger son's newly created portfolio:

www.adriankinyon.com

Exactly. One more year...

Here is what I just put under "Professional Service" in my annual self-evaluation:

"I wrote about a dozen referee reports last year for various journals. I stopped keeping careful records of these because what is even the point?"

I am fine today.[citation needed]

It's well known to those who know about it

"This is a well known folk result."

I just thanked an LLM for its response, so now I have to walk into the sea. It's OK, I had a good life.

Picture of a Schediphone, a rare brass instrument with two bells.

Check out the Schediphone. No idea what it would sound like.

I've always been partial to Densely Defined Operator myself

"Whereof one cannot speak, thereof one must post." -- Wittgenstein, updated translation

This is not Ganesan's proof, which is not quite 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.

4/4

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
3/4

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).

2/4

About a month ago, I posted

Ganesan's Theorem: If R is a commutative ring with exactly n > 0 zero divisors, then |R| ≤ (n+1)^2.

In this thread I would like to prove it.

(Conventions: R does not necessarily have a unity; 0 itself is not a zero divisor.)

1/4

Just like any tree

The reason scientific publications need a lot of coauthors is so that the starters can take rest days

I can honestly say I've never seen it before

I trust we've all read the classic "If the IRS had discovered the quadratic formula." www.amherst.edu/system/files...

Fun stuff! Had you already discussed earlier which elements in Z_n have square roots or was that just another part of the whole messiness?

That both of these things are true gobsmacked me when I was a student:
* Modules are a special case of abelian groups because every module is just an abelian group with extra structure.
* Modules are a generalization of abelian groups because every abelian group is a Z-module.

Ugh, taxes

It certainly looks very much like him! And the time period is consistent with his travels.

www.irishamerica.com/2020/05/paul...