Subject: On the recent expulsion of Jack Chua from the housing system

As you can imagine, the Save Jack effort has been quite strong. Thankfully, housing has extended the time during which he can appeal the decision (presumably due to the end-of-quarterness). There's a petition going around, and Scott, who's in Student Government, has sent to the listhost the email addresses of various officials we can complain to. So far I've used complaints.uchicago.edu, and emailed Katie Callow-Wright (Director of Housing), Jim Wessel (Assistant Director), and the Student Ombudsman. He also suggested Susan Art, Dean of Students, but I didn't write her because she already knows of me with, according to my Hum teacher, an attitude problem, and, more directly, a person who forgets to schedule meetings when he's supposed to and then, when he actually does, doesn't pay attention to whether or not he can actually attend them, and ends up missing them. He also suggested Bill and Fiona, but I think they know well enough what the house thinks.

-Sniffnoy

Let R be a ring where a^7=a forall a. Then R is commutative.

First, what we know from what we've done before: 42a=0 ∀a. Idempotents are central. a^6 is idempotent ∀a. (From now on, if I don't quantify a variable, I am just implicitly universally quantifying it, OK?) R/2R is commutative as (7+1)/2=4 and if R a ring where a^4=a ∀a, then R commutative. Thus 21a central.

OK. So, (10a^3+11a^6) is idempotent. a^6 central, so 10a^3 central. 21a^3 central so a^3 central. (a+a^2)^3=a^3+3a^4+3a^5+a^6 central, so 3a^4+3a^5 central. Plugging in a^5, 3a+3a^2 central. Now this gets us 3xy=3yx like in the n=4 case by considering x(3(x+y)+3(x+y)^2) (It's the same argument but with the coefficient 3 stuck on everything.) So 3a central.

Now, let's look at R/3R. (a-a^3)^3=0, so a=a^3 - but we already have it for n=3, so R/3R commutative, 14a central, a central. Done. (This part used to be much messier.)

Wait... this gets me thinking... take n (greater than 1, of course) and take a p st p-1,p|n-1. (Of course, we won't so often have such p, but...). Then letting k=(n+p-1)/p, we get (a-a^k)^p=0, so a-a^k=0, a=a^k. So if it's true for k, then in the case of n it's true for R/pR (as I did above - n=7, p=3 - and as in fact I've done in general with odd numbers, p=2.) This is actually really useful. Well, maybe. OK, it'll be occasionally useful. (At first I thought this proved it for all n, or at least a lot of n, then I realized my conditions were wrong. :P ) I doubt there are many n for which it could be proven entirely this way, but it's still a useful lemma. Question: For what numbers could it be proven entirely this way? Well, 2*3*7*43+1=1807 works: If it's true for 43, 259, 603, and 904, then it's true for 1807. Amazing. :P

So, OK, I don't immediately get it for any other n, but I do have a useful lemma and I have the n=7 case. Which isn't a whole lot, but...

-Sniffnoy

The list so far:
2,3,4,5,6,7
2^n+2
p^n+1 where p≡3,5 mod 8