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sniffnoyFirst, 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
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
no subject
Date: 2006-03-15 07:59 pm (UTC)no subject
Date: 2006-03-16 12:35 am (UTC)no subject
Date: 2006-09-19 04:32 pm (UTC)