Let be a ring which is not necessarily commutative or with identity element. Suppose that there exists an integer such that and for all
Prove that for all and hence is commutative.
Firstly note that holds. Thus, I shall prove that , which will prove the result.
Let us forget about rings for a moment and look at . This is clearly equal to . So,
. As this is always a whole number and as then either is even with or and .
Thus, , . This expands to .
Therefore, . Inserting this into our ring we see that this is actually just equal to , as required.
Thus, by Problem 36, every element is central and so the ring is commutative.
I was unsure if this was valid or not - surely is just the polynomial? Although I used 1s when I expanded it the expanded version does not use a 1 (I took the time to say when we were exiting and re-entering the realm of our ring). I used 1s in my expansion of it, but if someone had taken the time to expand it properly without removing the -term they would get the same answer (as it is just a polynomial) but their method would still be valid. Thus, although my method is invalid w.r.t. the ring the answer is valid, and thus my solution is still valid?
EDIT: I have just picked up another problem with my solution. It should be at the start, not just an . Although because everything becomes zero it doesn't really matter.
i probably wasn't very clear in my previous post. in your proof you said which is clearly not correct even if we assume the ring has 1. the correct one is
anyway, since there is nothing to prove if n = 0, we may assume that as you mentioned for all therefore, since for all
we have: