I want to know if this set is a field.

For a ring to be a field, Wikipedia defines the requirement that every non zero element be invertible. In this set, every element that is not a factor of zero is invertible, so it is slightly different from the wikipedia requirement. ]]>

Having claimed victory over the devil, I find myself amusingly playing devil’s advocate to try and find a flaw in my paper. The binomial theorem works for negative integers. Does it work when one of the integers is 0?

My problem is to analyse the sum of two pth powers of integers, p an odd prime. We will do it with residues of these integers modulo p^k. The integers :a:b: are prime to p, but their sum has a factor p, and possibly a facot with higher powers of p.

One stratagem generally employed is to express

the sum as ((a+b)-b)^p +b^p and expand out the first term with the binomial theorem. The b^p terms cancel out leaving a factored expression (a+b)() and you prove the () factor has a single factor p.

Now I can prove what a and b must be for all the various combinations, but my factorisation is of the form (a+b)() where () does not have the factor p. The devil points this out, but I reply I can take my solution :a:b: mod p^k and exponentiate it using the stratagem and the result will be a residue expression mod p^k of (a+b)p(). Is this also valid if a+b=0 mod p^k, i.e. the stratagem forms (0)p() mod p^k? If not, I need to add another section. ]]>

My last effort is as follows.

I form a+b residues mod p^2 on a 1:1 basis from other residues mod p^2. I end up with 10 say mod p^2. 5 of them don’t add to zero and can be rejected. The other 5 add to zero and are formed by adding two residues mod p^2. I can factor these as p.(a+b) where a+b is one of the rejected a+b pairs.

Since I am not factorising an integer, but a residues sum, I contend that the (a+b) factor in p.(a+b) is a residue mod p. I cannot make it an (a+b) residue mod p^2, and therefore I maintain my 1:1 relationship with a+b mod p^2.