Let n = pq, p and q not necessarily primes. The following are necessary conditions for the Frobenius map to be a homomorphism:
These are all found by expanding (1+1)^n, (p*1 +1)^n, and (q*1+1)^n by the binomial theorem. What I find rather interesting is that (p*1 + q*1)^n = (p*1)^n + (q*1)^n, again found by expanding the binomial theorem.
]]>Characteristic also makes sense for rings with zero-divisors, so that's a good question to ask Jane. The proof of the theorem involves use of the binomial theorem along with the fact that if d is not 0 or p, then p divides p choose d. This simplifies the expression the binomial theorem provides considerably, leading directly to the conclusion. This fact does not hold if p is not prime, which would lead one to believe that the theorem doesn't hold in cases where p is not prime. The following counter-example justifies that thought:
In the integers modulo 4, (1+1)^4 = 2^4 = 16 = 0 (mod 4). However, 1^4 + 1^4 = 1 + 1 = 2.
While this counter example shows that there are rings of non-prime characteristic where the theorem does not hold, is this true for all rings with non-prime characteristic? This I'm not sure of, but I believe you can find a proof based off the fact that any such ring must have zero divisors. In particular, if the characteristic of the ring is n = pq, then (p*1)(q*1) = (pq)*1 = n*1 = 0, and since p, q < n we have that p*1 and q*1 must both be nonzero.
]]>By the way, the definition requires the characteristic of the ring
to be prime. What if the characteristic is a non-prime positive integer? will still be a homomorphism, wont it? ]]>This property has important implications in Field theory and Galois theory.
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