Frobenius Endomorphism - Definition

Definition

Let R be a commutative ring with prime characteristic p (an integral domain of positive characteristic always has prime characteristic, for example). The Frobenius endomorphism F is defined by

F(r) = rp

for all r in R. Clearly this respects the multiplication of R:

F(rs) = (rs)p = rpsp = F(r)F(s),

and F(1) is clearly 1 also. What is interesting, however, is that it also respects the addition of R. The expression (r + s)p can be expanded using the binomial theorem. Because p is prime, it divides p! but not any q! for q < p; it therefore will divide the numerator, but not the denominator, of the explicit formula of the binomial coefficients

for 1 ≤ k ≤ p − 1. Therefore the coefficients of all the terms except rp and sp are divisible by p, the characteristic, and hence they vanish. Thus

F(r + s) = (r + s)p = rp + sp = F(r) + F(s).

This shows that F is a ring homomorphism.

In general, F is not an automorphism. For example, let K be the field F_p(t), that is, the finite field with p elements together with a single transcendental element. We claim that the image of F does not contain t. We will prove this by contradiction: Suppose that there is an element of K whose image under F is t. This element is a rational function q(t)/r(t) whose p'th power (q(t)/r(t))p equals t. This makes, which is impossible. So F is not surjective and hence not an automorphism. It is also possible for F to be non-injective. This occurs if and only if R has a nilpotent element ≠0. (An example is F_p/Tp, where F(T)=0, but T≠0.)