The Frobenius endomorphism (and the maps it induces on cohomology) plays a very important rôle in modern mathematics. Despite this it is rather easily described.

Proposition Let K be a field of characteristic p, for a prime number p. Then the function F:K-->K defined by F(a)=ap is a ring homomorphism. If K is a finite field then F is an automorphism.

Definition The ring homomorphism in the proposition is called the Frobenius endomorphism.

Before we prove the proposition we need:

Lemma The binomial coefficient pCn is divisible by p for 0<n<p.

Proof: pCn=p!/n!(p-n)!=p(p-1)..(p-n+1)/n!. Now since n<p none of its prime factors divide p. The same goes for all the other factors of n!. So as the binomial coefficient is an integer n! must divide into (p-1)...(p-n+1) and we see that pCn is divisible by p.

Proof of the proposition:

• F(1)=1p=1.
• F(ab)=(ap)p=apbp=F(a)F(b).
• F(a+b)=(a+b)p= ap+pap-1b+...+ pCr.arbp-r+...+pabp-1+bp. By the previous lemma all the binomial coefficients except the first and last are divisible by p and hence zero in F. It follows that F(a+b)=ap+bp=F(a)+f(b).
Note that F is always injective (this is true for any ring homomorphism from a field to itself). Thus if F is a finite field then by the pigeon-hole principle it must be a bijection, as is required.

Its worth noting a special case of this. For the field Zp of integers modulo p the Frobenius is the identity map that fixes every element. This follows by Fermat's little theorem. The same idea as the proof of the theorem shows that in the ring Zp[x] the Frobenius is a homomorphism. Note that F(f(x))=f(xp).

Log in or register to write something here or to contact authors.