Frobenius endomorphism

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)=a^p 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 ^pC_n is divisible by p for 0<n<p.

Proof: ^pC_n=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 ^pC_n is divisible by p.

Proof of the proposition:

• F(1)=1^p=1.
• F(ab)=(ap)^p=a^pb^p=F(a)F(b).
• F(a+b)=(a+b)^p= a^p+pa^p-1b+...+ ^pC_r.a^rb^p-r+...+pab^p-1+b^p. 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)=a^p+b^p=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 Z_p 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 Z_p[x] the Frobenius is a homomorphism. Note that F(f(x))=f(x^p).

 

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