Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $F$ be a field of characteristic $p$ and $F$ has infinitely many elements. Prove that the map $\sigma: \alpha\mapsto \alpha^p$ is a field endomorphism but not necessarily automorphism. How to prove?

share|cite|improve this question
As it has not been mentioned: the Frobenius is always an automorphism if $F$ is finite. That's why the examples below are all infinite. – Sebastian Dec 19 '13 at 15:22
up vote 3 down vote accepted

The first part is standard, as you will know that $(a + b)^{p} = a^{p} + b^{p}$ in $F$.

Now for an example in which $\sigma$ is not onto, let $K$ be the field with $p$ elements, and consider the field of rational functions $F = K(x)$. Now prove that $x$ is not in the image of $\sigma$.


Write, by way of contradiction, $$x = \left( \frac{f(x)}{g(x)}\right)^{p},$$ with $f(x), g(x) \in K[x]$. You get $f(x)^{p} = x g(x)^{p}$. Compare degrees.

share|cite|improve this answer


If you've studied finite fields, you should know the "Freshman's dream theorem," which should supply some inspiration. This will help you see why the map is additive.

For the example where it is not onto, consider the rational polynomial ring $F_p(x)$, and what taking powers of elements does in this field.

share|cite|improve this answer

The Frobenius is a homomorphism because of $\sigma(\alpha+\beta)=(\alpha+\beta)^p=\alpha^p+\beta^p=\sigma(\alpha)+\sigma(\beta)$, and $\sigma(\alpha\beta)=(\alpha\beta)^p=\alpha^p\beta^p=\sigma(\alpha)\sigma(\beta)$. The kernel of $\sigma$ is an ideal in $F$, hence zero or $F$. Since $\sigma\neq 0$, we have $\ker(\sigma)=0$, and $\sigma$ is injective. But it need not be surjective.

share|cite|improve this answer

Hint $\ $ The classical proof that $\sqrt[p]q$ is irrational (via Rational Root Test) extends to the quotient field $K$ of any UFD $R,$ showing that $X^p\! - q$ has no roots in $K$ for prime $q\in R$, e.g. $R = \Bbb Z_p[x]$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.