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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?

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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

4 Answers 4

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 $\sigma$.

Hint

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.

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Hint:

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.

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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.

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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. $R = \Bbb Z_p[x]$.

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