Field of characteristic p. automosphism.

[caffeinemachine]

Let $F$ be a field of characteristic $p \neq 0$. Let $K$ be an extension of $F$. Define $T= \{ a \in K : a^{p^n} \in F \text{ for some } n \}$.
a) Prove that $T$ is a subfield of $K$.
b) Show that any automorphism of $K$ leaving every element of $F$ fixed also leaves every element of $T$ fixed.

ATTEMPT:Part (a) is easy after observing that $(a+b)^{p^m}=a^{p^m}+b^{p^m}$.

Now part (b). Let $\phi : K \rightarrow K$ be an automorphism with $\phi(x)=x, \, \forall x \, \in F$.

NOTATION: $\phi^2(a)= \phi(\phi(a))$, $\phi^3(a)=\phi(\phi(\phi(a)))$ and so on.

Now consider the special case when $a \in T-F$ with $a^p \in F$. We need to show that $\phi(a)=a$.

Since $a^p \in F$ we have $\phi(a^p)=a^p$.

Thus $[\phi(a)]^p = a^p$. This leads to $[\phi^r(a)]^p=a^p$ and also to the conclusion that $a \in T \Rightarrow \phi(a) \in T$.

Consider $\phi(a), \phi^2(a), \ldots, \phi^{p+1}(a)$.

If these are all distinct then the polynomial $x^p - a^p \in K[x]$ will have $p+1$ distinct roots. Since this is impossible thus some two of
the elements are same. This leads to the conclusion that $\exists r \in \mathbb{Z}^+$ such that $\phi^r(a)=a$.

Now we need to show that the minimum value of such an $r$ is one.

How do I proceed from here?

 

[jvicens]

To prove that the minimum value of $r$ is one, we can assume that $r>1$ and reach a contradiction.

Assuming $r>1$, we have $\phi^r(a)=a$ and $\phi^r(a^p)=a^p$. Since $\phi$ is an automorphism, we can apply it to both sides of the equation to get $\phi^{r+1}(a^p)=\phi(a^p)=a^p$.

Now, since $a^p \in F$, we have $\phi^{r+1}(a^p)=\phi(a^p)=a^p$. We can continue this process and show that $\phi^{r+k}(a^p)=a^p$ for all $k \in \mathbb{Z}^+$.

This means that $a^p$ is a fixed point of $\phi$ for all positive powers of $r$, which contradicts the fact that $r$ is the minimum value for which $\phi^r(a)=a$.

Therefore, our assumption that $r>1$ is false and we can conclude that $r=1$, which means that $\phi(a)=a$.

Since this holds for all $a \in T-F$, we can conclude that $\phi$ leaves every element of $T$ fixed.