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?