# Field of characteristic p. automosphism.

• May 6th 2012, 08:18 AM
abhishekkgp
Field of characteristic p. automosphism.
Let $\displaystyle F$ be a field of characteristic $\displaystyle p \neq 0$. Let $\displaystyle K$ be an extension of $\displaystyle F$. Define $\displaystyle T= \{ a \in K : a^{p^n} \in F \text{ for some } n \}$.
a) Prove that $\displaystyle T$ is a subfield of $\displaystyle K$.
b) Show that any automorphism of $\displaystyle K$ leaving every element of $\displaystyle F$ fixed also leaves every element of $\displaystyle T$ fixed.

ATTEMPT:

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

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

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

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

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

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

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

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

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

How do I proceed from here?