abhishekkgp
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- Jan 23, 2012
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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 \(\displaystyle a^p \in F\). 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?
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 \(\displaystyle a^p \in F\). 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?