Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $ q = p^r$. Let $F$ be the splitting field of $X^q - X$. Let $\phi : F \to F$ be the Frobenius automorphism $\phi(x) = x^p$. Then let $F' \subseteq F$ be the fixed field of $ < \phi^r >$.

Claim: $x \in F'$ iff $\phi^r(x) = x $.

Why is the claim true? I can see one direction easily (if $\phi^r(x) = x$, then $x$ is fixed by $\phi^r$). But what about the other direction? If $x \in F'$, then $x$ is fixed by some $\phi^{kr}$. Why must it be fixed by $\phi^r$?

Thanks

share|improve this question
3  
You misunderstand the definition of "fixed field." It means that every element of the subgroup fixes an element of the subfield (whereas you seem to think it means some element fixes it). –  Qiaochu Yuan Jan 19 '12 at 20:05
1  
Ah, that makes sense. Thanks –  Matt Jan 19 '12 at 20:19
add comment

1 Answer

up vote 2 down vote accepted

If $\phi^r(x) = x$, then $x$ is fixed by $\phi^r \in \langle \phi^r \rangle$, so $x \in F'$.

Conversely, if $x \in F'$, then $x$ is fixed by everything in $\langle \phi^r \rangle$, and so in particular is fixed by $\phi^r$ i.e. $\phi^r(x) = x $.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.