How Do Automorphisms Affect Elements in Non-Prime Subfields of Finite Fields?

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SUMMARY

The discussion centers on the effects of field automorphisms on elements within non-prime subfields of finite fields, specifically examining F_{2^{29}}, F_{2^{58}}, and F_{2^{116}}. It is established that any automorphism of F_{2^{58}} must map F_{2^{29}} to itself due to the normality of finite extensions. Consequently, elements in F_{2^{58}} that are not in F_{2^{29}} remain in F_{2^{58}} \ F_{2^{29}} and do not shift to F_{2^{116}}. This conclusion is supported by the properties of automorphisms being one-to-one and onto.

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isaiah
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I am trying to figure out the effect of a field automorphism on a field with a non prime subfield.

Say for example F_{2^{29}}, F_{2^{58}} and F_{2^{116}}

Let \alpha \in F_{2^{58}}\F_{2^{29}}

Under {\sigma}^{i}, 1 \le i \le 58 do we get any case where \alpha becomes an element of F_{2^{29}} ?

If not why not since the orbit of \alpha under this automorphism will be 58.

Does it mean that the other elements shift to F_{2^{116}}?

Thanks in advance.

Isaiah.
 
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Remember that a finite extension E/F is normal iff every automorphism of E sends F to itself. Since any finite extension of finite fields is Galois, any automorphism of F_2^58 must send F_2^29 to itself. Does that answer your question?
 
Now what really happens to the elements of F_2^58\F_2^29 (ie exclude the ones in F_2^29)? Do they go to F_2^29 or do they remain in F_2^58\F_2^29?

Thanks,

Isaiah.
 
Well, an automorphism is one to one and onto, so if f:A->A is one to one and onto, and f(B)=B, we must have f(A\B)=A\B, mustn't we?
 

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