# Automorphisms of Finite Fields

1. Sep 7, 2005

### isaiah

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 $$F_{2^{116}}$$?

Isaiah.

Last edited: Sep 8, 2005
2. Sep 7, 2005

### Palindrom

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?

3. Sep 8, 2005

### isaiah

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.

4. Sep 8, 2005

### Palindrom

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?