Abstract Algebra: Proving E=F(a^p)

dabien
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Homework Statement


Let F be a field of characteristic p>0 and let E = F(a) where a is separable over F. Prove that E=F(a^p).


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The Attempt at a Solution


I know that maybe show how mod F(a) = mod F(a^p) or something around there.
 
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So, since the characteristic is p, a prime E is generated by some field F adjoined some set of elements, and for each minimal polynomial associated to these elements there are no duplicate roots.

Now, there are two cases to consider E is finite or E is infinite. If E is finite then you're done, since E is iso to Z mod p^k. If E is infinite (for example Z4(u) where u is some transcendental element) then you have some work to do. Consider what frobenius endomorphism
http://en.wikipedia.org/wiki/Frobenius_endomorphism
Tells about the minimal polynomial of a.
 
icantadd said:
E is iso to Z mod p^k.
No it's not. Such an isomorphism can only exist when k=1 and E is a field whose cardinality is p.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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