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

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SUMMARY

The discussion focuses on proving that if F is a field of characteristic p>0 and E = F(a) where a is separable over F, then E equals F(a^p). The participants explore the implications of the Frobenius endomorphism and consider two cases: finite and infinite extensions of E. For finite extensions, the isomorphism holds true, while for infinite extensions, additional analysis is required. The conclusion emphasizes that isomorphism can only exist under specific conditions related to the cardinality of E.

PREREQUISITES
  • Understanding of field theory and separable extensions
  • Knowledge of Frobenius endomorphism in algebra
  • Familiarity with minimal polynomials and their properties
  • Concept of isomorphism in the context of field extensions
NEXT STEPS
  • Study the properties of separable extensions in field theory
  • Learn about the Frobenius endomorphism and its applications
  • Explore minimal polynomials and their role in field extensions
  • Investigate the structure of finite fields and their cardinalities
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Mathematics students, algebraists, and researchers in field theory looking to deepen their understanding of separable extensions and the implications of field characteristics.

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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.
 

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