Adjoining Elements to Finite Fields: Finding Generators

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SUMMARY

The discussion focuses on the challenge of finding a generator for the multiplicative group within the field F_5(2^{1/4}). The participant successfully identified a generator for the case of F_5(2^{1/2}) as 2 + √2 but is uncertain about generalizing this result. The consensus is that determining a generator for the cyclic group in F_5(2^{1/4}) is complex and may require testing all elements for cyclicity. The notation for the ring should be expressed as ℱ_5[X]/(X^4-2) for clarity.

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Homework Statement



I have the field [tex]F_5[/tex] and I adjoin some square root of 2 , say [tex]2^{1/4}[/tex]. Is there a way to see that the multiplicative group inside [tex]F_5(2^{1/4})[/tex] is cyclic and find the generator?

Homework Equations


The Attempt at a Solution



I did the [tex]F_5(2^{1/2})[/tex] case and think the generator is [tex]2+\sqrt{2}[/tex]. But don't know how this generalizes..
Thanks!
 
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I don't really like the exponent notation. You should write your ring as [itex]\mathbb{F}_5[X]/(X^4-2)[/itex].

Finding a generator for the cyclic group is a quite difficult problem and still an active problem of research. I fear that the only solution is to test all the elements and see whether they are cyclic.
 
micromass said:
I don't really like the exponent notation. You should write your ring as [itex]\mathbb{F}_5[X]/(X^4-2)[/itex].

Finding a generator for the cyclic group is a quite difficult problem and still an active problem of research. I fear that the only solution is to test all the elements and see whether they are cyclic.

Really?:cry::cry: Even the fact that [tex]\mathbb{F}_5[/tex] itself is cyclic does not help..?
 

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