Adjoining Elements to Finite Fields: Finding Generators

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The discussion centers on the challenge of finding a generator for the multiplicative group in the field F_5(2^{1/4}). The user successfully identified a generator for the case of F_5(2^{1/2}) as 2 + √2 but is unsure how to extend this to the fourth root. It is noted that finding a generator for the cyclic group is a complex problem and remains an area of active research. The suggestion is made to represent the ring as ℱ_5[X]/(X^4-2) for clarity. Ultimately, the consensus is that testing all elements may be necessary to determine cyclicity.
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Homework Statement



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

Homework Equations


The Attempt at a Solution



I did the F_5(2^{1/2}) case and think the generator is 2+\sqrt{2}. 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 \mathbb{F}_5[X]/(X^4-2).

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 \mathbb{F}_5[X]/(X^4-2).

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 \mathbb{F}_5 itself is cyclic does not help..?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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