Looking for a creative or quick method for finding roots in GF(p^n)

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SUMMARY

This discussion focuses on finding explicit roots of irreducible polynomials of degree 3 over GF(3) within the field GF(27). The polynomial x27 - x is factored into monic polynomials of degrees 1 and 3, totaling 11 factors. The user seeks methods to find roots for 7 irreducible polynomials constructed from GF(3)/<x3 + x2 + 2> without resorting to exhaustive enumeration of elements. Two relevant papers are cited for further exploration of polynomial factorization techniques.

PREREQUISITES
  • Understanding of Galois theory
  • Familiarity with finite fields, specifically GF(3) and GF(27)
  • Knowledge of polynomial factorization techniques
  • Ability to work with isomorphisms in field theory
NEXT STEPS
  • Study the paper "Polynomial Factorization" from MIT for advanced techniques in polynomial roots finding.
  • Research Galois theory applications in finite fields, focusing on irreducible polynomials.
  • Explore the properties of GF(3) and its extensions to better understand field constructions.
  • Investigate alternative methods for root finding in finite fields beyond brute force enumeration.
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Mathematicians, cryptographers, and computer scientists interested in Galois theory, finite field applications, and polynomial factorization methods.

kmitza
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TL;DR
I have been revising some calculation skills in Galois theory and I ran into a problem. if we have an irreducible polynomial of degree 3 over GF(3) and we construct GF(27) as GF(3)/<f> and we have another such polynomial g(x) how do we find it's explicit roots in GF(27)
I am going to give up a bit more on the given problem. We start with polynomial ## x^27 -x ## over GF(3)[x] and we factorize it using a well known theorem it turns out it factorises into the product of monic polynomials of degree 1 and 3, 11 of them all together.

We then choose one of those polynomials and consider ## GF(3)/<x^3+x^2+2> ## and we're asked to find roots for the other 7 irreducible polynomials of degree 3 in GF(27) constructed this way and give isomorphisms between ## GF(3)/<f> ## and ## GF(3)/<g> ## for each of them.

It's not hard to see how to give the isomorphism but I am struggling with finding the explicit roots. I am wondering if anyone knows of a good method for doing this that isn't just plugging in all 27 elements the first time and then 24 and so on...

I know that there are probably computer methods for doing this but I am not looking for those.
 
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Fix use {} around 27 for Latex.
 
mathman said:
Fix use {} around 27 for Latex.
Thanks for the tip!
 
kmitza said:
Summary:: I have been revising some calculation skills in Galois theory and I ran into a problem. if we have an irreducible polynomial of degree 3 over GF(3) and we construct GF(27) as GF(3)/<f> and we have another such polynomial g(x) how do we find it's explicit roots in GF(27)

I am going to give up a bit more on the given problem. We start with polynomial ## x^27 -x ## over GF(3)[x] and we factorize it using a well known theorem it turns out it factorises into the product of monic polynomials of degree 1 and 3, 11 of them all together.

We then choose one of those polynomials and consider ## GF(3)/<x^3+x^2+2> ## and we're asked to find roots for the other 7 irreducible polynomials of degree 3 in GF(27) constructed this way and give isomorphisms between ## GF(3)/<f> ## and ## GF(3)/<g> ## for each of them.

It's not hard to see how to give the isomorphism but I am struggling with finding the explicit roots. I am wondering if anyone knows of a good method for doing this that isn't just plugging in all 27 elements the first time and then 24 and so on...

I know that there are probably computer methods for doing this but I am not looking for those.
I have found two papers about it.
https://www.ams.org/journals/mcom/1...-1969-0257039-X/S0025-5718-1969-0257039-X.pdf
https://people.csail.mit.edu/dmoshkov/courses/codes/poly-factorization.pdf

I once met von zur Gathen, so I would start with the second one, but the first one is on the reference list of the second.
 

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