Irreducible Polynomials over Finite Fields

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SUMMARY

The discussion focuses on the construction of an irreducible polynomial F(x) over the finite field GF(q) with degree nm, derived from two irreducible polynomials f(x) and g(x) with coprime degrees n and m. The roots of F(x) are expressed as products of the roots of f(x) and g(x), specifically \(\alpha^{q^i}\beta^{q^j}\). The polynomial F(x) can be explicitly defined using the resultant of f(x) and g(x), with methods including matrix determinants and the Euclidean algorithm. The conversation highlights the complexity of expressing F(x) without directly referencing the roots.

PREREQUISITES
  • Understanding of irreducible polynomials over finite fields
  • Familiarity with the concept of resultants in polynomial algebra
  • Knowledge of finite field theory, specifically GF(q)
  • Basic proficiency in polynomial composition and transformations
NEXT STEPS
  • Research the computation of resultants in polynomial algebra
  • Study Theorem 4.2 and related formulas from the referenced paper
  • Explore efficient algorithms for polynomial multiplication and composition
  • Investigate special classes of polynomials that allow for faster computations than O(n^3 log^2(n))
USEFUL FOR

Mathematicians, computer scientists, and researchers working in algebra, particularly those focused on finite fields and polynomial theory.

burritoloco
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Hi, yet another question regarding polynomials :). Just curious about this.

Let f(x), g(x) be irreducible polynomials over the finite field GF(q) with coprime degrees n, m resp. Let \alpha , \beta be roots of f(x), g(x) resp. Then the roots of f(x), g(x), are \alpha^{q^i}, 0\leq i \leq n-1, and \beta^{q^j}, 0\leq j \leq m-1.

Question: What is the irreducible polynomial over GF(q) of degree nm with roots \alpha^{q^i}\beta^{q^j} where 0\leq i \leq n-1, and 0\leq j \leq m-1. Can you define such polynomial explicitly in terms of just f(x) and g(x) without the roots appearing in the formula?

Note: The last sentence/question is what really interests me as the following is the required polynomial (but defined in terms of the roots of f(x))

F(x) = \prod_{i=0}^{n-1}\alpha^{mq^i}g\left(\alpha^{-q^i}x\right)

Thank you!
 
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It depends somewhat on just what you mean by "formula" and "appearing in".

I know that F can be expressed as a resultant. The formula you give is presumably comes from one of the methods of computing resultants. There exist other ways to compute resultants, such as as the determinant of a matrix, or a Euclidean algorithm-like method.

You could always find a way to write down the system of equations that literally expresses that \alpha \beta is a root of F(x). Then F would be the solution!
 
Ah! The resultant formulas I was recalling are formulas 4.3 and 4.4. (Appearing just before theorem 4.3)
 
Thanks Hurkyl. I'll probably be coming up with more questions as my project goes along :). Feel free to answer whenever you can, haha. Great to have Physics Forums around.
 
Hmm, resultants and Thm 4.2 seem to be general methods on any two polynomials f(x), g(x) to obtain F(x). What if there are special classes of polynomials for which this computation could be done totally differently and much faster than the O(n^3 log^2(n)) in their paper? But what is this all good for anyways lol.
 
I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...

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