How do I factorize x^{6}-1 over GF(3)? I know that

Kaiya Hardin 2022-04-25 Answered
How do I factorize x61 over GF(3)? I know that the result is (x+1)3(x+2)3, but I'm unable to compute it myself.
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Answers (2)

Cynthia Herrera
Answered 2022-04-26 Author has 16 answers
Step 1
If n=± is a multiple of p then over GF(p) one has
xn1=(xm1)p
so the problem reduces swiftly to the case where n is co' to p.
If p is not a factor of n then over an algebraic closure of GF(p)
xn1=k=0n1(xζj)
where ζ is a primitive n-th root of unity. One makes this into a factorization over GF(p) by combining conjugate factors together. For each k, the polynomial
(xζk)(xζpk)(xζp2k)(xζpr1k)
has coefficients in, and is irreducible over, GF(p) where r is the least positive integer with prkk (mod n)
Using this, it's easy to work out the degrees of the irreducible factors of xn1, but to find the factors themselves needs a bit more work, using for instance Berlekamp's algorithm.
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Simone Ali
Answered 2022-04-27 Author has 18 answers
Step 1
Factoring the polynomials xn1 is very different from factoring general polynomials, since you already know what the roots are; they are, in a suitably large finite extension, precisely the elements of order dividing n. Since you know that the multiplicative group of a finite field is cyclic, the conclusion follows from here.
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