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

Asher Olsen 2022-03-25 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)

Pubephenedsjq
Answered 2022-03-26 Author has 11 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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Matronola3zw6
Answered 2022-03-27 Author has 10 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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For simple ones, like quadratic equations, I can usually find the minimum point and give a correct answer.
But take, for example:
f ( x ) = 3 2 x + 1 , x > 0
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I am so confused with the whole process of finding the ranges of functions, including those above as samples, that I can't even quite explain what or why.
Can someone please, in a very step-by-step process, detail exactly what steps you would take to get the ranges of the above functions? I tried substituting x-values (such as 0), and came up with f ( x ) > 3, but that was mostly guesswork -- also, f ( x ) > 3 is incorrect.
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