How do I factorize ${x}^{6}-1$ over GF(3)? I know that the result is $(x+1)}^{3}{(x+2)}^{3$ , but I'm unable to compute it myself.

Kaiya Hardin
2022-04-25
Answered

How do I factorize ${x}^{6}-1$ over GF(3)? I know that the result is $(x+1)}^{3}{(x+2)}^{3$ , but I'm unable to compute it myself.

You can still ask an expert for help

Cynthia Herrera

Answered 2022-04-26
Author has **16** answers

Step 1

If$n=\pm$ is a multiple of p then over $GF\left(p\right)$ one has

$x}^{n}-1={({x}^{m}-1)}^{p$

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)

${x}^{n}-1=\prod _{k=0}^{n-1}(x-{\zeta}^{j})$

where$\zeta$ 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

$(x-{\zeta}^{k})(x-{\zeta}^{pk})(x-{\zeta}^{{p}^{2}k})\cdots (x-{\zeta}^{{p}^{r-1}k})$

has coefficients in, and is irreducible over, GF(p) where r is the least positive integer with${p}^{r}k\equiv k$ (mod n)

Using this, it's easy to work out the degrees of the irreducible factors of${x}^{n}-1$ , but to find the factors themselves needs a bit more work, using for instance Berlekamp's algorithm.

If

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)

where

has coefficients in, and is irreducible over, GF(p) where r is the least positive integer with

Using this, it's easy to work out the degrees of the irreducible factors of

Simone Ali

Answered 2022-04-27
Author has **18** answers

Step 1

Factoring the polynomials${x}^{n}-1$ 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.

Factoring the polynomials

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Module with matrix multiplication

Define the rational$3\times 3$ -matrix

$A:=(\begin{array}{ccc}1& 4& 1\\ 0& 1& 0\\ 0& 1& 2\end{array})$

We then define a$\mathbb{Q}\left[X\right]$ -module on $V={\mathbb{Q}}^{3}$ such that

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Argue if${V}_{A}=\mathbb{Q}[X](\begin{array}{c}1\\ 0\\ 1\end{array})\oplus \mathbb{Q}[X](\begin{array}{c}0\\ 1\\ -1\end{array})$ .

Define the rational

We then define a

where

Argue if

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