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.

Asher Olsen
2022-03-25
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

Pubephenedsjq

Answered 2022-03-26
Author has **11** 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

Matronola3zw6

Answered 2022-03-27
Author has **10** 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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I know that the matrix representation of the linear transformation:

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${x}_{2}={X}_{2}-\lambda {X}_{1}$

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$\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]=\left[\begin{array}{cc}1& 2\lambda \\ -\lambda & 1\end{array}\right]\left[\begin{array}{c}{X}_{1}\\ {X}_{2}\end{array}\right]$

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How can it be expressed in matrix form to get it's inverse?

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${x}_{2}={X}_{2}-\lambda {X}_{1}$

is:

$\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]=\left[\begin{array}{cc}1& 2\lambda \\ -\lambda & 1\end{array}\right]\left[\begin{array}{c}{X}_{1}\\ {X}_{2}\end{array}\right]$

but what if I have the non-linear transformation:

${x}_{1}={X}_{1}+2\lambda {X}_{2}{X}_{1}^{2}$

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How can it be expressed in matrix form to get it's inverse?

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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:

$f(x)=\frac{3}{2x+1},x>0$

and

$g(x)=\frac{1}{x}+2,x>0$

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.

Also, is there an outline I can follow -- even for the thinking process, like check if A, check if B -- that would work every time?

But take, for example:

$f(x)=\frac{3}{2x+1},x>0$

and

$g(x)=\frac{1}{x}+2,x>0$

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.

Also, is there an outline I can follow -- even for the thinking process, like check if A, check if B -- that would work every time?