1

I am reading Geisel's tutorial$^{\color{red}{\star}}$ on Reed-Solomon codes, in which it is tested whether the polynomial $f(x) = x^4 + x + 1$ is irreducible. From pages 15 and 16,

enter image description here

As a first step, the polynomial $f(x) = x^4 + x + 1$ is evaluated at 0 (zero). The result is 1. The conclusion is 'Therefore, $(x + 0) = (x - 0)$ is not a factor (0 is not a root)'.

My questions are about this conclusion:

  • How can $(x + 0) = (x - 0)$ be derived from the preceding text?
  • What is meant by 'a factor' here? A reduced part of the polynomial?
  • I understand that 0 (zero) is not a root, a root of a polynomial should result in 0 (zero) when evaluated. Why does this prove that $(x + 0) = (x - 0)$ is not a factor?
  • Why does this conclusion eliminate one possible factor of the polynomial?

Hoping for someone patient to answer these questions.

$\color{red}{\star}$ William A. Geisel, Tutorial on Reed-Solomon Error Correction Coding [PDF], NASA Technical Memorandum 102162, NASA, August 1990.

Rodrigo de Azevedo
  • 1
ʕ ᵔᴥᵔ ʔ
  • 143
  • 3
    You asked "what is a factor" twice. You've factored polynomials before, like $x^2-1=(x+1)(x-1)$, so you know what a factor is. As you should learn in (elementary) algebra before abstract algebra, $r$ is a root of a polynomial if and only if $(x-r)$ is a factor. If you've ruled out $x-0$ as a possible factor, then you've eliminated at least one possible factor. – anon Mar 30 '22 at 19:17
  • @arctictern: thank you. I found this: https://en.m.wikipedia.org/wiki/Factor_theorem. Is this the theorem you are mentioning? – ʕ ᵔᴥᵔ ʔ Mar 30 '22 at 20:30
  • 1
    Yes, I was talking about the factor theorem. – anon Mar 31 '22 at 14:44

1 Answers1

1

To address your second and fourth bullet points, it is a general fact for a polynomial ring over a field, $F[x]$: If $p(x)\in F[x]$ have root $r$ show that $x-r$ is a factor of $p(x)$ The converse of that statement is easily seen to be true also.

To address your first and third bullet points (the same thing?) when you express a polynomial as the product of two other polynomials, each of the two may be called a factor.

Usually we're interested in the case when the factors have degree 1 or more, so that we've really decomposed the original polynomial into a product.

They repeat the same logic with $1$, demonstrating $(x-1)$ is not a factor off the polynomial either.

At the same time, this establishes the polynomial doesn't have a factor of degree 3 (because the other factor would have to be something like $(x-r)$, and you already eliminated that.)

To show that the polynomial is irreducible, there's only one more case to rule out: the possibility that it's a product of two polynomials of degree $2$.

PS also if it wasn't obvious, since you're working over $GF(2)$, you have that $p=-p$ for every polynomial... that is, addition and subtraction in this ring are the same thing. This explains the sign changes you were apparently asking about.

rschwieb
  • 153,510
  • Thanks, I understand the way of proving now. Where you wrote "it is a general fact for a polynomial ring over a field, F[x]: If p(x)∈F[x] have root r show that x−r is a factor of p(x)", this seems like the factor theorem. Mentioned in the comment above. Is this true? – ʕ ᵔᴥᵔ ʔ Apr 02 '22 at 18:24
  • "PS also if it wasn't obvious, since you're working over GF(2), you have that p=−p for every polynomial... that is, addition and subtraction in this ring are the same thing." Because of the modulo 2 operation, which is performed after every 'normal' calculation, right? – ʕ ᵔᴥᵔ ʔ Apr 02 '22 at 18:46
  • 1
    Yes and yes. ${}{}{}{}{}$ – rschwieb Apr 02 '22 at 20:57