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I am trying to solve this question for my abstract algebra class.

Let $f = X^6 - 2X^5 + 3X^4 - 2X^3 + 3X^2 - 2X + 2 \in \mathbb{Z}[X]$.

Either prove that $f$ is irreducible over $\mathbb{Z}$ or find all irreducible factors of $f$.

I know the answer is $(X^2 - 2X + 2)(X^2 - X + 1)(X^2 + X + 1)$.

However, I have no idea how we came up with this factorization. I have read this post, but I don't understand:

  1. Why we assume the factors are monic polynomials?
  2. Can we apply this to all kinds of polynomials, over any field?

I would also appreciate any different point of view, or any different solution.

Zek
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  • For (1), remember that you're working in $\Bbb Z[X]$. – Karl Nov 14 '23 at 08:29
  • One guess might be to write it as $(X^6+3X^4+3X^2+2)-2X(X^4+X^2+1)$. Then you can quickly check that in this case you are lucky: $X^6+3X^4+3X^2+2=(X^2+2)(X^4+X^2+1)$, giving the initial factorisation $(X^2-2X+2)(X^4+X^2+1)$. – Andrew Hubery Nov 14 '23 at 08:46
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    In general, factoring polynomials over $\mathbb Z[X]$ is hard. – Peter Nov 14 '23 at 09:01
  • @Karl I'm still a bit lost, is it because of the Rational Root Theorem? I don't see how they correlate. – Zek Nov 14 '23 at 13:14
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    @Zek no, it's because all the coefficients have to be integers, and $f$ is monic. Think about what happens to the leading coefficients when you multiply polynomials. – Karl Nov 14 '23 at 14:55
  • @Karl Yes, thank you, I can't believe I missed that. – Zek Nov 14 '23 at 14:57

1 Answers1

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By the Rational Root Theorem there is no linear factor. For testing quadratic factors we expand $(x^4+ax^3+bx^2+cx+d)(x^2+rx+s)$ and compare the coefficients with $f$. This gives easy equations in $a,b,c,d,r,s\in \Bbb Z$. Note that we obtain for example $ds=2$, which implies up to signs either $d=1$ and $s=2$ or vice versa. This is one reason that we can solve these equations easily. Finally, we repeat this for the quartic polynomial. So we obtain the complete factorization.

Dietrich Burde
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