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I am trying to factorize the polynomial

$6x^8+2x^7+6x^4+6x^3+8x^2+8x+2$ over $\mathbb{Z}[X]$ and $\mathbb{Q}[X]$

I only have the following tools at my disposal:

  1. Eisenstein's criterion.
  2. Reduction of coefficients modulo prime ideals (e.g., working over finite fields $\mathbb{F}_p$).

I firstly took 2 as a common factor and worked on the polynomial $3x^8+x^7+3x^4+3x^3+4x^2+4x+1$. I tried applying Eisenstein's criterion directly, but it doesn't seem to work for any prime $p$. Then I reduced the polynomial modulo $p=2$, obtaining $x^8+x^7+x^4+x^3+\overline{1}$. I checked that this polynomial could be written as $(x^3+x^2+\overline{1})(x^5+x^2+\overline{1})$, but this doesn't tell me anything. Could someone guide me through the correct approach to factorize this polynomial using these tools?

Any help or detailed explanation would be appreciated.

Also, I got some problems with this other polynomial $3x^8+x^7+3x^3+4x+1$, but I guess it can be done in a similar way to the previous one.

Martin Brandenburg
  • 181,922
Peru Barranco
  • 19

1 Answers1

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The polynomial is irreducible. It is enough to prove that $g := 3x^8+x^7+3x^4+3x^3+4x^2+4x+1$ is irreducible over $\mathbb{F}_7$. The normed version is $f := g/3 = x^8 + 5 x^7 + x^4 + x^3 - x^2 - x + 5$. Rabin's test of irreducibility tells us that $f$ is irreducible iff the following is true:

  • $f$ divides $x^{7^8}-x$
  • $\mathrm{gcd}(f, x^{7^4}-x) = 1$

This could be done by hand, simply by using polynomial division and Euclid's algorithm. This is essentially linear algebra in terms of the coefficients. But here, I would not recommend to do this by hand. A computer algebra system can do this tedious task for us, which is good since this computation will not bring us any mathematical insight anyway. Here is some GAP code.

gap> x := Indeterminate(GF(7),"x");
gap> f := x^8 + 5 * x^7 + x^4 + x^3 - x^2 - x + 5;
gap> Gcd(x^(7^8) - x, f) = f;    # true
gap> Gcd(x^(7^4) - x, f) = f^0;    # true

Or just cheat:

gap> Factors(f);

Maybe there are alternative approaches that actually work by hand in a reasonable amount of time. But the issue is that clever solutions are not "scalable". The general problem of factoring polynomials over finite fields (in smallish degrees) is solved by various algorithms, see the Wikipedia article, and computer algebra systems have implemented these algorithms. Once you have understood why they work, you can use them.

PS: $g$ is also irreducible modulo other primes.

gap> for p in [1..1000] do if IsPrime(p) then x := Indeterminate(GF(p),"x"); f := 3*x^8 + x^7 + 3*x^4 + 3*x^3 + 4*x^2 + 4*x+1; if Length(Factors(f))=1 then Print(p,",");fi;fi;od;

outputs the prime numbers

7,11,43,97,157,283,359,367,397,431,439,499,563,617,691,739,773,787,929,941,947,971
Martin Brandenburg
  • 181,922
  • Modulo $5$, the given polynomial is equivalent to a CUBIC polynomial ($x^3+x^2+x=2$) and for cubics the fact that don't have a root in the prime field, is the same that irreducible (because if reducible then a factor is linear). What is wrong with this solution? – Ataulfo Jan 24 '25 at 18:26
  • You keep repeating your claim which doesn't make any sense. You cannot reduce the degree. Please ask a separate question if you want to clarify this attempt. Apparently you are not in the role of giving an answer here. – Martin Brandenburg Jan 24 '25 at 19:04
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    Of course this (so-called Rabin) irreducibility test is already described in many prior answers (e.g. here) so this answer adds nothing new. It should have been (at most) a comment, not an answer (also since it is not one of the "only tools at OP's disposal"). – Bill Dubuque Jan 25 '25 at 18:55
  • Maybe I have made a mistake but it is not a "gross" one that would make me the object of a disparaging sufficiency. What I know is that if the degree of $f(x)$ is greater than or equal to $p$ then we have $f(x)=(x^p-x)Q(x)+R(x)$ i.e. $f(x)=R(x)$ in $\mathbb F_p[x]$ so there are practically no polynomials of degree greater than $p-1$ in $\mathbb F_p[x]$. – Ataulfo Jan 25 '25 at 21:38