2

I think the first step is determining whether this polynomial is irreducible or not. But it seems that this is not an Eisenstein polynomial (and the substitutions $X\rightarrow Y - 1$ or $X\rightarrow Y + 1$ don't help).

I know that this polynomial is reducible mod $2$, because $X^5 + 5 X^3+5 X^2+4 X+1 \equiv X^5 + X^3+X^2+X+1 \equiv (X + 1)^3 (X^2 + X + 1) \mod{2}$. But I don't see how this would help, because we usualy show that something is irreducible modulo a prime.

Also computing the discriminant for a $5$-th degree polynomial seems too much work to me.

Via the rational root theorem I have checked that it does not have rational roots.

Jens Wagemaker
  • 1,334
  • 2
    It is irreducible modulo five. – Jyrki Lahtonen May 30 '18 at 11:59
  • Are you familiar with Dedekind's theorem? If so, then factorization modulo $7$ is very revealing as it implies that the order of the Galois group $G$ is divisible by six, and that $G$ is not a subgroup of the alternating group $A_5$. – Jyrki Lahtonen May 30 '18 at 12:06
  • @JyrkiLahtonen Am I correct with the following argument: since it is irreducible mod $5$, G has a 5-cycle (Dedekind's theorem), since it reduces in irreducible factors of degree 2 and degree 3 in mod 2, G has a 2-cycle. Now as 5 is a prime, $S_5$ is generated by an arbitrary 2-cycle and 5-cycle. So we find that $G = S_5$. – Jens Wagemaker May 30 '18 at 12:12
  • That factor $(x+1)$ modulo two is repeated (in other words, $2$ is a factor of the discriminant of the monomial basis), so you cannot apply Dedekind (because it assumes that there are no repeated factors). OTOH modulo seven we have $$f(X)=(X^2-X+3)(X^3+X^2+3X-2).$$ – Jyrki Lahtonen May 30 '18 at 12:26
  • @JyrkiLahtonen, so we can apply my argument, but then with the 2-cycle obtained from Dedekind applied to mod 7? And how did you know/find the factorization mod 7? – Jens Wagemaker May 30 '18 at 12:35
  • Polynomial $g(x)=x^5 f(1/x)$ satisfies $g(6)=14221$ which is a prime, so Cohn's criterion could be used (or even the original polynomial satisfies $f(10)=105541$ is a prime). Also it should be irreducible modulo $3$ as well. – Sil May 30 '18 at 22:46

0 Answers0