Which way I can determine whether the polynomial $x^5 + 5x^2 +1$ is irreducible over $\mathbb Q$ or not?

Mod $p$ Irreducibility Test and Eisenstein's criterion not applicable here.

Which way I should proceed?

  • $\begingroup$ You mean you cant use the mod P irreducibility or that you tried without success? $\endgroup$ – AHandsomeAlien Jun 29 '17 at 4:29
  • 6
    $\begingroup$ Reduce modulo $2$. Show that your new polymomial has no roots in $\Bbb Z_2$ and show that your reduced polynomial can't be written as $q(x)p(x)$ where $q$ has degree $2$ and $p$ has degree $3$. $\endgroup$ – Rocket Man Jun 29 '17 at 4:31
  • $\begingroup$ No. Actually I tried. But then after reducing by mod5 polynomial becomes x^5 + 1 . Then I dont know how should I check if x^5 +1 is irreducible in Z5 or not. $\endgroup$ – user458361 Jun 29 '17 at 4:32
  • 2
    $\begingroup$ I second AJ Stas's advice. Reduce modulo 2. The polynomial $x^5+x^2+1$ has no zeros in $\Bbb{Z}_2$, and it is not divisible by the unique (!) quadratic irreducible either. $\endgroup$ – Jyrki Lahtonen Jun 29 '17 at 4:41
  • $\begingroup$ By the way, $x^5 + 1$ is not irreducible mod $5$ because $-1$ is a root. $\endgroup$ – cat Jun 29 '17 at 7:29

Try $x=y-1$. It should help by the Eisenstein's criterion.

  • $\begingroup$ Yeah it worked!! Thanks Dear!! $\endgroup$ – user458361 Jun 29 '17 at 4:47
  • $\begingroup$ @user458361 You are welcome! $\endgroup$ – Michael Rozenberg Jun 29 '17 at 4:59

$7^5+5\cdot 7^2+1 = 17053$ which is a prime number. Thus the polynomial is irreducible by Cohn's irreducibility criterion.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.