2
$\begingroup$

I need to show that $x^4+4x^3+6x^2+9x+11$ is irreducible in the integers.

First, I tried to apply Eisenstein's irreducibility criterion by shifting $x$ to $x+\alpha$. However, I can't think of any shift to apply that would fit the criterion.

Next, I tried using polynomial division. If it were reducible, this polynomial would have either a linear or a quadratic factor. It has no integer roots, so I tried to divide by a factor $x^2+ax+b$ and require that the remainder be zero; however, this yields two very difficult equations, I'm not sure how to prove that they have no integer solution.

Any hints?

$\endgroup$
4
  • 7
    $\begingroup$ Just a thought, but considering that $(x+1)^4 = x^4 + 4x^3 + 6x^2 + 4x + 1$, maybe we can say let $y = x+1$ and get $x^4 + 4x^3 + 6x^2 + 9x + 11 = (x+1)^4 + 5x + 10 = (x+1)^4 + 5(x+1) + 5 = y^4 + 5y + 5$, and then start from there? Just thought it'd be nice to have fewer terms if possible. $\endgroup$
    – 2012ssohn
    Feb 23 '14 at 18:14
  • $\begingroup$ Hint: Consider the polynomial modulo $2$; if it is reducible, it stays reducible modulo any number too. $\endgroup$ Feb 23 '14 at 18:18
  • 3
    $\begingroup$ @2012ssohn ... Actually, Eisenstein applies to $y^4+5y+5$. Huh. $\endgroup$
    – Andrea
    Feb 23 '14 at 18:21
  • $\begingroup$ @PeterKošinár I didn't know that, that's a very useful trick! $\endgroup$
    – Andrea
    Feb 23 '14 at 18:22
1
$\begingroup$

To push it from unanswered queue:

  1. Notice that given polynomial is monic, hence by Rational Root Theorem, all rational solutions are integer, namely $\pm1,\pm11$. Also, Descartes Rule of Signs assures that there are no positive roots, making our list $-1,-11$, and its now easy to verify both won't work.

From comments:

  1. if a polynomial is irreducible over a finite field, it is irreducible over $\mathbb Q$, too, and check if its irreducible in $\mod 2$, as Peter Košinár pointed out.

  2. As 2012sshon pointed out, let $y=x+1$ and as OP pointed out, then use Eisenstein's criteria.

$\endgroup$

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.