Which criterion (test) one can use in order to prove that $x^4+x^3+x^2+3x+3 $ is irreducible over ring $\mathbb{Z}$ of integers ?

Neither of Eisenstein's criterion and Cohn's criterion cannot be applied on this polynomial. I know that one can use factor command in Wolfram Alpha and show that polinomial is irreducible but that isn't point of this question.

  • $\begingroup$ Is semi brute force OK? If so, there is an easy argument. It actually doesn't care very much about the coefficient of $x$, as long as integer root is avoided. $\endgroup$ – André Nicolas Nov 10 '11 at 5:35
  • $\begingroup$ @AndreNicolas,Actually I would like to know if there is some good computational-algebra algorithm $\endgroup$ – Peđa Terzić Nov 10 '11 at 5:52
  • $\begingroup$ @There are good canned procedures for computing Galois groups. Overkill, probably. $\endgroup$ – André Nicolas Nov 10 '11 at 6:53

Let your polynomial be $f$. Clearly it has no linear factor, since it has no root in $\mathbb{Z}$. Hence if it factors, it factors as the product of two irreducible quadratics $f_1, f_2$.

Now looking mod $2$ we get a factorization $f=x^4+x^3+x^2+x+1=f_1f_2$ in $\mathbb{F}_2[x]$. Now since $f$ has no root mod $2$, $f_1$ and $f_2$ are also irreducible quadratics in $\mathbb{F}_2[x]$. But the only irreducible quadratic in $\mathbb{F}_2[x]$ is $x^2+x+1$. This would imply $x^4+x^3+x^2+x+1 = (x^2+x+1)^2$ in $\mathbb{F}_2[x]$, which is false.

  • 2
    $\begingroup$ +1 Another way of seeing that $x^4+x^3+x^2+x+1$ is irreducible in $F_2[x]$ is to observe that it is a factor of $(x^5-1)=(x-1)(x^4+x^3+x^2+x+1)$. Thus any zero would a root of unity of order five. But $q=16$ is the lowest power of two with the property that $5\mid q-1$, so those roots of unity live in $GF(16)$. Thus their primitive polynomial has to be quartic. Your solution may be simpler. $\endgroup$ – Jyrki Lahtonen Nov 10 '11 at 6:54

Hint: If a quartic is reducible it has either a linear factor or a quadratic factor. It is easy to check that your polynomial above has no linear solutions, and you can work out a contradiction if you assume that it can be factored into the product of two (monic) quadratics.


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