The question is True or False: "If a polynomial $f$ has integer coefficients and is irreducible over $\mathbb{R}$, then it is irreducible over $\mathbb{Z}$."

The solution manual claims this is True but I think that has to be a mistake. $2x^2+2$ is irreducible over $\mathbb{R}$ but 2 is not a unit in $\mathbb{Z}$ so $2(x^2+1)$ would be consider a non trivial factorization over $\mathbb{Z}$ right?

  • 8
    $\begingroup$ Reducibility of a polynomial in this case refers to breaking it down into a product of polynomials of lower degrees. $\endgroup$ Aug 10 at 11:43
  • 1
    $\begingroup$ It's a bit sad that "Irreducible in the ring $\Bbb Z[x]$" and "Irreducible polynomial over $\Bbb Z$" don't necessarily mean the same thing. But such is life, unfortunately. $\endgroup$
    – Arthur
    Aug 10 at 12:11
  • 2
    $\begingroup$ Recall that an element $r$ of a unique factorization domain (UFD) is irreducible if and only if the equation $r = st$ implies that $s$ is a unit or $t$ is a unit. But the polynomial $2x^2 + 2 = 2(x^2 + 1)$ is not irreducible in $\mathbb Z[x]$ because neither $2$ nor $x^2 + 1$ is a unit in $\mathbb Z[x].$ $\endgroup$ Aug 10 at 17:43
  • 1
    $\begingroup$ @Arthur Wait, what? $\endgroup$ Aug 11 at 0:39
  • 1
    $\begingroup$ @Dylan r should be non-zero and a non-unit $\endgroup$ Aug 11 at 0:41

Your Answer

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

Browse other questions tagged or ask your own question.