Let $f(x) \in \mathbb{Z}[x]$ be a polynomial of degree $\geq 2$. Then choose correct

a) if $f(x)$ is irreducible in $ \mathbb{Z}[x] $ then it is irreducible in $ \mathbb{Q}[x] $.

b) if $f(x)$ is irreducible in $ \mathbb{Q}[x] $ then it is irreducible in $ \mathbb{Z}[x] $.

(1) is definitely true, for (2) $f(x)=2(x^2+2)$ clearly irreducible over $\mathbb{Q}[x]$

But I am confused about whether $f(x)$ is irreducible over $\mathbb{Z}[x]$ or not? According to Gallian, as 2 is non unit in $\mathbb{Z}$, $f(x)$ is reducible over $\mathbb{Z}[x]$, (2) is false.

But definition of irreducible polynomial on Wikipedia says a polynomial is reducible if it can be written as product of non constant polynomials hence $f(x)$ is irreducible over $\mathbb{Z}[x]$ accordingly (2) is true .

  • 2
    $\begingroup$ You may find Gauss' Lemma useful here. $\endgroup$ Jul 1, 2016 at 5:33
  • 2
    $\begingroup$ @Merlin I found this'' Irreducibility statement: A non-constant polynomial in Z[X] is irreducible in Z[X] if and only if it is both irreducible in Q[X] and primitive in Z[X].'' SO can i say for primitive polynomial (b) option hold not for $f(x)=2(x^2+2)$ $\endgroup$ Jul 1, 2016 at 5:50
  • $\begingroup$ If content is one , then i think statement 2 is true. $\endgroup$
    – blue boy
    Nov 13, 2018 at 5:13

2 Answers 2


You are totally correct, (1) is true and (2) is false. The statement you quote from Wikipedia is only true, if the coefficients come from a field.

  • $\begingroup$ Will you please give me precise definition of irreducible polynomial ? Thank you very much for your precious time $\endgroup$ Jul 1, 2016 at 4:58
  • 2
    $\begingroup$ It is just the definition of an irreducible element: It cannot be decomposed in two non-units. $\endgroup$
    – MooS
    Jul 1, 2016 at 4:59

Consider the polynomial $p(x)=3x+3$. Since the coefficients are integer $p(x)$ belongs to $\mathbb{Z}[x] \subset\mathbb{Q}[x] $. We can rewrite it as $3(x+1)$, but now: $3$ is a unit in $\mathbb{Q}$ since it is inveritble, then the polynomial is irreducible, but $3$ is not inveritble in $\mathbb{Z}$, so the factorization above show that the polynomial is reducible as product of irreducible element in $\mathbb{Z}[x]$. So the statement 2 is false.

  • $\begingroup$ As is stated here: mathonline.wikidot.com/… The degree of the factors should be less than that of the original function. I have no idea how that would work for only an x, but why is 3 a factor and in the link I send you its not allowed? $\endgroup$
    – user304120
    Jul 14, 2019 at 13:58
  • $\begingroup$ The definition applies to fields like $\Bbb Q$. Indeed, viewing $3x+3$ as a polynomial in $\Bbb Q[x]$, $3$ is not a factor but a unite. The picture is different if you work with rings which are not fields, like $\Bbb Z$. In this case, $3$ is not invertibile in $\Bbb Z$, hence it is a factor of your polynomial. $\endgroup$
    – InsideOut
    Jul 14, 2019 at 14:04

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