# Irreducible polynomial over $\mathbb{Q}$ implies polynomial is irreducible over $\mathbb{Z}$

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 .

• You may find Gauss' Lemma useful here. – SquirtleSquad Jul 1 '16 at 5:33
• @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 $f(x)=2(x^2+2)$ – slow but keen learner Jul 1 '16 at 5:50
• If content is one , then i think statement 2 is true. – blue boy Nov 13 '18 at 5:13

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.
• 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. – InsideOut Jul 14 '19 at 14:04