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 .

Solution 1:

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.

Solution 2:

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.