abstract algebra – Irreducibility in $mathbb{Z}[x]$ implies irreducibility in $mathbb{Q}[x]$?

The question is

Show that there is no irreducible monic polynomial of degree 5 in $mathbb{Z}(x)$ that is irreducible in the field $mathbb{Q}(sqrt{2})$.

My attempt: Let $f(x)$ is an irreducible monic polynomial of degree 5. Let $alpha$ is a root of $f(x)$. Thus $(mathbb{Q}(alpha):mathbb{Q})=5$.

Since $f(x)$ is reducible over $mathbb{Q}(sqrt{2})$, then

$5= (mathbb{Q}(alpha):mathbb{Q})=(mathbb{Q}(alpha):mathbb{Q}(sqrt{2})). (mathbb{Q}(sqrt{2}):mathbb{Q})=(mathbb{Q}(alpha):mathbb{Q}(sqrt{2})). 2$

This is contradiction because $2$ does not divide $5$.

In the first line I take $f(x)$ is irreducible over $mathbb{Q}$. Is it true irreducibility in $mathbb{Z}(x)$ implies irreducibility in $mathbb{Q}(x)$?