# 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 $$mathbb{Q}(x)$$?