What is the conductor of $ \mathbb Q(\sqrt 2 )/\mathbb Q?$ $4\mathbb Z$ or $8 \mathbb Z?$

Question

By execise 6.8 of Childress's Class Field Theory, $\mathbb f(\mathbb Q(\sqrt 2)/ \mathbb Q)=8\mathbb Z$. But considering the norm of the local field about the place correspondent with 2, it should be $4\mathbb Z$ and there is contraction to the ordering theorem.