We can calculate the discriminant of an algebraic number field $\mathbb{Q}(\sqrt{d})$, with $d$ not a square by: $$ disc(\mathbb{Q}(\sqrt{d}))=\det(\sigma_i(\alpha_j))^2 $$ (ref: p. 18 "Number Fields" Marcus)
where $\sigma$ is in the the Galois group of $\mathbb{Q}(\sqrt{d})$ over $\mathbb{Q}$, which must be empty for $\mathbb{Q}(\sqrt{d})=\mathbb{Q}$, and $\alpha$ which is the integral basis of $\mathbb{Q}(\sqrt{d})$, which must be $\{1\}$ for $\mathbb{Q}(\sqrt{d})=\mathbb{Q}$. I suppose then the $disc(\mathbb{Q})=1$ as the determinant of an $0\times 0 $ matrix is $1$. Or am I mistaken? Can't find any on the particular case.