Here's my attempt at calculating the discriminant of $\mathbb{Q}(\sqrt{-13})$. The $\mathbb{Z}$-basis of its ring of integers $R=\mathbb{Z}(\sqrt{-13})$ is $\{e_1,e_2\}:=\{1,\sqrt{-13}\}$
Its complex embeddings are given by $\sigma_1(\sqrt{-13})=\sqrt{-13}$ and $\sigma_2(\sqrt{-13})=-\sqrt{-13}$.
Therefore, its dicriminant is the $[\det(\sigma_i(e_j))]^2$. Which is $-52$. My lecturers notes say the discriminant is $-4{.}13$. Please help me understand where I went wrong
Nothing is wrong, $-4 \times 13=-52$.