Checking solution on discriminant of $\mathbb{Q}(\sqrt{-3})$ I am reviewing for an exam, and am trying to work out the discriminant of the number field $\mathbb{Q}(\sqrt{-3})$, from the definition. Because $-3\equiv 1 (\text{mod}\; 4),$ we know the answer must be $-3$. However, when attempting to do it from the definition, I am getting mixed up. Here's my work. I'm using the fact that the discriminant of the number field with basis (as a $\mathbb{Z}$-module) $\{b_1,...,b_n\}$ and complex embeddings $\{\sigma_1,...,\sigma_n\}$ is $\left[\det(M_{ij})\right]^2,$ where $M_{ij}$ is the matrix whose $ij$ entry is $\sigma_i (b_j)$. Here, our basis is $\{1,\sqrt{-3}\},$ and our complex embeddings are $\{\sigma_1,\sigma_2\}$, where $\sigma_1$ is the identity, and $\sigma_2$ is the conjugate map. So, our matrix should be
$$\begin{pmatrix}
\sigma_1(1) & \sigma_1 (\sqrt{-3})\\
\sigma_2(1) & \sigma_2 (\sqrt{-3})
\end{pmatrix}$$
$$=\begin{pmatrix}
1 & \sqrt{-3}\\
1 & \sqrt{-3}
\end{pmatrix}$$
the determinant of this matrix is $-2\sqrt{-3}$, and the square of it is $-12$.
Where have I gone wrong? I should get $-3$ as the answer, not $-12$.

$\{1,\sqrt{-3}\}$ is not a $\mathbb Z$-basis for $O_K$ since $x^2+x+1$ has root $w=\frac{-1+\sqrt{-3}}2$. This points to the correct basis $\{1,w\}$ and we get the matrix as
$$\begin{bmatrix}1&w\\1&\overline w=w^2\end{bmatrix}$$
with determinant $w^2-w=-2w-1=-\sqrt{-3}$ and square $-3$.
