My main issue is with the case when $d \equiv 1 \mod 4$ is a squarefree integer. In this situation I know that the ring of integers in $\mathbb{Q}(\sqrt{d})$ is $\mathbb{Z}[(\sqrt{d} + 1)/2]$. Now, we observe that $\mathbb{Z}[(\sqrt{d} + 1)/2]$ has a $\mathbb{Z}$-basis given by $\{1, (1 + \sqrt{2})/2\}$ (I suspect that this is my mistake).
Here I will take the definition of trace to be the sum of the galois conjugates. The Galois group of $\mathbb{Q}(\sqrt{d})/\mathbb{Q}$ is $\mathbb{Z}/2\mathbb{Z}$, generated by the automorphism sending $\sqrt{d} \mapsto -\sqrt{d}$. As such, $\operatorname{Tr}(1) = 2$ and $\operatorname{Tr}((1 + \sqrt{d})/2) = 1/2 + 1/2 = 1$.
Similarly, we see that $$\left(\frac{1 + \sqrt{d}}{2} \right)^2 = \frac{1 + 2\sqrt{d} + d}{2}$$ has trace $1 + d$. Hence, the discriminant is the determinant of the matrix $$\begin{bmatrix}
2 & 1\\ 1 & 1 + d
\end{bmatrix}$$ which is $2d + 1$. Now, Wikipedia states that this discriminant is $d$, which leads me to question my solution.
Where is my mistake?
Thanks!
EDIT: I'll add that I'm defining the discriminant to be $\operatorname{det}(\operatorname{Tr}(x_ix_j))$ where $\{x_i\}$ is an $\mathbb{Z}$-basis of the ring of integers. The alternative definition, which defines it to be $[\operatorname{det}(x_i^{\sigma_j})]^2$ gives the solution more clearly but I am nevertheless unsettled by the fact that my above solution is wrong.
Best Answer
You didn't square $\frac{1+\sqrt{d}}{2}$ correctly. The denominator should be 4 which means the trace is $(1+d)/2$. If you replace $1+d$ with $(1+d)/2$ in your matrix, you get a determinant of $d$ as expected.