Let $n=[L:K]$ and write $g(X)=\prod_{j=1}^n(X-\alpha_j)$. Lagrange interpolation, followed by summing over the conjugates, shows that the trace-dual of the power basis $1,\alpha,\ldots,\alpha^{n-1}$ is contained in $g'(\alpha)^{-1}\mathcal O_K[\alpha]$ and has the same determinant. Hence
$$
\mathfrak D_{L/K}^{-1}=\frac1{g'(\alpha)}\mathcal O_L,
\qquad
\mathfrak D_{L/K}=(g'(\alpha)).
$$

For $L=\mathbb Q(\sqrt3)$, the <ring of integers of a quadratic field> is $\mathbb Z[\sqrt3]$. Taking $g=X^2-3$ gives
$$
\mathfrak D_{L/\mathbb Q}=(2\sqrt3).
$$
For $L=\mathbb Q(\sqrt5)$, use $\alpha=(1+\sqrt5)/2$ and $g=X^2-X-1$. Then
$$
\mathfrak D_{L/\mathbb Q}=(2\alpha-1)=(\sqrt5).
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 136 / 1 / a / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-136/1/a/ii/solution