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

The inverse different isIt is an ${\mathcal{O}}_L$-submodule of $L$. Choose an integral basis $e_1,\dots ,e_n$ of a finite-index free submodule of ${\mathcal{O}}_L$. Nondegeneracy of the trace pairing gives a dual $K$-basis $e_1^{\ast },\dots ,e_n^{\ast }$, and the codifferent lies between two finitely generated full ${\mathcal{O}}_K$-lattices obtained from these bases. It is therefore a fractional ${\mathcal{O}}_L$-ideal.

Every algebraic integer has integral trace, so ${\mathcal{O}}_L\subseteq {\mathfrak{D}}_{L/K}^{-1}$. Consequently its inverseis contained in ${\mathcal{O}}_L$. It is thus an integral ${\mathcal{O}}_L$-ideal, called the different ideal.

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 $1,\alpha ,\dots ,{\alpha }^{n-1}$ is contained in $g^{\prime }(\alpha {)}^{-1}{\mathcal{O}}_K[\alpha ]$ and has the same determinant. Hence

For $L=\mathbb{Q}(\sqrt{3})$, the ring of integers of a quadratic field is $\mathbb{Z}[\sqrt{3}]$. Taking $g=X^2-3$ givesFor $L=\mathbb{Q}(\sqrt{5})$, use $\alpha =(1+\sqrt{5})/2$ and $g=X^2-X-1$. Then

The ring $\mathbb{Z}[T]$ is Noetherian and integrally closed, but it is not a Dedekind domain because it has Krull dimension two. Concretely, the nonzero prime ideal $(2)$ is properly contained in the prime ideal $(2,T)$ and is therefore not maximal.

The element $T$ belongs to the fraction field of $R=\mathbb{C}[T^2,T^3]$ and is integral over $R$, since it satisfies the monic polynomial $X^2-T^2$. But $T\notin R$. Thus $R$ is not integrally closed domain and hence is not a Dedekind domain.

The ring $\mathbb{Z}[{\zeta }_3]$ is the full ring of integers of the quadratic number field $\mathbb{Q}(\sqrt{-3})$. Every ring of integers of a number field is a Dedekind domain, so this ring is Dedekind.