dedekind domain
Idea
Locally, at least away from prime ideals, a Dedekind domain looks like a discrete valuation ring. In particular, they are locally PIDs.
Definition
Let $\mathcal{O}$ be a Noetherian integrally closed domain, and $k$ be its field of fractions.
The following are equivalent characterizations of a Dedekind domain:
- Any nonzero prime ideal is maximal
- For any nonzero prime ideal $\mathfrak{p}\subset\mathcal{O}$, the local ring $\mathcal{O}_\mathfrak{p}$ is a d.v.r.
- For any fractional ideal $I\subset k$, we have $I^{-1}I=\mathcal{O}$.
Notice (1) is also a property shared by Artinian rings.
Properties
Let $\mathcal{O}$ be a Dedekind domain
- The fractional ideals $I\subset k$ form a commutative group under multiplication, denoted $\mathcal{F}(\mathcal{O})$, which is freely generated by the prime ideals of $\mathcal{O}$.
- If $I\subset k$ is a fractional ideal, then $$I=\prod_\mathfrak{p}\mathfrak{p}^{v_\mathfrak{p}(I)}$$ (which is a finite product)
- For all $\mathfrak{p}$, $$I\mathcal{O}\mathfrak{p}=(\mathfrak{p}\mathcal{O}\mathfrak{p})^{v_\mathfrak{p}(I)}$$
- Suppose $N\subset K$ is an arbitrary $\mathcal{O}$-submodule. Then $$N=\bigcap_\mathfrak{p}N_\mathfrak{p}$$
- The integral closure $\mathcal{O}_K$ is a Dedekind domain.
- Suppose $\mathfrak{p}\subset\mathcal{O}$ is a prime ideal. Then $(\mathcal{O}K)\mathfrak{p}$ is the integral closure of $\mathcal{O}_\mathfrak{p}$ in $K$, i.e. $$(\mathcal{O}K)\mathfrak{p}=(\mathcal{O}_p)_K$$
- For any prime ideal $\mathfrak{p}\subset\mathcal{O}$, there exists a prime ideal $\mathfrak{B}\subset\mathcal{O}_K$
lying over it, i.e. $\mathfrak{B}\mid\mathfrak{p}$.
- Conversely, for any nonzero prime ideal $\mathfrak{B}\subset\mathcal{O}_K$, we have $\mathfrak{B}\cap\mathcal{O}\neq(0)$.
- (compare to property 3 in complete dvr’s at discrete valuation ring) $$n=\sum e_if_i$$ (see also residue class degree, ramification index)
Proofs
2 Suppose $I\subset\mathcal{O}$ is a nontrivial nonzero ideal. There exists a maximal (hence prime) ideal $\mathfrak{p}_1\subset\mathcal{O}$ such that $I\subset\mathfrak{p}_1$. Then $I\mathfrak{p}_1^{-1}\subset\mathcal{O}$. (Recall $\mathfrak{p}_1^{-1}\coloneqq{x\in k\mid x\mathfrak{p}_1\subset\mathcal{O}}$.) Note also that $I\subset I\mathfrak{p}_1^{-1}$ since $I\subset\mathfrak{p}_1\subset\mathfrak{p}_1^{-1}$ since $\mathfrak{p}_1\subset\mathcal{O}$.
If $I\mathfrak{p}_1^{-1}$ is nontrivial, it is contained in a maximal ideal $\mathfrak{p}_2$ such that $I\mathfrak{p}_1^{-1}\mathfrak{p}_2^{-1}\subset\mathcal{O}$ and $I\mathfrak{p}_1^{-1}\subset I\mathfrak{p}_1^{-1}\mathfrak{p}_2^{-1}$. We thus obtain a chain $$I\subset I\mathfrak{p}_1^{-1}\subset I\mathfrak{p}_1^{-1}\mathfrak{p}_2^{-1}\subset\cdots$$ Because all of these are in $\mathcal{O}$ which is Noetherian, this chain stabilizes hence comes to a trivial ideal in a finite number of steps.