Showing explicitly by using the valuation that DVRs are Dedekind domains.

Question

We know that DVRs are Dedekind domains by the characterization of Dedekind domains in terms of localizations at maximal ideals. There are many nice abstract ways to prove that DVRs are Dedekind domains. This is not what this question is about. I would like to see the following:

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Can we see explicitly how the existence of a valuation map into $\mathbb Z$ ensures our ring is a Dedekind domain?

Here are the definitions I am using:

• $A$ is a dedekind domain if every non zero ideal $\mathfrak{a}$ has a factorization into primary ideals, $\mathfrak{a}=\prod q_i$ where the $q_i$.

• $A$ is DVR if it is a valuation ring and the valuation map has image in $\mathbb Z$.

How can we see that every non zero ideal has a factorization into primary ideals using just that we have a valuation map in $\mathbb Z$.

Answer 1

The ideals of a DVR are all of the form $(\pi^n)$ where $\pi$ is any element with valuation $1$ (let's assume that the image of $v$ is $\mathbb{Z}$ and not $k\mathbb{Z}, k >1$).

To see this, suppose $I$ is an ideal and pick $a \in I$ with minimal valuation $n$. Then $v(a/\pi^n) = 0$ so $a/\pi^n$ is a unit, so $\pi^n \in I$. This means that $(\pi^n) \subseteq I$.

Conversely, if $b \in I, b \neq 0$ then $v(b) \ge v(a)$ so $v(b/\pi^n) \ge 0$. Therefore $b/\pi^n \in R$ and therefore $b = (b/\pi^n) \cdot \pi^n \in (\pi^n)$.

So given that the ideals of $A$ are all of the form $(\pi^n)$, we immediately get unique factorization of (fractional) ideals because $(\pi^n) = (\pi)^n$.

The ideal $(\pi)$ is maximal since $x \in (\pi) \iff v(x) \ge 1$ so $A - (\pi) = \{x : v(x) = 0\} = A^\times$.