# Equivalent characterizations of discrete valuation rings

Let $$R$$ be a commutative ring with identiy, then the following are equivalent:

1. $$R$$ is a DVR
2. $$R$$ is a local Euclidean domain that is not a field.
3. $$R$$ is a local PID that is not a field.
4. $$R$$ is a local Dedekind domain that is not a field.
5. $$R$$ is a UFD with a unique irreducible element up to a unit.
6. There is a non-nilpotent non-unit element $$\pi \in R$$ such that every $$a \in R \setminus\{0\}$$ has a unique expression $$a = u \pi^n$$, where $$u \in R^\times$$ and $$n\in \mathbb{N}$$.
7. $$R$$ is a Noetherian valuation ring and not a field.
8. $$R$$ is a Noetherian local ring with a principal maximal ideal generated by a non-nilpotent element.
9. $$R$$ is a regular local Noetherian ring of dimension $$1$$.
10. $$R$$ is a local Noetherian domain that is not a field such that every non-zero ideal is a power of the maximal ideal.

Here is what I tried so far:

$$(1.) \Rightarrow (2.)$$ Every valuation ring is local. Suppose $$v$$ is a discrete valuation on the fraction field of $$R$$ such that $$v^{-1}(\mathbb{N}) \cup \{0\} = R$$. We claim that $$v$$ is actually also a Euclidean function for $$R$$. Let $$x,y \in R$$, $$y \neq 0$$, if $$\frac{x}{y} \in R$$, we have $$x = y \frac{x}{y} + 0$$, so we are done. If $$\frac{x}{y} \notin R$$, then $$0 > v(\frac{x}{y}) = v(x)- v(y)$$, so $$v(y) > v(x)$$. Now we have $$v(x) = v(y + (x - y)) \geq \min(v(y), v(x - y))$$, but as $$v(y) > v(x)$$, it must be the case that $$v(x) \geq v(x-y)$$. Now we write $$x = 1\cdot y + (x - y)$$ and $$v(y) > v(x) \geq v(x-y)$$.

$$(2.) \Rightarrow (3.)$$ Every Euclidean domain is a PID.

$$(3.) \Rightarrow (4.)$$ Every PID is a Dedekind domain.

$$(3.) \Rightarrow (5)$$ Every PID is a UFD and every irreducible element generates a maximal ideal in a PID and every maximao ideal is generated by an irreducible element (As $$R$$ is not a field, the maximal ideals are non-zero). As all these maximal ideals must coincide, there is exactly one irreducible element up to a unit.

$$(5.) \Rightarrow (6.)$$ A domain does not contain nontrivial nilpotents. So we can take $$\pi$$ to be the unique irreducible element.

$$(1.) \land (3.) \Rightarrow (7.)$$ Every PID is Noetherian.

$$(7.) \Rightarrow (3.)$$ Every valuation ring is a local Bezout domain. Noetherian Bezout domains are PIDs.

$$(4.) \Rightarrow (10.)$$ Let $$R$$ be a local Dedekind domain that is not a field. Then $$R$$ is one-dimensional, so every nonzero prime ideal is maximal, but there is only one maximal ideal as $$R$$ is local, so there is only one non-zero prime ideal. As $$R$$ is Dedekind, every non-zero ideal is a product of prime ideals, thus every non-zero ideal is a power of the maximal ideal.

$$(8.) \Rightarrow (9.)$$ Let the maximal ideal $$m$$ be generated by $$x$$, then the image of $$x$$ generates $$m/m^2$$. If $$m/m^2 = 0$$, $$m = m^2$$, but then $$m = 0$$ by Nakayama's lemma, so $$m/m^2 \neq 0$$, thus $$\operatorname{dim}_{R/m}m/m^2 = 1$$. It follows from Krull's principal ideal theorem that $$\operatorname{dim}R= 1$$.

$$(9.) \Rightarrow (8.)$$ Let $$m$$ be the maximal ideal and $$\bar{x}$$ be a generator of $$m/m^2$$ and let $$x$$ be a preimage of $$\bar{x}$$ under the natural quotient map. We then have $$m = m^2 + (x)$$, so $$m = (x)$$ by a corollary to Nakayama's lemma. Now we need to show that $$x$$ is not nilpotent. As $$\operatorname{dim}R = 1$$, $$m$$ properly contains a minimal prime ideal $$p$$. If $$x$$ was nilpotent, then $$x \in p$$, but this implies $$m \subset p$$, which is impossible.

$$(10.) \Rightarrow (8.)$$ Let $$m$$ be the maximal ideal. Note that we cannot have $$m^2 = m$$ or else $$m = 0$$ by Nakayama's lemma, but $$R$$ is not a field. Choose $$x \in m \setminus m^2$$, then $$(x)$$ is a power of $$m$$, but by our choice of $$x$$ the only possibility is that $$m = (x)$$. As $$R$$ is a domain, $$(x)$$ is not nilpotent.

$$(8.) \Rightarrow (6.)$$ Let $$m = (\pi)$$ be the maximal ideal. We first show that every nonzero $$x_0 \in R$$ has an expression of the required form. First, if $$x_0$$ is a unit, then we are done. If $$x_0$$ is not a unit, then $$x_0 \in m$$, so $$x_0 = x_1 \cdot \pi$$. Then, do the same for $$x_1$$. If $$x_1$$ is a unit, then we are done, if $$x_1$$ is not a unit $$x_1 \in m$$, so we can write $$x_1 = x_2 \cdot \pi$$ etc. To see that this process must terminate, we note that if it doesn't, we have $$x \in m^n$$ for all $$n\in \mathbb{N}$$ which contradicts Krull's intersection theorem. Now that every nonzero element is of the form $$u\pi^n$$ we have to show uniqueness. If we take two non-zero elements write them as $$u_{1}\pi^n$$ and $$u_{2}\pi^k$$, then their product $$u_{1}u_{2}\pi^{n+k}$$ is non-zero, as $$\pi$$ is not nilpotent. This shows that $$R$$ is an integral domain, thus the multiplicative monoid is cancellative, from this, the uniqueness follows easily, if we have $$u_1 \pi^n = u_2 \pi^k$$ Assume wlog that $$n \geq k$$, then we have $$u_1 = u_2 \pi^{n-k}$$. Now if $$n > k$$, the LHS is a unit while the RHS is not, which is absurd. Thus $$n = k$$ and $$u_1 = u_2$$.

$$(6.) \Rightarrow (1.)$$ $$R$$ is an integral domain by the same argument as in $$(8.) \Rightarrow(6.)$$ Now define a valuation on $$R$$ via $$v(u\pi^n)=n$$ We have $$v(u_1\pi^n\cdot u_2 \pi^k) = n+k = v(u_1\pi^n)+v(u_2\pi^k)$$ Assume wlog that $$k \leq n$$, then $$v(u_1\pi^k+u_2\pi^n)=v( (u_1+u_2\pi^{n-k})\pi^k) \geq k = \min(v(u_1\pi^k),v(u_2\pi^n))$$. Extend $$v$$ to the fraction field of $$R$$ by setting $$v(\frac{a}{b})=v(a)-v(b)$$, then it is obvious that $$R$$ is the valuation ring of $$v$$.

My question is, is this all correct? Do you know any other characterization of DVRs? If so, why is it equivalent? For example, wikipedia has the following

$$R$$ is a (edit: Noetherian) domain that is not a field, and every nonzero fractional ideal of $$R$$ is irreducible in the sense that it cannot be written as finite intersection of fractional ideals properly containing it.

But I have no idea how to show that it is equivalent.

• Depending on your definition of a Dedekind domain, you have not included 'one-dimensional normal local domain.
– MooS
Mar 22, 2017 at 16:26
• math.stackexchange.com/q/4140714/688539 Jun 26, 2022 at 22:25

One direction is clear: If $(R, \pi)$ is a DVR, every fractional ideal is of the form $R\pi^{n}$ with $n \in \mathbb Z$, thus they are linearly ordered and in particular the intersection of two of those will always be the smaller one.
For the other direction, the property clearly implies that $R$ is local, because the intersection of two maximal ideals is a proper subset of both maximal ideals. More generally, the set of all fractional ideals is linearly ordered, because $I \cap J \in \{I,J\}$. In particular the set of all sub-vectorspaces of $\mathfrak m/\mathfrak m^2$ is linearly ordered, this of course enforces it to be one-dimensional. Hence $\mathfrak m$ is principal.