# 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 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 '17 at 16:26

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.