# Valuation Ring with value group $\Gamma=\mathbb{Z} \oplus \mathbb{Z}$

Let $\Gamma=\mathbb{Z}\oplus \mathbb{Z}$ be the free abelian group with two generators, lexicographically ordered. That is, $(a,b)\geq (a^{'},b^{'})$ iff either $a>a^{'}$ or $a=a^{'} \mathrm{and}\, b \geq b^{'}$. Prove or disprove: there exists a Noetherian valuation ring with value group $\Gamma$.

Here we take a value group in the following sense: let $A$ be a valuation ring of the field $K$. The group $U$ of units of $A$ is a subgroup of the multiplicative group $K^{*}$ of $K$. Let $\Gamma=K^{*}/U$. If $\alpha, \beta \in \Gamma$ are represented by $x,y \in K$ define $\alpha \geq \beta$ to mean $xy^{-1} \in A$, where this defines a total ordering which is compatible with group structure (i.e. $\alpha \geq \beta \Rightarrow \alpha\omega \geq \beta\omega$ for all $\omega \in \Gamma$). Namely, $\Gamma$ is a totally ordered abelian group. (via Atiyah-Macdonald)

A friend proposed this question to me a little while back. There doesn't seem to be that much to it, but I'm not sure how to get the result. Any help would be appreciated! Thanks!

• Are you sure your question doesn't come from here? math.harvard.edu/~jsuh/pset_7.pdf
– ff90
Nov 8, 2013 at 1:14
• I don't think you should assume that. It's a natural enough question, as $\mathbb Z \oplus \mathbb Z$ is the first example of a totally ordered abelian group that isn't a subgroup of $\mathbb R$. Nov 12, 2013 at 14:34
• @neilme The identical wording of the question just stands out. You do have a point though...
– ff90
Nov 15, 2013 at 23:49

First of all, I think you mean "$(a,b) \geq (a', b')$ iff either (1) $a>a'$ or (2) $a=a'$ and $b>b'$".
As for a construction, I seem to remember that the following works. Let $k$ be a field; let $X,Y$ be indeterminates, let $A = k[X,Y]$, and let $L$ be the fraction field of $A$. Define $v: L^\times \rightarrow \Gamma$ as follows. One can translate the lexicographic ordering given into an ordering on the monomials in $A$. That is, $X^k Y^l > X^p Y^q$ iff $(k,l) > (p,q)$ in $\Gamma$. Then for any nonzero polynomial $f \in A$, if $m = X^k Y^l$ is the "biggest" monomial in the unique reduced form of $f$, then set $v(f) := (k,l)$. Extend $v$ to $L^\times$ by setting $v(f/g) =v(f) - v(g)$. Verify that this is well-defined (independent of how one writes the fraction, as long as $f$, $g$ have no additively cancelable terms); then let $R := \{0\} \cup\{f \in L^\times \mid v(f) \geq (0,0)\}$. I think this is the desired valuation ring.
EDIT: I just saw the word "Noetherian" in your question that I hadn't noticed before. You can't make a Noetherian valuation ring with a value group other than $0$ or $\mathbb Z$. Comparability of ideals in the ring, along with the fact that all ideals are finitely generated, forces every ideal to be principal. (Basically, if $I=(a_1, \dotsc, a_n)$, then since all pairs of ideals are comparable, there must be some $a_i$ such that $I \subseteq (a_i)$, whence $I=(a_i)$.) Then the principalness of the maximal ideal along with Nakayama's lemma and the Krull intersection theorem forces every nonzero ideal to be a power of the maximal ideal.
Namely, say $M=(\pi)$ is the maximal ideal of $R$, where $R$ is a Noetherian valuation ring that is not a field. Let $I$ be any nonzero ideal of $R$. Then by the Krull intersection theorem, there is some $n$ such that $I \subseteq M^n$ but $I \nsubseteq M^{n+1}$. I claim that $\pi^n \in I$. To see this, let $a \in I \setminus M^{n+1}$. Then since $a \in M^n$, we have $a = u \pi^n$ for some $u \in R$, but since $a \notin M^{n+1} = (\pi^{n+1})$, we must have $u \notin (\pi) = M$. Hence $u$ is a unit, so $u^{-1} \in R$, so $\pi^n = u^{-1}a \in (a) \subseteq I$. Thus, $(\pi^n) \subseteq I \subseteq M^n = (\pi^n)$, so $I = (\pi^n)$. Thus, every nonzero element of $R$ is of the form $u \pi^k$, $u$ a unit of $R$, $k \in {\mathbb Z}_{\geq 0}$.
It is easy to see now that $K=R[\pi^{-1}]$ is a field, hence it must be the fraction field of $R$. Moreover, every nonzero element of $K$ is of the form $u \pi^n$, where $u$ is a unit of $R$ and $n \in \mathbb Z$; for any such element, both $u$ and $n$ are uniquely determined. Then the map $v: K^\times \rightarrow \mathbb Z$ given by $v(u \pi^n) = n$ satisfies the conditions of a valuation, and clearly $R = \{x \in K \mid v(x) \geq 0\}$. Hence, $R$ is a valuation ring with value group $\mathbb Z$.
• The end of your first line should be "...and $b\geq b'$", since you certainly want $(a,b)\geq (a,b)$. Nov 14, 2019 at 9:59