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!
 A: 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 $\mathbb Z$.
