1

Here it is claimed that:

The dimension of a valuation ring is equal to the rank of its value group.

I couldn't find confirmation of this statement so I'm trying to prove it myself. Here is what I have so far.

Let $R$ be a valuation ring with (surjective) valuation $v \colon Q(R)^\times \to \Gamma$, so that $\Gamma$ is its value group. Suppose that $\Gamma$ has rank $r$, so it has a maximal linearly independent subset $\{z_1, \ldots, z_r\}$. Then $\Gamma$ has at least $r + 1$ isolated (convex) subgroups (a subgroup such that if it contains $a$, $b$ then it contains all $x$ in between $a \le x \le b$): the trivial subgroup and each of the subgroups $\langle{z_1}\rangle, \ldots, \langle{z_r}\rangle$.

By exercise 32 in chapter 5 of Atiyah-Macdonald "Introduction to Commutative Algebra", there is a bijective correspondence between the set of prime ideals of $R$ and the set of isolated subgroups of $\Gamma$. Therefore there are at least $r + 1$ prime ideals. Since in a valuation ring ideals are linearly ordered by inclusion, this means that there is a chain of prime ideals of length at least $r$: $$ (0) = \mathfrak{p}_0 \subsetneq \mathfrak{p}_1 \subsetneq \cdots \subsetneq \mathfrak{p}_r $$ Therefore the dimension of $R$ is at least $r$.

Can you help me verify and complete the proof, namely show there are exactly $r + 1$ isolated subgroups of $\Gamma$?

Anakhand
  • 3,075
  • I do not know what an isolated subgroup is. However, your value group is finitely generated and abelian, thus, it is isomorphic to $$\mathbb{Z}^n \oplus \bigoplus_{j=1}^m \left(\mathbb{Z}/a_j\mathbb{Z}\right).$$ This might make it easier to classify your isolated subgroups. – Severin Schraven Dec 03 '23 at 23:24

1 Answers1

2

Suppose $\dim R=r$. Then the chain you've written consisting of $r+1$ prime ideals must be all of the prime ideals of $R$, since ideals of $R$ are totally ordered by inclusion. Thus $R$ has exactly $r+1$ prime ideals and so $\Gamma$ has exactly $r+1$ isolated subgroups.

TY Mathers
  • 19,533