A example of a commutative chain ring

Question

We say a commutative ring $R$ is a chain ring whenever its ideals form a chain with respect to inclusion. I am looking for a chain ring with Krull dimension two? Thank you for any help.

Answer 1

By applying the Krull-Kaplansky-Jaffard-Ohm theorem to a totally ordered group of suitable size, you can create valuation domains with whatever Krull dimension you wish.

The special case (first proved by Krull himself) of the theorem says that every totally ordered abelian group can be realized by the divisibility group of a valuation domain, and later this was expanded to say that every lattice-ordered abelian group can be realized as the divisibility group of a Bezout domain.

Answer 2

From Bourbaki, Commutative Algebra, ch. 5, Valuations:

The group $G=\mathbf Z\times\mathbf Z$, lexicographically ordered, has height $2$. We can associate a valuation on a field $K$ with value group $G$ in the following way:

• $K=\bigl\{(x_{m,n})_{(m,n)\in G}\mid \operatorname{supp}\bigl((x_{m,n})\bigr)\enspace\text{is well-ordered}\bigr\}$. One checks componentwise addition is well defined on $K$, and it can be endowed with a multiplication (analogous to the definition of multiplication in formal power series rings). For these operations $K$ is a field.
• The valuation $v$ of an element $(x_{m,n})_{(m,n)\in G}$ is defined to be the smallest element in $\operatorname{supp}\bigl((x_{m,n})\bigr)$

The valuation ring $V=\bigl\{(x_{m,n})_{(m,n)\in G}\;\big\vert\; v\bigl((x_{m,n})\bigr)\ge (0,0)\bigr\}$ is the required height $2$ valuation domain.