# Examples of perfect Valuation rings, of finite Krull dimension, with non-finitely generated maximal ideal

Let $$R$$ be a ring of prime characteristic $$p>0$$. Then we can consider the Frobenius map $$F_R : R \to R$$ given by $$F_R(x)=x^p$$. Let us call $$R$$ to be perfect if $$F_R$$ is an isomorphism.

My question is: What are some large class of examples of perfect Valuation rings, of finite Krull dimension, with non-finitely generated maximal ideal ? Like how to construct such rings in general ?

I am interested only in the case of non-finitely generated maximal ideal because I know that perfect local rings with finitely generated maximal ideals are fields.

For any field $$K$$ of characteristic $$p$$ and any non-trivial valuation $$v:K\to G$$ (to an ordered abelian group) then $$O_v = \{ a \in K, v(a)\ge 0\}$$ is a valued ring and $$R=O_v^{1/p^\infty} = \{ a \in \overline{K}, \exists m, a^{p^m} \in O_v\}$$ is a perfect valued ring.

Example : $$O_v = \Bbb{F}_p[[x]], R= \Bbb{F}_p[[x]][x^{1/p^\infty}]$$.

Any perfect valued ring is of this form.

If the maximal ideal of a valued ring is finitely generated then the value group is a finitely generated $$\Bbb{Z}$$-module so $$a \mapsto a^p$$ can't be surjective.

For the Krull dimension, the proper prime ideals of a valuation ring correspond to subsets $$S$$ of the value group satisfying $$s > 0, nt > s \implies t \in S$$. Thus the Krull dimension depends on the "depth" of the value group.

For example the non-archimedian valuation $$v(x^n y^m) = n+m\epsilon\in \Bbb{Z+\epsilon Z}$$ on $$k[x,y]$$, the order is $$n +m\epsilon \ge 0$$ iff $$n> 0$$ or $$n=0,m\ge 0$$, extend $$v$$ to $$Frac(k[x,y])$$ then the valuation ring $$O_v = \{ a \in k(x,y), v(a) \ge 0\}= x k(y)[x]_{(x)}+k[y]_{(y)}$$ has 3 prime ideals $$\{0\}, (x)$$ and $$(x,y)$$.

• So basically your $R$ is the colimit of $F: O_v \to O_v \to O_v \to ...$ ... could you add why every perfect valuation ring appears this way ... I mean if you take the limit of $O_v$ along Frobenius map then again you get a perfect ring , not sure whether it is a valuation ring or not ... also why is your $R$ not a field ?
– user
Sep 11, 2019 at 1:56
• Because all its elements have $\ge$ valuation. A nonarchimedian valuation on a field always extend to the algebraic closure and in a unique way to $K^{1/p^\infty}$ in characteristic $p$. If $R$ is a perfect valued ring then $R=R^{1/p^\infty}$ and $R =\{ a \in Frac(R),v(a)\ge 0\}$ so they all appear this way. The Krull dimension is superflous since the only prime ideals are $\{0\}$ and $m$. Maybe you meant a perfect local domain (for example $\Bbb{F}_p[[x,y]][x^{1/p^\infty},y^{1/p^\infty}]$) instead of perfect valued ring in which case it is more complicated. Sep 11, 2019 at 2:11
• So you're saying every perfect valuation ring has Krull dimension $1$ ? Unfortunately I don't see that ... could you elaborate why ?
– user
Sep 11, 2019 at 2:16
• A ring is called a Valuation ring if it is an integral domain and all its ideals are totally ordered ... en.m.wikipedia.org/wiki/Valuation_ring
– user
Sep 11, 2019 at 2:19
• Also, why is your $R$ a Valuation ring ?
– user
Sep 11, 2019 at 2:20