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.


1 Answer 1


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 $(x,y)$.

  • $\begingroup$ 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 ? $\endgroup$
    – user
    Commented Sep 11, 2019 at 1:56
  • $\begingroup$ 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. $\endgroup$
    – reuns
    Commented Sep 11, 2019 at 2:11
  • $\begingroup$ So you're saying every perfect valuation ring has Krull dimension $1$ ? Unfortunately I don't see that ... could you elaborate why ? $\endgroup$
    – user
    Commented Sep 11, 2019 at 2:16
  • $\begingroup$ 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 $\endgroup$
    – user
    Commented Sep 11, 2019 at 2:19
  • $\begingroup$ Also, why is your $R$ a Valuation ring ? $\endgroup$
    – user
    Commented Sep 11, 2019 at 2:20

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