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Let $K$ be a field, and let $S$ be the set of local subrings of $K$. Put an order $\leq$ on $S$ where $A \leq B$ when $A \subset B$ and the maximal ideal of $A$ is sent into the maximal ideal of $B$. Then the valuation subrings of $K$ are the maximal elements with respect to this order.

Valuation rings don't have to be maximal under $\subset$ though. It seems like those are the rank-$1$ valuation rings, i.e. DVR's. This question is to confirm this. Actually here is what I want to ask:

Claim 1: Let $S$ be the set of valuation rings of $K$. Are the maximal such valuation rings discrete valuation rings?

and the related claim:

Claim 2: Let $S$ be the set of subrings of $K$ whose fraction field is $K$. Put an order $\leq$ on $S$ where $A \leq B$ when $A \subset B$. Then the discrete valuation subrings of $K$ are the maximal elements with respect to this order.

It seems like 1 is clear since if it is not a DVR (suppose it has value group $\Lambda$) then we can take a special $\lambda \in \Lambda$ with negative valuation and replace $A$ with $A[a]$ where $a \in K$ has value $\lambda$. Edit: actually this only works for valuation rings with finitely generated value group, if at all.

Could someone confirm whether this claim holds?

Ronald J. Zallman
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  • What about fields that don't admit discrete valuations ? – Captain Lama Apr 02 '19 at 02:03
  • Hmm... what is one of those? It seems from Zorn's lemma that a field always has a valuation ring when it has a proper subring. Oh, I guess it could be a field and not a DVR. – Ronald J. Zallman Apr 02 '19 at 02:13
  • A finite field is one of those. – Ronald J. Zallman Apr 02 '19 at 02:13
  • I will change the question to something more reasonable: we suppose its fraction field is the whole field $K$. – Ronald J. Zallman Apr 02 '19 at 02:13
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    What do you think of this: an algebraically closed field cannot have a non-trivial discrete valuation. But it can have non-trivial valuations. – Captain Lama Apr 02 '19 at 02:18
  • That's pretty interesting and it does show that the claim is false. How mysterious, since I really thought this would hold. Do you know an example and what the value group looks like. Can the value group ever be finitely generated? – Ronald J. Zallman Apr 02 '19 at 02:22
  • Ok I've thought about it, and that seems right. I think complications come from having a value group like $\mathbb{Z}^\omega = (n_1, n_2, ... )$ with some dictionary or reverse dictionary order. I think one of those should be able to be maximal. – Ronald J. Zallman Apr 02 '19 at 02:31

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