Let $R,S$ be discrete valuation rings inside the same fraction field $K$. Suppose $R\neq S$. Suppose $t\in R$ is the uniformizer, i.e. the generator of its unique maximal ideal. Is it true that $t\not\in S$?

I'm not sure whether this is even true. My intuition says it must be, but I am not able to prove it. I know that this is implied by the statement that the units in $R$ are units in $S$, but I'm not able to prove this either.

Edit: Actually I don't really need this statement in full generality. It would be enough for me to know whether this is true in the case where $R$ and $S$ contain a field $k$ such that $K/k$ is finitely generated with transcendence degree $1$.


Inside $\mathbb{Q}$ consider the localizations $R_p=\mathbb{Z}_{(p)}$ at the various primes $p$. The ring $R_p$ is a DVR with uniformizer $p$.

Then every prime $q$ belongs to $R_p$ for every $p$ but it is a unit whenever $p\neq q$.

For an example closer to your Edit consider the field of rational functions $\mathbb{Q}(t)$ and the localizations of the polynomial ring $\mathbb{Q}[t]$ at the various maximal ideas.

  • $\begingroup$ Ah, thank you very much. $\endgroup$ – asdq Jan 22 '18 at 12:56

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