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Feb 7, 2019 at 18:35 vote accept Will Chen
Feb 7, 2019 at 16:31 answer added Will Sawin timeline score: 13
Feb 7, 2019 at 11:58 comment added Jason Starr Let $L=K(\{s_i:i\in I\})$ be a maximal purely transcendental subextension of $K((t))/K$. Denote the integer ring of $K$ by $\mathfrak{o}_K$, and let $\mathfrak{p}$ be an ideal over $p\mathbb{Z}$. For the infinite polynomial ring $S=\mathfrak{o}_K[\{s_i:i\in I\}]$, the ideal $\mathfrak{p}S$ is a height one prime. The localization $S_{\mathfrak{p}S}$ is a DVR. For the algebraic field extension $K((t))/L$, there is a valuation ring $R$ dominating $S_{\mathfrak{p}S}$ whose fraction field equals $K((t))$. By Krull-Akizuki, you can choose $R$ to be a colimit of DVRs, probably not a DVR itself.
S Feb 7, 2019 at 10:07 history suggested user26857 CC BY-SA 4.0
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S Feb 7, 2019 at 10:07
Feb 6, 2019 at 6:55 history asked Will Chen CC BY-SA 4.0