Lemma 10.160.6. Let $p$ be a prime number. Let $k$ be a field of characteristic $p$. There exists a Cohen ring $\Lambda$ with $\Lambda /p\Lambda \cong k$.

Proof. First note that the $p$-adic integers $\mathbf{Z}_ p$ form a Cohen ring for $\mathbf{F}_ p$. Let $k$ be an arbitrary field of characteristic $p$. Let $\mathbf{Z}_ p \to R$ be a flat local ring map such that $\mathfrak m_ R = pR$ and $R/pR = k$, see Lemma 10.159.1. By Lemma 10.97.5 the completion $\Lambda = R^\wedge$ is Noetherian. It is a complete Noetherian local ring with maximal ideal $(p)$ as $\Lambda /p\Lambda = R/pR$ is a field (use Lemma 10.96.3). Since $\mathbf{Z}_ p \to R \to \Lambda$ is flat (by Lemma 10.97.2) we see that $p$ is a nonzerodivisor in $\Lambda$. Hence $\Lambda$ has dimension $\geq 1$ (Lemma 10.60.13) and we conclude that $\Lambda$ is regular of dimension $1$, i.e., a discrete valuation ring by Lemma 10.119.7. We conclude $\Lambda$ is a Cohen ring for $k$. $\square$

Comment #6049 by Mark on

It might be better to add a citation for the sentence "Then clearly $R$ is a discrete valuation ring." I think it's regular+dimension 1 (Tag 00PD (3)). (At least this was not clear to me at first glance, I wondered quite a while around noetherianness.)

Comment #6186 by on

Thanks for catching this! No you can't get this immediately from Lemma 10.119.7 because indeed you don't know that it is Noetherian (a priori). And indeed right now I don't see why $\bigcap p^nR = 0$ (with $R$ as in the proof). I fixed it by completing first. Unfortunately now the proof is rather long and I feel it doesn't do justice to what's really going on. Anway, the changes can be viewed here.

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