Given a perfect field $F$ of prime characteristic the ring of Witt Vectors $W(F)$ is a discrete valuation ring. For example, $W(\mathbb{F}_p)$ is the ring of $p$-adic integers. Is it possible to embed an arbitrary unramified discrete valuation ring of mixed characteristic into a Witt Vector ring of a perfect field?
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1$\begingroup$ Any witt ring is unramified over Z_p, so I would guess no. $\endgroup$– Dima SustretovCommented Jun 6, 2015 at 14:07
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$\begingroup$ @Dima Do you, by ramified, mean that the residual characteristic $P$ lies in the square power of the maximal ideal of the discrete valuation ring? If so, no problem, I should modify the question as: Is it possible to embed an arbitrary unramified discrete valuation ring into a Witt Vector ring of a perfect field? $\endgroup$– AuroraCommented Jun 6, 2015 at 14:17
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5$\begingroup$ Notation: In the context of Witt vectors, the field with $p$ elements should definitely not be denoted by $\mathbb Z_p$! $\endgroup$– ACLCommented Jun 6, 2015 at 17:01
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1$\begingroup$ @ACL, the "error" has been fixed! $\endgroup$– KConradCommented Jun 6, 2015 at 18:24
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1 Answer
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I think this is possible, if I understand your question correctly. If $R$ is a discrete valuation ring of mixed characteristic $(0,p)$ with residue field $k$ and maximal ideal $pR$, then $R$ is a Cohen ring for $k$. Cohen rings are unique up to (generally non-unique) isomorphism, and there is a construction of the Cohen ring for any field $k$ of characteristic $p$ which realizes the ring as a subring of $W(k)$. Now $W(k)$ is a subring of $W(k^{p^{-\infty}})$, where $k^{p^{-\infty}}$ is the perfect closure of $k$.
answered Jun 6, 2015 at 16:40