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I am trying to prove a claim in curves, and the example below shall contradict it.

Is there a DVR, $(A,\mathfrak{m})$, of char $0$ such that the residue field $\mathcal{k}(A) = A/\mathfrak{m}$ is a non-perfect field?

Basically I would like to find an example of a local ring extension $A\to B$ with both $(A,\mathfrak{m}_A)$ and $(B,\mathfrak{m}_B)$ being DVRs in some characteristic such that

  • $\mathfrak{m}_AB = \mathfrak{m}_B$

  • $K(B)$ is a separable extension of $K(A)$

  • $\mathcal{k}(B)$ is NOT a separable extension of $\mathcal{k}(A)$

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  • $\begingroup$ The characteristic $p$ example you're thinking of is $\mathbb F_p(x)[[t]]$ right? $\endgroup$ – RKD Aug 4 '17 at 19:31
  • $\begingroup$ Sorry I changed the question. I am thinking of $\mathbb{F}_p(x)[t]_{(t)}$. $\endgroup$ – Grobber Aug 4 '17 at 19:33
  • $\begingroup$ Yeah, that makes sense. Mine is just the completion of yours at the prime ideal. $\endgroup$ – RKD Aug 4 '17 at 19:38
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Asking for $\mathbb{F}_p(x)$ as the residue field immediately suggests $\mathbb{Q}(x)$ as the characteristic zero lift. The DVR you're looking for has

$$ A = \mathbb{Z}[x]_{(p)} \qquad \qquad \mathfrak{m} = p A$$

Explicitly, $A$ consists of all rational functions with integer coefficients for which $p$ does not divide the denominator.

We can verify this meets the conditions by noting:

  • $A$ is a unique factorization domain where every element is a unit times an integer power of $p$
  • The residue field is $\mathbb{Z}[x]_{(p)}/p \cong \mathbb{F}_p[x]_{(p)} \cong \mathbb{F}_p[x]_{(0)} \cong \mathbb{F}_p(x)$

Take care to note $A \neq \mathbb{Z}_{(p)}[x]$; I often see people get the notation mixed up.

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  • $\begingroup$ Just to add, if you wanted a complete example of this, we could just take $\hat A = \mathbb Z_p(x)$ right? $\endgroup$ – RKD Aug 4 '17 at 20:06
  • $\begingroup$ @Ravi: Assuming you mean rational functions whose Laurent series have $\mathbb{Z}_p$-integral coefficients, I don't think the completion of $(A,\mathfrak{m})$ works out that simply. $\sum p^n x^{n^2}$ should converge (?) in the completion, but it's not a rational function! This issue is complicated and I'm not sure what to expect is the right answer. $\endgroup$ – user14972 Aug 4 '17 at 20:12
  • $\begingroup$ Oh you are right. Indeed, that series should converge because the difference of two partial sums $S_n-S_m$ (which are just polynomials) will have $\mathfrak m$-adic valuation $\min(m,n)$ in the original ring $A$ so it is a Cauchy sequence. $\endgroup$ – RKD Aug 4 '17 at 20:23

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