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Dec 9, 2018 at 20:03 comment added Piotr Achinger More precisely, the variant with $p$-adics would be: $A=\mathbb{Z}_p[x]$, $I=(p)$ and for $a = c/p^n \in \mathbb{Q}_p$ with $c\in\mathbb{Z}^\times_p$ and $n>0$, the algebra $A[1/(p^n x-c)]$.
Dec 9, 2018 at 19:51 comment added Piotr Achinger Sorry, I misunderstood the question. But I think you can make a variant of the above counterexample work with $A = \mathbb{Z}_p[t]$ and $a \in \mathbb{Q}_p\setminus \mathbb{Z}_p$. However, in your particular case ($\mathbb{Z}_{(p)}$ rather than the $p$-adics), it seems that the rings are countable, so there are only countably many f.p. etale algebras over them, no?
Dec 9, 2018 at 19:37 comment added user132229 I see your point. However, in my question $A$ is smooth over some Noetherian henselian valuation ring with uniformizer $x$ (I should have made it clear the valuation ring is non-archimedean: at least I always mean that), and the ideal $I$ should be the extension of the principal maximal ideal of the valuation ring, ie. $xA$. In my question, I am thinking about $R = \mathbf{Z}_{(p)}^h$, $A$ smooth over $R$, $I = pA$. I have edited my final question to only include this case
Dec 9, 2018 at 19:24 history answered Piotr Achinger CC BY-SA 4.0