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Let $A$ be a ring, $I\subset A$ a finitely generated ideal.

The henselianization $A^h$ of $A$ along $I$ is the universal $A$-algebra that is henselian along $I$ and can be presented as a direct limit of étale ring maps that are the identity on mod $I$ fibers:

$$A^h = \varinjlim_{s\in S} A_s$$

where $A\to A_s$ is étale and such that $A/I\to A_s/I$ is the identity, and $S$ is an index set.

When $A$ is smooth over the Noetherian henselian valuation ring $R = \mathbf{Z}_{(p)}^h$ and $I = pA$ is principal, can $S$ be arranged to be a countable set?

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No.

Take $A = \mathbb{C}[x]$ and $I=(x)$. Suppose that $A^h$ is the direct limit of a system of etale algebras $A_i$ such that $A_i/(x) \cong \mathbb{C}$. We can assume that each $A_i$ is finitely presented.

For $a\in \mathbb{C}\setminus \{0\}$, consider the algebra $A[1/(x-a)]$. Then:

(1) for every nonzero $a\in \mathbb{C}$, the map $A\to A^h$ factors through $A[1/(x-a)]$ (by universal property of henselization)

(2) for every index $i$, there are only finitely many nonzero $a\in \mathbb{C}$ for which $A\to A_i$ factors through $A[1/(x-a)]$ (because the image of ${\rm Spec}(A_i) \to {\rm Spec}(A)$ is open and dense).

This shows that the index set has to have cardinality at least equal to the cardinality of $\mathbb{C}$.

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  • $\begingroup$ 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 $\endgroup$
    – user132229
    Commented Dec 9, 2018 at 19:37
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    $\begingroup$ 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? $\endgroup$
    – Piotr Achinger
    Commented Dec 9, 2018 at 19:51
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    $\begingroup$ 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 $A[1/(p^n x-c)]$. $\endgroup$
    – Piotr Achinger
    Commented Dec 9, 2018 at 20:03

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