# Henselianizations over countable index sets

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?

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 $$\mathbb{C}$$.

• 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 – user132229 Dec 9 '18 at 19:37
• 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? – Piotr Achinger Dec 9 '18 at 19:51
• 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 $A[1/(p^n x-c)]$. – Piotr Achinger Dec 9 '18 at 20:03