Diagonal morphism of henselization is an open immersion?

Question

Let $(R,\mathfrak{m})$ be a local ring, denote by $R \rightarrow R^h$ its henselization. Write $S = \operatorname{Spec} R$ and $S^h = \operatorname{Spec} R^h$. Is it true that the diagonal morphism $\Delta \colon S^h \rightarrow S^h \times_S S^h$ is an open immersion? I am reading the chapter by Orgogozo and Vidal in "Courbes semi-stables et groupe fondamental en geometrie algebrique", where this is used (without proof) in the proof of Lemme 5.4.

I have tried the following. Write $$R^h = \varinjlim_{(A,\mathfrak{q})} A,$$ with the limit running over the pairs $(A,\mathfrak{q})$, where $A$ is an étale algebra over $R$, and $\mathfrak{q} \in \operatorname{Spec} A$ is such that $\kappa(\mathfrak{q}) = R/\mathfrak{m}$ (TAG 04GN). Then $$ S = \varprojlim_{(A,\mathfrak{q})} \operatorname{Spec} A,$$ and $$S^h \times_S S^h = (\varprojlim_{(A,\mathfrak{q})} \operatorname{Spec} A) \times_R (\varprojlim_{(A,\mathfrak{q})} \operatorname{Spec} A) = \varprojlim_{(A,\mathfrak{q})} (\operatorname{Spec} A \times_R \operatorname{Spec} A).$$ Now the diagonal morphism $\Delta \colon S^h \rightarrow S^h \times_S S^h$ is the $\varprojlim$ of the individual diagonal morphisms $\operatorname{Spec} A \rightarrow \operatorname{Spec} A \times_R \operatorname{Spec} A$. Each of them is an open immersion, because $A$ is étale over $R$, in particular, $A$ is unramified over $R$, and hence we can apply TAG 02GE. Can I somehow deduce from this that $\Delta$ is itself an open immersion? I tried to apply the theory in EGA IV.3, Section 8, but it is not obvious to me how the setup there is applicable to this problem. Any help would be greatly appreciated!

Answer 1

This is not true.

As pointed out by R. van Dobben de Bruyn in the comments, $\Delta$ is an open immersion if and only if its ideal $\operatorname{ker}( R^h \otimes_R R^h \to R^h)$ is finitely generated.

Since $R^h$ is the filtered colimit of the $A \otimes_R A$ for $A$ pointed étale algebras over $R$, any finite set of generators must be contained in some $A \otimes_R A$ and thus in $\operatorname{ker}(A \otimes_R A \to A)$.

Since every element of $\operatorname{ker}(A \otimes_R A \to A)$, when sent to $R^h \otimes_R R^h$, certainly lies in $\operatorname{ker}( R^h \otimes_R R^h \to R^h)$, if the kernel is generated by elements of $A$ then in fact

$$\operatorname{ker}( R^h \otimes_R R^h \to R^h) = \operatorname{ker}(A \otimes_R A \to A) \otimes_{ (A \otimes_R A)} (R^h \otimes_R R^h)$$

which implies

$$ (R^h \otimes_R R^h) \otimes_{(A \otimes_R A)} A = R^h.$$

The left hand side is the colimit over pointed étale algebras $B$ over $A$ of $(B\otimes_R B) \otimes_{ (A \otimes_R A)} A = B\otimes_A B$ and thus is $A^h \otimes_A A^h$. Since $A^h$ is $R^h$, this is saying that $A^h \otimes_A A^h = A^h$, which is absurd unless $A = A^h$, for example by restricting to the generic point and noting that this map is never an isomorphism for any nontrivial algebra over a field.

So this only happens if some finitely generated étale algebra over $R$ is already Henselian, which does not happen in the case of interested.

(This was partially inspired by the strategy of Angelo in the comments.)