Clarification of answer to "Connectedness of the spectrum of a tensor product" I recently came across this interesting question here on Mathematics Stackexchange:

Connectedness of the spectrum of a tensor product

Its only answer is excellent, but there is one part of it that I don't understand. The answerer claims that $i_g(X)\cap i_h(X)\neq\varnothing$ if and only if $gh^{-1}$ belongs to the inertia subgroup $I_x$ of some $x\in X$. First of all I believe that should be $h^{-1}g$ in stead of $gh^{-1}$, but more pertinently it seems to me that it is sufficient that $h^{-1}g$ belongs to the decomposition group $G_x$ of some $x\in X$. After all, if $i_g(x)=i_h(y)$ for some $x,y\in X$ then $x=y$ and $g(x)=h(y)=h(x)$, so $h^{-1}g\in G_x$.
Although this is no problem for the validity of the proof, it does leave me wondering what I am misunderstanding, if anything at all.

EDIT: I see now that if $h^{-1}g\in I_x$ then $gh^{-1}=h(h^{-1}g)h^{-1}\in I_{hx}$, so my first concern is solved. But my main point of confusion, that I believe it to be sufficient for $h^{-1}g$ (or $gh^{-1}$) to belong to a decomposition group in stead of an inertia group, remains.
 A: The set-theoretic fibre of $X \times_S X$ over $s \in S= \operatorname{Spec} \mathbf Z$ is not the product $X_s \times X_s$ of the set-theoretic fibre $X_s$ of $X \to S$ above $s$ with itself. Instead it is the fibre product $X_s \times_{\operatorname{Spec} \kappa(s)} X_s$, i.e. the spectrum of the tensor product $\mathcal O_F/p \otimes \mathcal O_F/p$. It's easiest understood via its universal property: a $k$-point of $X \times_S X$ is a pair of morphisms $f_1, f_2 \colon \operatorname{Spec} k \to X$ whose compositions with $X \to S$ agree.
Now if $p \in X \times_S X$ is a point in $i_g(X)$, this means that the map $\operatorname{Spec} \kappa(p) \to X \times_S X$ factors through
\begin{align*}
i_g \colon X &\to X \underset S\times X \\
x &\mapsto (x,g(x)).
\end{align*}
In other words, the components $f_1, f_2 \colon \operatorname{Spec} \kappa(p) \to X$ are related by $f_2 = g \circ f_1$. Both projections $i_g(X) \to X$ are isomorphisms, so if $x_1 = f_1(p)$ and $x_2=f_2(p)$, then $f_i \colon \operatorname{Spec} \kappa(p) \to X$ factors through an isomorphism $\operatorname{Spec} \kappa(p) \stackrel\sim\to \operatorname{Spec} \kappa(x_i)$, which we denote by $\bar f_i$. The elements $x_1, x_2 \in X$ are related by $g(x_1) = x_2$, so $g$ also induces an isomorphism $\operatorname{Spec} \kappa(x_1) \stackrel\sim\to \operatorname{Spec} \kappa(x_2)$, which we denote $\bar g$. Then we see $\bar f_1 = \bar g \circ \bar f_2$.
If $p$ is both in $i_g(X)$ and $i_h(X)$, then $\bar f_1 = \bar g \circ \bar f_2 = \bar h \circ \bar f_2$, so we conclude that $\bar g = \bar h$ as isomorphisms $\operatorname{Spec} \kappa(x_1) \stackrel\sim\to \operatorname{Spec} \kappa(x_2)$, i.e. that $gh^{-1} \in I_{x_2}$. Conversely, if $gh^{-1} \in I_{x_2}$, then $\bar g = \bar h \colon \operatorname{Spec} \kappa(x_1) \stackrel\sim\to \operatorname{Spec} \kappa(x_2)$, so we conclude that $p \in i_g(X) \cap i_h(X)$.
