my questions relates to the following talk of Tsuji:
https://www.youtube.com/watch?v=2brDj26phP0
At around 10:30 of the video, Tsuji is interrupted by a man stating that his construction does not work if $k$ is not algebraically closed. I do not understand this remark at all.
For those of you that do not want to have a look at the video here is a quick summary of the part in question:
The assumptions are as follows: $K$ a complete valuation field of mixed characteristic (0,p) with ring of integers $\mathcal{O}_K$ uniformizer $\pi$ and residue field $k$ (algebraically closed). $A$ an semistable algebra i.e. $$ A \underset{étale}{\leftarrow} \mathcal{O}_K[T_1,\cdots, T_c, \cdots, T_d]/(T_1 T_2 \cdots T_c-\pi)$$ let $\mathcal{K}=Frac(A)$ and $\overline{\mathcal{K}}$ an algebraical closure with $\overline{K} \subset \overline{\mathcal{K}}$. For every finite extension $\mathcal{K} \subset \mathcal{L} \subset \overline{\mathcal{K}}$ denote by $A_{\mathcal{L}}$ the integral closure of $A$ in $\mathcal{L}$. Define the set $$ S:=\big\{\mathcal{L} \, \vert \, A_{\mathcal{L}}[\frac{1}{T_1\cdots T_d}] /A[\frac{1}{T_1\cdots T_d}] \, \text{étale}\big\}$$ such that $Spec(A)$ is connected and $Spec(A/\pi A)\neq 0$ and $Spec(A/\sum_{i \in I} T_iA)$ is irreducible or empty for all $I \subset\{1,\cdots,d\}$. Set $\mathcal{K}^{ur}=\cup_{\mathcal{L}\in S} \mathcal{L}$. He then states that $$ Gal(\mathcal{K}^{ur}/\mathcal{K}\overline{K})= \pi_1(A[1/T_1\cdots T_d] \otimes _{\mathcal{O}_K} \overline{K},Spec(\overline{K})).$$ Why is this not true if $k$ is not algebraically closed? Any help would be very much appreciated.
