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This is again false. The geometric interpretation is as follows: write $Y = \operatorname{Spec} S$ and $X_i = \operatorname{Spec} R_i$. Given étale morphisms $f_1 \colon Y \to X_1$ and $f_2 \colon Y \ …

Note that $\overline{X^*}$ is the scheme-theoretic image of $X^* \to \mathbf A^n_B$: if $X^*$ is reduced, this agrees with the reduced induced structure (see for instance exercise II.3.11 in Hartshorn …

As I wrote in my comment, these extensions are known in algebraic geometry as unramified, which can be tested computationally by the vanishing of $\Omega_{S/R}$. (In differential geometry, this means …

Like your other question, the answer to this one is related to miracle flatness: Theorem (Miracle flatness). Let $f \colon X \to Y$ be a finite dominant morphism of schemes with $Y$ regular. … Then miracle flatness says that $Y' \to X$ is flat. …

Let $U = X \setminus x_0$, which is smooth by assumption, so $\pi|_U$ is flat by miracle flatness. Now I claim that $\pi|_U \colon U \to \mathbf A^3$ does not have a flat compactification. … But $X$ is not Cohen–Macaulay, contradicting flatness of $\pi'$. $\square$ …

This is true, and here is one possible proof. There might be easier ones; however I suspect that it will always involve some algebra (not just geometry). Write $S = \operatorname{Spec} \mathbf Z_p$ a …

Since $\mathbf C[x]$ and $\mathbf C[y]$ are principal ideal domains, flatness is equivalent to torsion-freeness. …