Let $A$ be a non-noetherian, commutative $\mathbb{C}$-algebra and $X, Y$ be noetherian affine $\mathbb{C}$-schemes. Denote by $X_A:=X \times_{\mathbb{C}} \mbox{Spec}(A)$ and $Y_A:=Y \times_{\mathbb{C}} \mbox{Spec}(A)$. Let $m$ be a maximal ideal of $A$ and $A^\wedge$ the completion of $A$ with respect to $m$. Let $f:X_{A^\wedge} \to Y_{A^\wedge}$ be a morphism of finite type. Then, does there exist a morhism $g:X_A \to Y_A$ such that $f$ is obtained by the base change of $g$ under the natural morphism $A \to A^\wedge$?
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2$\begingroup$ It is not even true if $A$ is noetherian: take $A=\mathbb{C}[t]$, $\widehat{A}=\mathbb{C}[[t]]$, $X=\mathrm{Spec}(\mathbb{C})$, $Y=\mathrm{Spec}(\mathbb{C}[y])$, $f$ given by $\mathbb{C}[[t]][y]\to\mathbb{C}[[t]]$ sending $y$ to any $h\in\mathbb{C}[[t]]\smallsetminus\mathbb{C}[t]$. Geometrically, this is the section of the affine line over $\mathbb{C}[[t]]$ "with coordinate $y=h(t)$". $\endgroup$– Laurent Moret-BaillyMar 20, 2019 at 16:26
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$\begingroup$ @LaurentMoret-Bailly Thank you. I want to ask a follow-up. Suppose I have a morphism $f^\wedge:(X_A)^\wedge \to (Y_A)^\wedge$. Does there exist $g$ as above such that $f^\wedge$ is obtained as the completion of $g$? If I understand correctly, this holds in the noetherian case (completion of noetherian scheme is faithful). $\endgroup$– RonMar 22, 2019 at 21:26
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