# On resolution of singularities over an Artin ring

For a locally noetherian scheme $$X$$, Grothendieck conjectured that if $$X$$ is quasi-excellent then there is a proper birational map $$Y \to X$$ s.t. $$Y$$ is regular.

We now fix an Artin ring $$R$$ whose residue field $$k$$ is perfect. Let $$X$$ be a f.g. scheme over $$R$$. Under the assumption of resolution of singularities over $$k$$, the Grothendieck's conjecture is true for $$X$$.

Question. Under the assumption of resolution of singularities over $$k$$, is there a proper birational map $$Z \to X$$ such that $$Z$$ is smooth over $$R$$?

• What happens for $R=\mathbf{C}[\varepsilon]/(\varepsilon^2)$ and $X=\operatorname{Spec} R[x,y]/(xy-\varepsilon)$? Feb 25, 2021 at 10:30
• Simple proof: we show that there is no surjection $Y\to X$ (not necessarily proper or birational) with $Y$ smooth over $R$. We first check by hand that there is no section of $X\to \operatorname{Spec} R$ through the point $P=\{x=y=\varepsilon=0\}$. If $Y\to X$ is a surjection, we lift $P$ to a point $Q$ on $Y$. If $Y$ is also smooth over $R$, the map $Y\to \operatorname{Spec} R$ has a section through $Q$. Composing with $Y\to X$, we get a contradiction. Feb 25, 2021 at 10:34