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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 $R$?

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    $\begingroup$ What happens for $R=\mathbf{C}[\varepsilon]/(\varepsilon^2)$ and $X=\operatorname{Spec} R[x,y]/(xy-\varepsilon)$? $\endgroup$
    – Piotr Achinger
    Feb 25, 2021 at 10:30
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    $\begingroup$ 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. $\endgroup$
    – Piotr Achinger
    Feb 25, 2021 at 10:34

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