I have the following situation: $R,H$ schemes (can be assumed noetherian and of finite type) over a field $k$ which we can assume to be algebraically closed, with $H$ reduced, $Y\subset R\times \mathbb{P}_k^n$ an open subset, $p:Y\rightarrow R$ the restriction of the projection onto the first factor and $w:Y\rightarrow H$ a surjective formally smooth morphism. How can I show that $R$ is reduced? Thank you

  • $\begingroup$ Dear unknown, if R is the disjoint union of a point and a point with multiplicity two, H is a point, n=0, Y is the reduced point of (R cross P^0), and w an isomorphism, then your situation seems to hold but R is not reduced. Perhaps I am missing something, eg R connected? David $\endgroup$ – David Holmes Mar 15 '11 at 14:39
  • $\begingroup$ @David Holmes why $Y$ smooth$/ H$ should imply $R$ reduced? $\endgroup$ – unknown Mar 15 '11 at 14:44
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    $\begingroup$ Are you missing something in your question? You don't seem to use the morphism $p$ at all. As it stands it appears false: if $R$ is, say, a curve with an embedded point $P$, then take $Y$ to be the complement of $p^{-1}(P)$ in $R \times \mathbb{P}^1$, take $H=Y$ and $w$ the identity morphism. $\endgroup$ – Martin Bright Mar 15 '11 at 14:44
  • $\begingroup$ @David Holmes yes, can assume $R$ connected $\endgroup$ – unknown Mar 15 '11 at 14:46
  • $\begingroup$ @Martin Bright ok, what about if we assume also $p$ formally smooth or smooth? $\endgroup$ – unknown Mar 15 '11 at 14:51

Since $H$ is reduced and $Y$ is smooth over $H$ (I am assuming that everything is finite type over $k$, so smooth and formally smooth are the same) we see that $Y$ is reduced.

So the problem is the following: show that if $Y \subset R \times \mathbb P^n$ is open and reduced, and the projection $Y \to R$ is surjective (taking into account the remark to this effect in the comments), then $R$ is reduced.

Here is the proof: Let $x$ be a point of $R$, and let $y$ be a point of $Y$ lying over $x$. Recalling that $\mathbb P^n$ is the union of $n + 1$ open subsets isomorphic to $\mathbb A^n$, we may assume that $y \in R\times \mathbb A^n$ (for an appropriate choice of one of these $n+1$ copies). The stalk $\mathcal O_{Y,y}$ is then equal to a localization of $\mathcal O_{R,x}[x_1,\ldots,x_n]$. It is reduced by assumption, and so $\mathcal O_{R,x}$ is reduced. Since $x \in R$ was arbitrary, we see that $R$ is reduced.

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