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
  • 1
    $\begingroup$ 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.