formal smoothness versus reducedness

2011-03-15T14:03:10Z

Hi,

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

Answer by Emerton for formal smoothness versus reducedness

2011-03-15T16:16:59Z

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.

formal smoothness versus reducedness - MathOverflow

formal smoothness versus reducedness

Answer by Emerton for formal smoothness versus reducedness