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Let $X\rightarrow Y$ be a smooth morphism of schemes and consider the blowup of the fiber product along the diagonal, $W:=Bl_{\Delta}X\times_Y X$.

a.Does there exist a collection of smooth morphisms of schemes $X_i\rightarrow Y_i$ such that

1.the schemes $W_i:=Bl_{\Delta_i}X_i\times_{Y_i} X_i$ are irreducible and

2.there exist morphisms $W_i\rightarrow W$ ,such that $\{W_i\rightarrow W\}$ is an etale cover?

[Here $\Delta_i$ denotes the diagonal of $X_i\times_{Y_i} X_i$.]

b.Same question with the additional assumption that $Y$ is smooth.

In case $Y$ is singular, the answer in a. is negative, see Will Sawin's answer below.

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  • $\begingroup$ What are you denoting by $W$? $\endgroup$
    – dhy
    Commented Aug 28, 2014 at 0:05
  • $\begingroup$ Sorry, $W$ is the blowup of fiber product along the diagonal $W=Bl_{\Delta}X\times_Y X$, I'll edit the post. $\endgroup$
    – matthew
    Commented Aug 28, 2014 at 0:07
  • $\begingroup$ Can't one take $Y_i$ an etale cover of $Y$ and $X_i = X \times_Y Y_i$? Fiber product and blow-up are both etale-local constructions, so everything will just be a base change from $Y$ to $Y_i$, hence etale. $\endgroup$
    – Will Sawin
    Commented Aug 28, 2014 at 2:56
  • $\begingroup$ @Will Savin:Why are the resulting schemes irreducible? $\endgroup$
    – matthew
    Commented Aug 28, 2014 at 4:51
  • $\begingroup$ What does the index $i$ mean? $\endgroup$
    – Laurent Moret-Bailly
    Commented Aug 28, 2014 at 6:50

1 Answer 1

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Let $Y$ be the nodal cubic curve and let $X$ be its finite etale double cover. Then $X \times_Y X$ is just a union of two copies of the finite etale double cover. Blowing up changes nothing.

You have a slight problem here - the finite etale double cover is union of two $\mathbb P^1$ which intersect at $0$ and $\infty$. It has no etale cover by any irreducible schemes, because each irreducible scheme must map to only one of the two $\mathbb P^1$s, but any etale neighborhood of either $0$ or $\infty$ intersects both $\mathbb P^1$s.

So I think no $W_i$ work, which means the answer is no.

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  • $\begingroup$ Thank you, Will. I was originally interested in the case where $Y$ is smooth, which I didn't mention in the hypothesis. I'll edit the post to include this case, too. $\endgroup$
    – matthew
    Commented Aug 28, 2014 at 22:26

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