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For a flat morphism $f:X \rightarrow B$ and a sub scheme $Z$ of $B$ we know that the strict and total transforms of $X$ with respect to the blow up at $Z$ agree.

I want to know what happens when $f$ fails to be flat. Suppose that $f$ fails to be flat at some point $P$ and we blow up $P$, could the strict transform of $X$ agree with the total transform? It feels as if the non-flat locus of the total transform of $f$ should contain the entire exceptional divisor which in many cases would show that this is not the case. My first question is suppose we have $f:X \rightarrow B$ a morphism of varieties not flat at $P$ in $B$ and we blow up $P$ can the product $X \times_B Bl_P B$ be flat over $Bl_P B$?

In general suppose that we have a non flat morphism of varieties $f:X \rightarrow Y$ and a proper surjective morphism $p:Z \rightarrow Y$ could the product $Z\times_Y X$ ever be flat over $Z$?

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  • $\begingroup$ What precisely is your question? Do you just want an example? Consider the case where $Y$ is $\text{Spec}\ k[u,v]/\langle u^2,uv \rangle$, where $X$ is the closed subscheme $\text{Spec}\ k[u,v]/\langle u \rangle$, and where you blow up the ideal $\mathfrak{m} =\langle \overline{u},\overline{v} \rangle$ in $Y$. $\endgroup$
    – Jason Starr
    Commented Oct 4, 2017 at 12:03
  • $\begingroup$ Oh yes, I should include some sort of integrality constraints for the picture I had in mind to work. I'll edit the question appropriately. $\endgroup$
    – Lawrence Jack Barrott
    Commented Oct 4, 2017 at 13:15
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    $\begingroup$ If $Y$ is reduced, and if $X\to Y$ is not flat, then also $Z\times_Y X\to Z$ is not flat. This is local, but for simplicity, consider the case when $X\to Y$ is projective and $Y$ is connected. Then $X\to Y$ is flat if and only if the Hilbert polynomials of geometric fibers are constant, cf. Theorem 9.9, p. 261 of "Algebraic geometry" by Robin Hartshorne. If two geometric fibers of $X\to Y$ have distinct Hilbert polynomials, then the same holds for $Z\times_Y X \to Z$ for points of $Z$ mapping to the two points of $Y$. $\endgroup$
    – Jason Starr
    Commented Oct 4, 2017 at 13:57
  • $\begingroup$ Is the general result in EGA or somewhere else? I'm looking for the general case, eg $X$ may very well not be proper over $Y$. $\endgroup$
    – Lawrence Jack Barrott
    Commented Oct 4, 2017 at 17:32
  • $\begingroup$ It is Th'eor`eme 11.8.1, p. 159 of EGA IV_3. For $Y$ reduced and Noetherian, if $X\to Y$ is not flat, then there exists a closed point $y\in Y$ and a DVR $R$ in $\text{Frac}(Y)$ dominating the local ring $\mathcal{O}_{Y,y}$ such that the base change of $X$ by $i:\text{Spec}\ R \to Y$ is not flat over $R$. For every proper, birational morphism $f:Z\to Y$, by the valuative criterion of properness, the morphism $i$ factors through $f$, i.e., $i$ equals $f\circ g$. Since the further pullback by $g$ is not flat, the pullback of $X$ by $f$ is not flat. $\endgroup$
    – Jason Starr
    Commented Oct 4, 2017 at 19:22

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I am just posting my comments above as an answer. If $Y$ is allowed to be nonreduced, then there does exist a finitely presented, non-flat morphism $\pi:X\to Y$ and a proper, birational morphism $Z\to Y$ such that $Z\times_Y X \to Z$ is flat. One example is when $Y=\text{Spec}\ k[u,v]/\langle u^2,uv \rangle,$ when $X$ is the closed subscheme $\text{Spec}\ k[u,v]/\langle u \rangle,$ and when $Z$ is the blowing up of $Y$ at the ideal $\mathfrak{m}= \langle \overline{u},\overline{v} \rangle.$

If you assume that $Y$ is reduced, then the positive result follows from the "valuative criterion of flatness", Théorème 11.8.1 of EGA $IV_3.$ For simplicity, assume that $Y$ is integral. If $X\to Y$ is not flat, then there exists a point $y\in Y$ and a DVR $R$ with fraction field $\text{Frac}(Y)$ containing $\mathcal{O}_{Y,y}$ such that the base change of $\pi$ by $i:\text{Spec}\ R \to Y$ is not flat over $\text{Spec}\ R.$

For every proper, birational morphism $f:Z\to Y,$ by the valuative criterion of properness, there exists a morphism $g:\text{Spec}\ R \to Z$ such that $i$ equals $f\circ g.$ If the pullback of $\pi$ by $f$ is flat over $Z,$ then the further pullback along $g$ is flat over $\text{Spec}\ R.$ This contradicts that the pullback of $\pi$ by $i$ is not flat over $\text{Spec}\ R.$

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  • $\begingroup$ Thanks Jason, that is exactly what I was hoping for. I'll accept it. $\endgroup$
    – Lawrence Jack Barrott
    Commented Oct 5, 2017 at 16:13

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