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I have a very simple question, because I basically just need to know if a certain train of thought I've had is correct. My reference is Liu's book "Algebraic Geometry and Arithmetic Curves", in particular Proposition 8.1.15, and of course Hartshorne. Consider the following situation:

Let $f:W\to X$ be a morphism of locally Noetherian schemes. Let $\mathcal{I}$ be a quasi-coherent sheaf of ideals on $X$. Now, I will only require

            $\mathcal{K}\supseteq(f^{-1}\mathcal{I})\mathcal{O}_W=:\mathcal{J}$

to be a quasi-coherent sheaf of ideals on $W$ which contains the inverse image ideal sheaf. Let $\pi:\widetilde{X}\to X$ and $\rho:\widetilde{W}\to W$ denote the blowing-ups of $X$ and $W$ with respective centers $\mathcal{I}$ and $\mathcal{K}$. Then there exists a map $\widetilde{f}:\widetilde{W}\to\widetilde{X}$ such that

          $\begin{matrix} \widetilde{W} & \xrightarrow{\quad\widetilde{f}\quad} & \widetilde{X} \\ \hphantom{\scriptstyle\rho} \downarrow {\scriptstyle\rho} & {\scriptstyle\circlearrowleft} & \hphantom{\scriptstyle\pi} \downarrow {\scriptstyle\pi} \\ W & \xrightarrow{\quad f\quad} & X \end{matrix}$

This can be shown exactly as in Liu's book, but the more important point is this: It would seem to me that $(\rho^{-1}\mathcal{J})\mathcal{O}_{\widetilde{W}}$ is an invertible sheaf on $\widetilde{W}$, so I would also get uniqueness of $\widetilde{f}$.

My question is very simple: Have I missed anything or made some mistake? I am asking because both Hartshorne and Liu require $\mathcal{K}=\mathcal{J}$ in their respective propositions, but I see no reason why it could not be weakened to $\mathcal{K}\supseteq\mathcal{J}$.

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1 Answer 1

up vote 5 down vote accepted

Suppose $f={\rm Id}_X$, $X={\bf A}^3_{\bf C}$ (affine space of dimension $3$ over the complex numbers). Suppose that $\cal I$ is the sheaf of ideals of a smooth curve going through $0$ and that $\cal K$ is the sheaf of ideals of the point $0$ in ${\bf A}^3_{\bf C}$. Then the pull-back of ${\cal J}={\cal I}$ to $\widetilde{W}$ defines a subscheme $Z$ of $\widetilde{W}$ and the dimension of the intersection of $Z$ with the complement of the exceptional divisor of $\widetilde{W}$ is of dimension $1$ and thus $Z$ is not a Cartier divisor (ie $(\rho^{-1}J){\cal O}_{\widetilde{W}}$ is not an invertible sheaf).

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I see. My motivation was the following: I want to blow up $X$ in a subvariety $Y$ and $W$ in $f^{-1}(Y)$, so in my case $\cal K$ would be the radical of $\cal J$. Can I get it to work in this special case? –  Jesko Hüttenhain Aug 30 '11 at 13:13
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@Jesko Hüttenhain : I don't think so. I don't have an example handy, but in the example I give, you could replace the smooth curve by some wild infinitesimal neighborhood of $0$ (one, which is not regularly embedded in $X$, for instance); there is no reason why the pull-back of such a neighborhood should give a Cartier divisor on $\widetilde{W}$. Remember that a codimension one closed subscheme in a regular scheme is not necessarily a Cartier divisor (but it will be if it is reduced - see Prop. 1.12A in Hartshorne). –  Damian Rössler Aug 30 '11 at 15:58
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(continued) ... In fact an example is simply the following. Let $f$ and $X$ be as in my answer and let $\cal K$ be the sheaf of ideals of a reduced surface in ${\bf A}^3_{\bf C}$; let $\cal I$ be any closed subscheme, with the same support as $\cal K$, which is not a Cartier divisor. Then $\rho$ is an isomorphism and by construction $(\rho^{-1}J){\cal O}_{\widetilde{W}}$ is not an invertible sheaf. –  Damian Rössler Aug 30 '11 at 17:04

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