It is well-known that there is a universal characterization of blow-ups. However, this one is not so easy to use. According to Hartshorne-Theorem II.8.24. The blow-up of a nonsingular $X$ along a nonsingular subvariety $Y$ have three properties: 1. The blow-up $\tilde{X}$ is nonsingular. 2. The blow-up restricts to the exceptional divisor $\tilde{Y}$ is the projective bundle $P$ of the normal bundle of $Y$ in $X$. 3. The normal sheaf $N_{\tilde{Y}/\tilde{X}}$ corresponds to the $P(-1)$.

My question is that assume we have a projective birational morphism $f: \tilde{X}\to X$ with exceptional locus $Y$ on $X$. We assume both $X$ and $Y$ are nonsingular as before (even projective). Let $\tilde{Y}$ be the inverse image sheaf of $Y$. If $f$ satisfies the above three properties, must it be the blow-up of $X$ along $Y$? Especially, I want to see an example for a non-blowup having property 1,2. Thank you so much.


1 Answer 1


Example (property 1 fails, but property 2 is satisfied)

Look for $f$ as the blow up of an ideal sheaf $\mathscr I$, so $\widetilde X=\mathrm{Proj}_X(\oplus_d \mathscr I^d)$. Then the pre-image of the subscheme $Z\subset X$ defined by the ideal $\mathscr I$ is given by $\widetilde Z=\mathrm{Proj}_Y(\oplus_d \mathscr{I^d/I^{d+1}})$. Now if $X$ is Cohen-Macaulay and $Z$ is a complete intersection in $X$, (i.e., $\mathscr I$ is generated by a regular sequence), then $\mathscr{I/I^2}$ is locally free and $\mathscr{I^d/I^{d+1}}\simeq \mathrm{Sym}^d(\mathscr{I/I^2})$ and hence $\widetilde Z\simeq \mathbb P(\mathscr{I/I^2})$.

Property #3 is kind of a red herring. The $(-1)$-twist is almost automatic, it comes from the construction of the blow up of $\mathscr I$.

Finally, here is a simple concrete example: Let $X$ be a plane (or any smooth surface) and $\mathscr I=(x^2,y^2)$ where $x,y$ are local coordinates at a point. The blow up will be the surface with a pinch point (locally around the interesting singularity defined by $x^2z=y^2$) with the singular line contracted to a point. I think it is relatively easy to check that this satisfies properties #2 and #3.

To round things up Mike Roth in the comments below gives a nice example of a blow up along a non-smooth subvariety such that the resulting variety is actually smooth.

  • $\begingroup$ As a complement to the construction above, here is an entertaining example of a blow up of a smooth variety along a non-smooth subvariety, such that the blow up is again smooth. Let $X$ be $\mathbb{A}^{n(n+1)}$ thought of as the space of $n\times (n+1)$ matrices, and let $Y$ be the closed locus of matrices of rank $\leq n-1$. Then the blow up $\widetilde{X}$ (of $X$ along $Y$) is smooth. Explictly it is the incidence correspondence $\Gamma = \{(M,v) | Mv =0 \} \subset \mathbb{A}^{n(n+1)} \times \mathbb{P}^{n}$, which we can see is smooth by viewing it as a fibration over $\mathbb{P}^{n}$. $\endgroup$
    – Mike Roth
    Dec 9, 2011 at 8:48
  • $\begingroup$ Thanks S$\'{a}$ndor for detailed answer. Unfortunately, I do ask for the nonsingular case (both $X$ and $Y$ are nonsingular as a scheme), which is also the assumption for the Theorem. We can also assume $f$ to be projective birational morphism, but I was not asking any Chow Lemma type statment but something very explicit. For the second question, the counterexamples I want to see should satisfy the property 1,2. I am sorry that I didn't mention property 1 before. I have updated the question. I am also new to MO. $\endgroup$
    – Jiarui Fei
    Dec 10, 2011 at 1:18
  • $\begingroup$ "As a scheme" is really a nonsense. Sorry. But do you think if we ask $\tilde{X}$ to be nonsingular, is there any chance to get some examples? $\endgroup$
    – Jiarui Fei
    Dec 10, 2011 at 2:07
  • $\begingroup$ Let $X$ be the plane (or a smooth surface), $g:W\rightarrow X$ the blowup of $X$ at a point $p$, and $h:\tilde X\rightarrow W$ the blowup at a point $q$ above $p$ (ie, such that $g(q)=p$). It seems to me that $g\circ h:\tilde X \rightarrow X$ satisfies the hypotheses of your claim (it does not satisfy #2 and #3 of the op) with $Y=\{p\}$, but it is not the blowup of $p$. What am I getting wrong? $\endgroup$
    – quim
    Dec 11, 2011 at 8:30
  • $\begingroup$ @quim: you're right, I think I secretly used more than what I stated.... $\endgroup$ Dec 11, 2011 at 8:37

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