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Claim Let $f:\widetilde X\to X$ be a projective birational morphism between smooth quasi-projective varieties over an algebraically closed field. Suppose that the image $Y$ of the exceptional set of $f$ in $X$ is also smooth. Then $f$ is the blow up of $X$ along $Y$.

Proof By the assumptions $f$ is the blow up of $X$ along an ideal sheaf $\mathscr I\subset \mathscr O_X$. Let $\mathscr J$ be the ideal sheaf of $Y$ and let $\pi: B\to X$ be the blow up of $X$ along $Y$ (i.e., along $\mathscr J$).

Since by assumption the support of $\mathscr O_X/\mathscr I$ is $Y$, it follows that $\mathscr J$ is the radical of $\mathscr I$ or in other words $Y$ is the reduced scheme of the scheme along which one has to blow up $X$ to get $f$. This implies that $\pi$ factors through $f$: $$\pi: B\overset{\sigma}\to \widetilde X\overset{f}\to X,$$i.e., $\pi=f\circ\sigma$ for an appropriate $\sigma:B\to\widetilde X$.

We may assume that $Y$ is connected and then if it is smooth it has to be irreducible.From the standard facts about blow ups along smooth varieties it follows that the exceptional set of $\pi$ is a single irreducible divisor on $B$ (it is a projective space bundle over the irreducible $Y$).

Now observe that if $\widetilde X$ is smooth (actually already if it is normal and has $\mathbb Q$-factorial singularities), then $\sigma$ is either an isomorphism or it has an exceptional divisor. The same holds for $f$, but there is only one divisor that is exceptional for $\pi$. Therefore either $\sigma$ or $f$ has to be an isomorphism. $\square$

Remark the proof suggests that even assuming that $\widetilde X$ is normal and has $\mathbb Q$-factorial singularities is enough for this characterization. Accordingly in the example below, when the ideal $(x^2,y^2)$ is blown up on the plane, then $\widetilde X$ is not normal and the blow up of the reduced point is the normalization (which happens to be smooth) of the pinch point surface $x^2z=y^2$.

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EDIT: This was

Claim Let $f:\widetilde X\to X$ be a projective birational morphism between smooth quasi-projective varieties over an answer when algebraically closed field. Suppose that the last line image $Y$ of the question said "Especially, I want to see an example for a non-blowup having property 2."

I think you should make your question more preciseexceptional set of $f$ in $X$ is also smooth. If Then $f$ is the blow up of $X$ along $Y$.

Proof By the assumptions $f$ is quasi-projective, the blow up of $X$ along an ideal sheaf $\mathscr I\subset \mathscr O_X$. Let $\mathscr J$ be the ideal sheaf of $Y$ and let $f:\widetilde X\to \pi: B\to X$ be the blow up of $X$ along $Y$ (i.e., along $\mathscr J$).

Since by assumption the support of $\mathscr O_X/\mathscr I$ is a projective birational morphism, then $f$ Y$, it follows that$\mathscr J$is a blow-up the radical of$\mathscr I$or in other words$Y$is the reduced scheme of the scheme along which one has to blow up$X$to get$f$. This implies that$\pi$factors through$f$: $$\pi: B\overset{\sigma}\to \widetilde X\overset{f}\to X,$$i.e.,$\pi=f\circ\sigma$for an appropriate ideal sheaf$\sigma:B\to\widetilde X$.I am guessing this We may assume that$Y$is not what you're afterconnected and then if it is smooth it has to be irreducible.(If you allow From the standard facts about blow ups along smooth varieties it follows that the exceptional set of fractional ideals$\pi$is a single irreducible divisor on$B$(it is a projective space bundle over the irreducible$Y$). Now observe that if$\widetilde X$is smooth (actually already if it is normal and has$\mathbb Q$-factorial singularities), then you can drop even the quasi-projective assumption$\sigma$is either an isomorphism or it has an exceptional divisor. Not assuming The same holds for$f$, but there is only one divisor that is exceptional for$\pi$. Therefore either$\sigma$or$f$is projective would make the question less interesting)has to be an isomorphism. Also$\square$Remark the way you phrased your question, I think there proof suggests that even assuming that$\widetilde X$is a simple example: take any blow ups normal and then choose a subscheme of has$X$with \mathbb Q$-factorial singularities is enough for this characterization. Accordingly in the same supportexample below, but different scheme structure than when the subscheme that was ideal $(x^2,y^2)$ is blown up .

Anywayon the plane, I will assume that you want an example that's then $\widetilde X$ is not a normal and the blow up centered at a smooth subvariety.

I think it of the reduced point is possible the normalization (which happens to give examples satisfying be smooth) of the pinch point surface $x^2z=y^2$.

Example (property #1 fails, but property 2 relatively easily.

Let's look is satisfied)

Look for $f$ as the blow up of an ideal sheaf $\mathscr I$. So , so $\widetilde X=\mathrm{Proj}_X(\oplus_d \mathscr I^d)$. Then the pre-image of the subvariety subscheme $Y\subset Z\subset X$ defined by the ideal $\mathscr I$ is given by $\widetilde Y=\mathrm{Proj}_Y(\oplus_d Z=\mathrm{Proj}_Y(\oplus_d \mathscr{I^d/I^{d+1}})$. Now if $X$ is Cohen-Macaulay and $Y$ 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 Y\simeq 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$. My example is not likely to satisfy property #1, that is$\widetilde X$is probably always singular if$Y$is. As far as finding an example that satisfies all 3 properties, I would not hold my hopes very high. If$\dim X=2$, then these properties just mean that we have a$(-1)$-curve and we know that that is always the blow up of a smooth point. In higher dimensions I would start by taking general hyperplane cuts to reduce to the case when the image is a point. Then Property #2 says that the preimage is a$\mathbb P^n$with normal bundle$\mathscr O_{\mathbb P^n}(-1)$so there is a good chance that one can prove that it is indeed a blow up. Then one would get (probably) that the original map is a blow up at general points and then possibly one can prove that then it has to be a blow To round things up everywhere, but I am not entirely sure about that. EDITMike Roth in the comments below gives a nice example of a blow up along a non-smooth subvariety such that the resulting variety is actually smooth. I don't think this satisfies the other conditions, but it might suggest that proving that those three conditions indeed characterize blow ups along smoothsubvarieties is probably not easy. 3 added 267 characters in body; deleted 119 characters in body EDIT: This was an answer when the last line of the question said "Especially, I want to see an example for a non-blowup having property 2." I think you should make your question more precise. If$X$is quasi-projective, and$f:\widetilde X\to X$is a projective birational morphism, then$f$is a blow-up of an appropriate ideal sheaf. I am guessing this is not what you're after. (If you allow blow ups of fractional ideals, then you can drop even the quasi-projective assumption. Not assuming that$f$is projective would make the question less interesting). Also the way you phrased your question, I think there is a simple example: take any blow ups and then choose a subscheme of$X$with the same support, but different scheme structure than the subscheme that was blown up. Anyway, I will assume that you want an example that's not a blow up centered at a smooth subvariety. I think it is possible to give examples satisfying property #2 relatively easily. Let's look for$f$as the blow up of an ideal sheaf$\mathscr I$. So$\widetilde X=\mathrm{Proj}_X(\oplus_d \mathscr I^d)$. Then the pre-image of the subvariety$Y\subset X$defined by the ideal$\mathscr I$is given by$\widetilde Y=\mathrm{Proj}_Y(\oplus_d \mathscr{I^d/I^{d+1}})$. Now if$X$is Cohen-Macaulay and$Y$is a complete intersection in$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 Y\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$. My example is not likely to satisfy property #1, that is$\widetilde X$is probably always singular if$Y$is. Finally, here is a simple concrete example: Let$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$) with the singular line contracted to a point. I think it is relatively easy to check that this satisfies properties #2 and #3. As far as finding an example that satisfies all 3 properties, I would not hold my hopes very high. If$\dim X=2$, then these properties just mean that we have a$(-1)$-curve and we know that that is always the blow up of a smooth point. In higher dimensions I would start by taking general hyperplane cuts to reduce to the case when the image is a point. Then Property #2 says that the preimage is a$\mathbb P^n$with normal bundle$\mathscr O_{\mathbb P^n}(-1)\$ so there is a good chance that one can prove that it is indeed a blow up. Then one would get (probably) that the original map is a blow up at general points and then possibly one can prove that then it has to be a blow up everywhere, but I am not entirely sure about that.

EDIT 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. I don't think this satisfies the other conditions, but it might suggest that proving that those three conditions indeed characterize blow ups along smooth subvarieties is probably not easy.

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