Let $X$ be a smooth projective variety over an algebraically closed field $k$. For any length $n$ ideal sheaf $P$ of $X$ (e.g $N$ different points $P=(P_1)(P_2)..(P_N)$), we can consider $Bl_{P}(X)$, the blow up of $X$ at $P$. Assume $Bl_{P}(X) \cong Bl_{Q}(X)$ for a length $n$ ideal sheaf $P$ and a length $m$ ideal sheaf $Q$ , what can we say about $P$ and $Q$ ? If neccessary, one can assume $P,Q$ are radical.

Note blow up at $I$ and $I^2$ are the same. So there is a trivial case i.e when there exists an automorphism $\phi:X \rightarrow X$ s.t $\phi^*(P^k)=Q^l$.

I am interested in some examples e.g projective spaces, abelian varieties, surfaces.

  • $\begingroup$ The question has more content and a different emphasis over non-algebraically closed fields. Let $P$ and $Q$ be reduced zero-dimensional subschemes of $X$ such that the blow ups are isomorphic. Are $P$ and $Q$ isomorphic schemes? $\endgroup$ – Evgeny Shinder Jan 6 '20 at 23:15
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    $\begingroup$ Note that in spite of blow ups of $I$ and $I^2$ being the same, ideals with the same radical as $I$ do not in general lead to isomorphic blow up. Indeed, if $I = (x,y)$ in the plane, then $I^2 \subset (x,y^2) \subset I$ but blowing up the intermediate ideal leads to a singular point (a node). $\endgroup$ – Evgeny Shinder Jan 6 '20 at 23:22

The answer in general is a mess, because isomorphisms between the blow-ups do not necessarily descend to automorphisms of $X$. If you take $X = \mathbb P^2$ and fix a general configuration $P$ of $n \geq 9$ points, then the set of $Q$ for which $Bl_P(X) \cong Bl_Q(X)$ is a countable union of subvarieties of $(\mathbb P^2)^n$, which is probably Zariski dense (though I'm not sure whether anybody has actually tried to prove this).

In some sense the point is that such blow-ups have infinitely many $(-1)$-curves on them, and so you can blow down to $\mathbb P^2$ in infinitely many different ways by choosing which curves to contract. The resulting configurations in $\mathbb P^2$ that you get as the images of the contracted curves do not simply differ by elements of $PGL(3)$, as you might hope.

If $X$ is not uniruled, things are probably easier. For example, if $X$ is abelian, any isomorphism $f : Bl_P(X) \to Bl_Q(X)$ descends in the obvious way to an isomorphism $g : X \to X$, and so the answer is that the blow-ups are isomorphic if and only if $P$ and $Q$ differ by some element of $Aut(X)$.

  • $\begingroup$ Just a comment on your last paragraph for those interested: If $X$ is smooth projective and has no rational curves, then every isomorphism $Bl_P(X) \to Bl_Q(X)$ descends to an isomorphism $X\to X$. $\endgroup$ – Ariyan Javanpeykar Jan 7 '20 at 20:00

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