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Let $X$ be a scheme and $I \subseteq \mathcal{O}_X$ be a quasi-coherent ideal of finite type. The blowing up $\mathrm{Bl}_I(X)$ has the following universal property: It comes with a morphism $p : \mathrm{Bl}_I(X) \to X$ such that $\langle p^*(I) \rangle \subseteq \mathcal{O}_{\mathrm{Bl}_I(X)}$ is invertible, and for every morphism $f : Y \to X$ such that $\langle f^*(I) \rangle \subseteq \mathcal{O}_Y$ is invertible, there is a unique morphism $\tilde{f} : Y \to \mathrm{Bl}_I(X)$ such that $f = p \tilde{f}$. In other words, $\mathrm{Bl}_I(X) \to X$ is a final object in the category of $X$-schemes which pull back $I$ to an invertible ideal.

First I thought that this implies that $\mathrm{Bl}_I(X)$ is a representing object of the functor

$\mathrm{Sch}^{\mathrm{op}} \to \mathrm{Set},~ Y \mapsto \{f \in \hom(Y,X) : \langle f^*(I) \rangle \subseteq \mathcal{O}_Y \text{ invertible}\},$

but this is wrong: This is not even a functor! Invertible ideals don't pull back to invertible ideals. If $H(Y)$ denotes the set above, then there is a map $\hom(Y,\tilde{X}) \to H(Y)$, which is injective, but far from being surjective.

Question. Which functor $\mathrm{Sch}^{\mathrm{op}} \to \mathrm{Set}$ does the blowing up represent? In other words, how can we simplify $\hom(Y,\mathrm{Bl}_I(X))$?

Since we have $\mathrm{Bl}_I(X) = \mathrm{Proj}_X \oplus_{n \geq 0} I^n$ , one can ask more generally what functor $\mathrm{Proj}_X \mathcal{A}$ represents when $\mathcal{A}$ is a sufficiently nice graded $\mathcal{O}_X$-algebra. This is indicated in EGA II, 3.7, but the condition that the partial morphism $r_{\mathcal{L},\psi}$ is defined everywhere is only made explicit for affine $X$ in loc. cit. Cor. 3.7.4. If $\mathcal{A}=\mathrm{Sym}(\mathcal{E})$ for some vector bundle $\mathcal{E}$, then the condition is just that $\psi$ (defined in loc. cit. 3.7.1) is surjective, so that the universal property of $\mathbb{P}(\mathcal{E})$ is very similar to the one of the usual projective space. But $\oplus_{n \geq 0} I^n$ obviously does not have this form, unless $I$ is flat or something like that.

In any case, there seems to be an injective map from

$$\{(f,\mathcal{L},\psi) : f \in \hom(Y,X) , \mathcal{L} \text{ line bundle on } Y, \psi : f^*(\mathcal{A}) \twoheadrightarrow \oplus_n \mathcal{L}^{\otimes n}\}$$

to $\hom(Y,\mathrm{Proj} \mathcal{A})$. EDIT: In Jason Starr's answer it is claimed that this is a bijection. Can someone give a reference in the literature for this observation?

In the note "Elementary Introduction To Representable Functors and Hilbert Schemes" by S. A. Stromme I have found the following amusing exercise:

"Let $X \to S$ be the blowing up of $S$ in some center $Y \subseteq S$. Try to the understand the point functor $h_{X/S}$. Then explain why it is hard to understand blow-ups."

I hope that this does not mean that we cannot understand the functor at all ...

EDIT. The universal property of Proj of a graded quasi-coherent algebra appears in section 16 of "Constructions of schemes" in the Stacks Project (here). It is the usual one when the graded algebra is generated in degree 1. I still wonder if there is any reference in the literature because I want to cite this.

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Since the blowing up is Proj of the blowup algebra $A$, which is generated in degree 1, the functor represented is the functor of invertible quotients $L$ of the pullback $f^*I$ of $I$ such that for every integer $n > 1$, the induced surjection $f^*\text{Sym}^n(I) \to L$ factors through $f^*(I^n)$. –  Jason Starr Mar 16 '12 at 13:04
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Thanks. Can you add this as an answer (it is not just a comment) and provide some details? –  Martin Brandenburg Mar 16 '12 at 13:12
    
The map arrow in row 9, isnt reverse? –  Buschi Sergio Mar 30 '12 at 16:42
    
The universal property of Proj appears in section 16 of "Constructions of schemes" in the Stacks Project. stacks.math.columbia.edu/tag/01NS But somehow it's more complicated. The represented functor $F$ is the union of some $F_d$, and $F_d$ is concerned with the graded algebra $\oplus_{n} A_{nd}$ ... –  Martin Brandenburg Sep 10 '13 at 8:31

1 Answer 1

For a graded algebra $\mathcal{A} = \oplus_d \mathcal{A}_d$ of $\mathcal{O}_X$-algebras whose associated graded pieces $\mathcal{A}_d$ are coherent and which is generated in degree $1$, the relative Proj, $P=\text{Proj}_X \mathcal{A}$ with its projection $p:P\to X$ comes with a natural invertible quotient $q:p^*\mathcal{A}_1 \to \mathcal{O}(1)$. Moreover, for every integer $d>0$, the induced surjection $\text{Sym}^d(q):p^*\text{Sym}^d(\mathcal{A}_1) \to \mathcal{O}(d)$ factors through the pullback of the natural surjection $\text{Sym}^d(\mathcal{A}_1) \to \mathcal{A}_d$. In fact this is the universal property of the pair $(p:P\to X,q:p^*\mathcal{A}_1 \to \mathcal{O}(1))$, i.e., for every $X$-scheme $f:T\to X$ and for every invertible quotient $r:f^*\mathcal{A}_1 \to \mathcal{L}$ such that every induced map $\text{Sym}^d(r):f^*\text{Sym}^d(\mathcal{A}_1) \to \mathcal{L}^{\otimes d}$ factors through $f^*\mathcal{A}_d$, there exists a unique $X$-morphism $h:T\to P$ such that $p\circ h$ equals $f$ and $h^*q$ equals $r$.

Since the blowing up is Proj of the blowup algebra $\mathcal{A}$, which is generated in degree 1, the functor represented is the functor of invertible quotients $\mathcal{L}$ of the pullback $f∗I$ of $I$ such that for every integer $d>0$, the induced surjection $f^∗\text{Sym}^d(I)\to \mathcal{L}$ factors through $f^*(I_d)$.

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Thanks! One can prove this as follows: 1) The case $X$ affine, $\mathcal{A}$ polynomial ring, is well-known (projective space). 2) Then deduce it for affine $X$, but $\mathcal{A}$ arbitrary; using some closed immersion. 3) By gluing, conclude the general case. # Is there a shorter proof? # Can you give a reference to this universal property in the literature? –  Martin Brandenburg Mar 17 '12 at 12:13
    
I also wanted to mention that this universal property implies the usual universal property of the blowing up: Namely, if $\langle f^* I \rangle \subseteq \mathcal{O}_Y$ is invertible, then one can show that there is an epimorphism $\mathcal{L} \to \langle f^* I \rangle$, which must be an isomorphism since both are invertible. This is compatible with $f^* I \to \mathcal{L}$. Thus, we end up with a unique $Y$-valued point of the blowing up. –  Martin Brandenburg Mar 17 '12 at 12:18
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The arrow map in the row nine, isnt reverse? –  Buschi Sergio Mar 30 '12 at 16:40
    
@Buschi: ? There is no arrow map in row 9. –  Martin Brandenburg Mar 30 '12 at 21:30

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