MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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. 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

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 $f^*(I_d)$.

share|cite|improve this answer
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
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.