Which functor does the blowing up represent? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T23:22:15Z http://mathoverflow.net/feeds/question/91357 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/91357/which-functor-does-the-blowing-up-represent Which functor does the blowing up represent? Martin Brandenburg 2012-03-16T08:51:23Z 2012-03-30T14:55:33Z <p>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 <code>$\langle p^*(I) \rangle \subseteq \mathcal{O}_{\mathrm{Bl}_I(X)}$</code> 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.</p> <p>First I thought that this implies that $\mathrm{Bl}_I(X)$ is a representing object of the functor</p> <p>$\mathrm{Sch}^{\mathrm{op}} \to \mathrm{Set},~ Y \mapsto \{f \in \hom(Y,X) : \langle f^*(I) \rangle \subseteq \mathcal{O}_Y \text{ invertible}\},$</p> <p>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.</p> <p><strong>Question.</strong> Which functor $\mathrm{Sch}^{\mathrm{op}} \to \mathrm{Set}$ does the blowing up represent? In other words, how can we simplify $\hom(Y,\mathrm{Bl}_I(X))$?</p> <p>Since we have <code>$\mathrm{Bl}_I(X) = \mathrm{Proj}_X \oplus_{n \geq 0} I^n$</code> , one can ask more generally what functor <code>$\mathrm{Proj}_X \mathcal{A}$</code> represents when $\mathcal{A}$ is a sufficiently nice graded <code>$\mathcal{O}_X$</code>-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.</p> <p>In any case, there seems to be an injective map from</p> <p>$$\{(f,\mathcal{L},\psi) : f \in \hom(Y,X) , \mathcal{L} \text{ line bundle on } Y, \psi : f^*(\mathcal{A}) \twoheadrightarrow \oplus_n \mathcal{L}^{\otimes n}\}$$</p> <p>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?</p> <p>In the note "Elementary Introduction To Representable Functors and Hilbert Schemes" by S. A. Stromme I have found the following amusing exercise:</p> <p>"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."</p> <p>I hope that this does not mean that we cannot understand the functor at all ...</p> http://mathoverflow.net/questions/91357/which-functor-does-the-blowing-up-represent/91388#91388 Answer by Jason Starr for Which functor does the blowing up represent? Jason Starr 2012-03-16T16:57:42Z 2012-03-16T16:57:42Z <p>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 $r$.</p> <p>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 $f^*(I_d)$.</p>