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Starting from an informal understanding of blowing-up as replacing a subscheme by the possible directions into it (or some more accurate formulation of this), how does one justify the definition of the blow-up of $X$ along $Y$ by the formula

$$ \mathrm{Bl}_Y(X) = \mathbf{Proj} \left( \mathrm{Rees}(\mathcal{I}) \right ) = \mathbf{Proj} \left ( \mathcal{O}_X \oplus \mathcal{I} \oplus \mathcal{I}^2 \oplus \cdots\right )?$$

Here $\mathcal{I}$ is the ideal sheaf defining $Y$ in $X$ (with appropriate conditions as needed).

I'm not used to thinking about what $\mathbf{Proj}$ looks like; although things like $ \mathbf{Proj}\left ( \mathrm{Sym} \left ( (\mathcal{O}_S^{n+1})^\vee \right )\right ) = \mathbb{P}^n_S$ are obviously quite intuitive.

What happens if I consider instead $ \mathbf{Proj}\left ( \mathrm{Sym}(\mathcal{I} ) \right )$? Is this a related object, and what's its geometric interpretation? (In some simple examples these are closely related, as in the case of the blowup of the plane at the origin where $\mathrm{Sym}(\mathcal{I}) = \mathrm{Rees}(\mathcal{I})$; see comments to rattle's answer.)

(I'm aware of the universal property of blow-ups where this description fits in rather easily, although my understanding of $\mathbf{Proj}$ is sufficiently lacking for me to understand that properly; but that's not really the question I'm asking.)

Will
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2 Answers2

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I had exactly the same question about how taking the Rees or symmetric algebra of an ideal would lead to diffrent geometric constructions, and wasn't satisfied by rattle's answer. Thought I might share my findings on this.

The essential difference between the two is the following : the Rees algebra corresponds to what is called the normal cone, ie the "set" of all normal directions through the subscheme, whereas considering the symmetric algebra leads to the normal bundle, ie the vector space generated by all normal directions. See the discussion of the tangent cone on wikipedia : http://en.wikipedia.org/wiki/Tangent_cone.

Now we always have a surjection $\mathrm{Sym}(\mathcal{I}) \to \mathrm{Rees}(\mathcal{I})$ which induce a canonical closed immersion of the cone in the bundle, which is quite intuitive. Observe that in the case of a smooth scheme, the normal directions already form a vector space, so that both algebras will coincide, as rattle mentioned in his answer.

Hence the reason why we define the blow-up by means of the Rees algebra is that we really want to replace the closed subscheme by all normal directions through it, and not by whatever vector space they would generate, as we would have done with the symmetric algebra.

Manu
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I can probably only satisfy your intuition when we are blowing up a nonsingular $k$-variety $X$ in a nonsingular subvariety $Y$. In this case, Theorem 8.24 in chapter II of Hartshorne's book shows that the exceptional divisor of the blow-up is actually $\mathbf{Proj}(\mathrm{Sym}(\mathcal{I}/\mathcal{I}^2))$, and $\mathcal{I}/\mathcal{I}^2$ is the conormal sheaf of $Y$ in $X$. Hence, we really replace $Y$ by all possible directions through it.

Note that $\mathbf{Proj}(\mathrm{Sym}(\mathcal{I}))$ does not usually make sense since $\mathcal{I}$ can only be locally free if it defines something in codimension one. In that case, it will be the same as the blow-up, namely the same as $X$ - you are changing nothing.

Alekos Robotis
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Jesko Hüttenhain
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  • OK, that's great. I just have two more questions: $$ $$
    1. Can you give a similar explanation for why the blowup is an isomorphism outside Y? $$ $$
    2. Why does $\mathbf{Proj} \left ( \mathrm{Sym}(\mathcal{I}) \right )$ not make sense if $\mathcal{I}$ doesn't define something in codimension 1? For instance $X = \mathrm{Spec}(K[X,Y])$, $I = (x,y)$, $\mathrm{Rees}(I) = \mathrm{Sym}_{K[X,Y]}(I) = K[X,Y,Z,W]/(XZ-YW)$ corresponds to the blowup of the plane at the origin (where $X$ and $Y$ have degree $0$, and $Z$ and $W$ have degree $1$).
     – Will Oct 17 '11 at 11:36
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    As far as (1) is concerned, just think of $I=(x_1,\ldots,x_r)\subseteq A$. Let $B$ be the localization of $A$ by the $x_i$, so the complement of $Y$ is the spectrum of $B$. If we localize $A[IT]$ by the $x_i$, we get to have an indeterminant $T=x_i^{-1}\cdot x_iT$, so we are looking at the proj of $B[T]$, which is exactly the same as the spec of $B$. – Jesko Hüttenhain Oct 17 '11 at 13:39
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    Regarding (2), can you explain why the symmetric algebra of $I$ is equal to $K[X,Y,Z,W]/(XZ-YW)$? My point was more that I only know the definition of $\mathbf{P} = \mathbb{Proj} \circ \mathrm{Sym}$ for locally free sheaves, and an ideal of a ring can only be a free module if it is principal. – Jesko Hüttenhain Oct 17 '11 at 13:39
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    In my comment on (1), to be formally correct, we may now localize $A[IT]$ in the $x_i^2T$, obtaining $T=(x_i^2T)^{-1}\cdot x_i^2T^2$. Localizing in the degree-zero elements $x_i$ makes no sense, but I cannot edit my comment to correct that mistake. – Jesko Hüttenhain Oct 17 '11 at 16:50
  • Sorry for taking so long to answer, I wasn't completely sure of my explanation. For the symmetric algebra of $I$, you use the fact that $(X,Y)$ forms a regular sequence, and so I think if you write down the Koszul complex you find that instead of just K[X,Y][Z,W], you have to quotient out by (XZ-YW). $$ $$

    Your explanation to (1) is great, thanks.

     – Will Oct 18 '11 at 10:26
  • In more details: from the exact sequence $$ 0 \to A \overset{\begin{pmatrix} -Y \ X \end{pmatrix}}{\longrightarrow} A^2 \overset{\begin{pmatrix} X & Y \end{pmatrix}}{\longrightarrow} I \to 0, $$ you get the exact sequence $$ A \otimes_A \mathrm{Sym}_A(A^2) \overset{\begin{pmatrix} -Y \ X \end{pmatrix}}{\longrightarrow} \mathrm{Sym}_A(A^2) \to \mathrm{Sym}_A(I) \to 0, $$ and that's what you want. – Will Oct 18 '11 at 10:59
  • That argument seems very solid, and thinking about it, I figured that the morphism of graded $A$-algeras $\mathrm{Sym}(I)\to\mathrm{Rees}(I)$ "defined" by $x\otimes y\mapsto xyT^2$ should always be well-defined and surjective. Howver, I have to be frank and admit that right now, I don't really know much more and don't want to speculate. I will look into it, though. – Jesko Hüttenhain Oct 18 '11 at 21:24