Blowing up an affine variety along an ideal

After having seen this example, we now describe a general blow-up of an affine variety along an ideal.

We start by considering an affine variety $X\subseteq \mathbb {A} ^{n}$$X\subseteq \mathbb {A} ^{n}$ and an ideal $I$$I$ of the coordinate ring ${\frac {\mathbb {C} }{\mathbb {I} (X)}}$${\frac {\mathbb {C} }{\mathbb {I} (X)}}$ of $X.$$X.$ The zero locus

$$Z(I)=\{\mathbf {x} \in X\colon f(\mathbf {x} )=0\;\forall f\in I\}$$

$$Z(I)=\{\mathbf {x} \in X\colon f(\mathbf {x} )=0\;\forall f\in I\}$$

defines a closed subvariety of $X$$X$ and the blow up of $X$$X$ along $Z$$Z$ will be a variety isomorphic to $X$$X$ everywhere away from $Z(I)$$Z(I)$. It is defined as follows:

Definition 2.1

Let $\{f_{0},f_{1},\ldots ,f_{r}\}$$\{f_{0},f_{1},\ldots ,f_{r}\}$ be a set of generators for $I$$I$. We define the blow-up of $X$$X$ at $I$$I$ to be

$$\operatorname {Bl} _{I}X={}{\overline {\{(\mathbf {x} ,[f_{0}(\mathbf {x} ):f_{1}(\mathbf {x} ):\ldots :f_{r}(\mathbf {x} )])\in \mathbb {A} ^{n}\times \mathbb {P} ^{r}\colon \mathbf {x} \in X\setminus Z(I)\}}},$$

$$\operatorname {Bl} _{I}X={}{\overline {\{(\mathbf {x} ,[f_{0}(\mathbf {x} ):f_{1}(\mathbf {x} ):\ldots :f_{r}(\mathbf {x} )])\in \mathbb {A} ^{n}\times \mathbb {P} ^{r}\colon \mathbf {x} \in X\setminus Z(I)\}}},$$

that is, it is the closure inside $\mathbb {A} ^{n}\times \mathbb {P} ^{r}$$\mathbb {A} ^{n}\times \mathbb {P} ^{r}$ of the graph of the morphism

$$F\colon X\setminus Z(I)\to \mathbb {P} ^{r},\quad \mathbf {x} \mapsto [f_{0}(\mathbf {x} ):f_{1}(\mathbf {x} ):\ldots :f_{r}(\mathbf {x} )].$$

$$F\colon X\setminus Z(I)\to \mathbb {P} ^{r},\quad \mathbf {x} \mapsto [f_{0}(\mathbf {x} ):f_{1}(\mathbf {x} ):\ldots :f_{r}(\mathbf {x} )].$$

 

We see that this is well-defined, since at any point $\mathbf {x}$$\mathbf {x}$ in $X\setminus Z(I)$$X\setminus Z(I)$ at least one of the $f_{i}$$f_{i}$ does not vanish. Note that $\operatorname {Bl} _{I}X$$\operatorname {Bl} _{I}X$ comes with a morphism $\pi \colon \operatorname {Bl} _{I}X\to X$$\pi \colon \operatorname {Bl} _{I}X\to X$ given by projection onto the first coordinate. We define

Definition 2.2

The exceptional locus of the blow up of $X$$X$ at $I$$I$ is defined to be $E=\pi ^{-1}(Z(I))$$E=\pi ^{-1}(Z(I))$.

 

Remark 2.1

You might be worried that the above notation doesn't make sense, at least not until we show that $\operatorname {Bl} _{I}X$$\operatorname {Bl} _{I}X$ does not depend on the choice of generators for $I$$I$. We do this in Proposition 2.1 below. Before that, we make the following observations:

1. Note that $\operatorname {Bl} _{I}X\setminus E$$\operatorname {Bl} _{I}X\setminus E$ is isomorphic to $X\setminus Z(I)$$X\setminus Z(I)$ since there the morphism $\pi$$\pi$ is invertible with inverse

$${\begin{aligned}\sigma \colon X\setminus Z(I)&\longrightarrow \operatorname {Bl} _{I}X\setminus E\\\sigma \colon \mathbf {x} &\longmapsto (\mathbf {x} ,[f_{0}(\mathbf {x} ):f_{1}(\mathbf {x} ):\ldots :f_{r}(\mathbf {x} )]).\end{aligned}}$$

$${\begin{aligned}\sigma \colon X\setminus Z(I)&\longrightarrow \operatorname {Bl} _{I}X\setminus E\\\sigma \colon \mathbf {x} &\longmapsto (\mathbf {x} ,[f_{0}(\mathbf {x} ):f_{1}(\mathbf {x} ):\ldots :f_{r}(\mathbf {x} )]).\end{aligned}}$$

1. Let us investigate more closely how the closure in Definition 2.1 is formed. For this purpose, we need to know how to express the Zariski

closed subsets of $\mathbb {A} ^{n}\times \mathbb {P} ^{r}$$\mathbb {A} ^{n}\times \mathbb {P} ^{r}$. We can embed $\mathbb {A} ^{n}\times \mathbb {P} ^{r}$$\mathbb {A} ^{n}\times \mathbb {P} ^{r}$ into $\mathbb {P} ^{n}\times \mathbb {P} ^{r}$$\mathbb {P} ^{n}\times \mathbb {P} ^{r}$ and there we know from the Segre embedding that the closed subsets are intersections of vanishing sets of polynomials of the form $f(x_{0},x_{1},\ldots ,x_{n},X_{0},X_{1},\ldots ,X_{r})$$f(x_{0},x_{1},\ldots ,x_{n},X_{0},X_{1},\ldots ,X_{r})$ which are bihomogeneous in the $x$$x$ and $X$$X$ variables, that is, they satisfy

$${\begin{aligned}f(\lambda x_{0},\lambda x_{1},\ldots ,\lambda x_{n},X_{0},X_{1},\ldots ,X_{r})&=\lambda ^{p}f(x_{0},x_{1},\ldots ,x_{n},X_{0},X_{1},\ldots ,X_{r})\\f(x_{0},x_{1},\ldots ,x_{n},\lambda X_{0},\lambda X_{1},\ldots ,\lambda X_{r})&=\lambda ^{q}f(x_{0},x_{1},\ldots ,x_{n},X_{0},X_{1},\ldots ,X_{r})\end{aligned}}$$

$${\begin{aligned}f(\lambda x_{0},\lambda x_{1},\ldots ,\lambda x_{n},X_{0},X_{1},\ldots ,X_{r})&=\lambda ^{p}f(x_{0},x_{1},\ldots ,x_{n},X_{0},X_{1},\ldots ,X_{r})\\f(x_{0},x_{1},\ldots ,x_{n},\lambda X_{0},\lambda X_{1},\ldots ,\lambda X_{r})&=\lambda ^{q}f(x_{0},x_{1},\ldots ,x_{n},X_{0},X_{1},\ldots ,X_{r})\end{aligned}}$$

for some $p,\,q\in \mathbb {N}$$p,\,q\in \mathbb {N}$ and all $\lambda \in \mathbb {C}$$\lambda \in \mathbb {C}$ (the pair $(p,q)$$(p,q)$ is then called the bidegree of $f$$f$). Thus a basis for the closed subsets of $\mathbb {A} ^{n}\times \mathbb {P} ^{r}$$\mathbb {A} ^{n}\times \mathbb {P} ^{r}$ are precisely vanishing sets of polynomials $f(x_{1},\ldots ,x_{n},X_{0},X_{1},\ldots ,X_{r})$$f(x_{1},\ldots ,x_{n},X_{0},X_{1},\ldots ,X_{r})$, which are homogeneous only in the $X$$X$-variables.

 

We now prove the independence of generators. This proof will become obsolete when we give a much more general definition of blow up, but it is useful for our current hands-on approach.

Proposition 2.1

Let $\{g_{0},g_{1},\ldots ,g_{s}\}$$\{g_{0},g_{1},\ldots ,g_{s}\}$ be another set of generators for $I$$I$ and let us temporarily write

$$\operatorname {Bl} _{(f_{0},\ldots ,f_{r})}X={}{\overline {\{(\mathbf {x} ,[f_{0}(\mathbf {x} ):f_{1}(\mathbf {x} ):\ldots :f_{r}(\mathbf {x} )])\in \mathbb {A} ^{n}\times \mathbb {P} ^{r}\colon \mathbf {x} \in X\setminus Z(I)\}}}$$

$$\operatorname {Bl} _{(f_{0},\ldots ,f_{r})}X={}{\overline {\{(\mathbf {x} ,[f_{0}(\mathbf {x} ):f_{1}(\mathbf {x} ):\ldots :f_{r}(\mathbf {x} )])\in \mathbb {A} ^{n}\times \mathbb {P} ^{r}\colon \mathbf {x} \in X\setminus Z(I)\}}}$$

and

$$\operatorname {Bl} _{(g_{0},\ldots ,g_{s})}X={}{\overline {\{(\mathbf {x} ,[g_{0}(\mathbf {x} ):g_{1}(\mathbf {x} ):\ldots :g_{s}(\mathbf {x} )])\in \mathbb {A} ^{n}\times \mathbb {P} ^{s}\colon \mathbf {x} \in X\setminus Z(I)\}}}.$$

$$\operatorname {Bl} _{(g_{0},\ldots ,g_{s})}X={}{\overline {\{(\mathbf {x} ,[g_{0}(\mathbf {x} ):g_{1}(\mathbf {x} ):\ldots :g_{s}(\mathbf {x} )])\in \mathbb {A} ^{n}\times \mathbb {P} ^{s}\colon \mathbf {x} \in X\setminus Z(I)\}}}.$$

Then we have that

$$\operatorname {Bl} _{(f_{0},\ldots ,f_{r})}X\cong \operatorname {Bl} _{(g_{0},\ldots ,g_{s})}X$$

$$\operatorname {Bl} _{(f_{0},\ldots ,f_{r})}X\cong \operatorname {Bl} _{(g_{0},\ldots ,g_{s})}X$$

 

Proof

For this proof we may think of the ideal $I$$I$ as lying inside $\mathbb {C}$$\mathbb {C}$ and containing the vanishing ideal of $X$$X$, so that $\{f_{0},\ldots ,f_{r}\}\subseteq \mathbb {C}$$\{f_{0},\ldots ,f_{r}\}\subseteq \mathbb {C}$ and $\{g_{0},g_{1},\ldots ,g_{s}\}\subseteq \mathbb {C}$$\{g_{0},g_{1},\ldots ,g_{s}\}\subseteq \mathbb {C}$. Thus we have polynomial relations $f_{i}~=~\sum _{j}h_{ij}g_{j}$$f_{i}~=~\sum _{j}h_{ij}g_{j}$ and $g_{l}=\sum _{m}k_{lm}f_{m}$$g_{l}=\sum _{m}k_{lm}f_{m}$ for some $\{h_{ij},k_{lm}\colon 0\leq i,m\leq r,\;0\leq j,l\leq s\}\subseteq \mathbb {C}$$\{h_{ij},k_{lm}\colon 0\leq i,m\leq r,\;0\leq j,l\leq s\}\subseteq \mathbb {C}$. We define the morphism

$${\begin{aligned}\phi \colon \operatorname {Bl} _{(f_{0},\ldots ,f_{r})}X&\longrightarrow \operatorname {Bl} _{(g_{0},\ldots ,g_{s})}X\\\phi \colon \left(\mathbf {x} ,[X_{0}:X_{1}:\ldots :X_{r}]\right)&\longmapsto \left(\mathbf {x} ,\left[\sum _{m=0}^{r}k_{0m}(\mathbf {x} )X_{m}:\ldots :\sum _{m=0}^{r}k_{sm}(\mathbf {x} )X_{m}\right]\right).\end{aligned}}$$

$${\begin{aligned}\phi \colon \operatorname {Bl} _{(f_{0},\ldots ,f_{r})}X&\longrightarrow \operatorname {Bl} _{(g_{0},\ldots ,g_{s})}X\\\phi \colon \left(\mathbf {x} ,[X_{0}:X_{1}:\ldots :X_{r}]\right)&\longmapsto \left(\mathbf {x} ,\left[\sum _{m=0}^{r}k_{0m}(\mathbf {x} )X_{m}:\ldots :\sum _{m=0}^{r}k_{sm}(\mathbf {x} )X_{m}\right]\right).\end{aligned}}$$

We need to check that this is well-defined, i.e. that $\left\{\sum k_{0m}(\mathbf {x} )X_{m},\ldots ,\sum k_{sm}(\mathbf {x} )X_{m}\right\}$$\left\{\sum k_{0m}(\mathbf {x} )X_{m},\ldots ,\sum k_{sm}(\mathbf {x} )X_{m}\right\}$ cannot vanish simultaneously on $\operatorname {Bl} _{(f_{0},\ldots ,f_{r})}X$$\operatorname {Bl} _{(f_{0},\ldots ,f_{r})}X$. Consider the set of polynomials $\left\{\psi _{i}=X_{i}-\sum _{j=o}^{s}h_{ij}\left(\sum _{m=0}^{r}k_{jm}X_{m}\right)\colon 0\leq i\leq r\right\}$$\left\{\psi _{i}=X_{i}-\sum _{j=o}^{s}h_{ij}\left(\sum _{m=0}^{r}k_{jm}X_{m}\right)\colon 0\leq i\leq r\right\}$. These are homogeneous in the $X$$X$-variables and vanish on

$$\{(\mathbf {x} ,[f_{0}(\mathbf {x} ):f_{1}(\mathbf {x} ):\ldots :f_{r}(\mathbf {x} )])\in \mathbb {A} ^{n}\times \mathbb {P} ^{r}\colon \mathbf {x} \in X\setminus Z(I)\}.$$

$$\{(\mathbf {x} ,[f_{0}(\mathbf {x} ):f_{1}(\mathbf {x} ):\ldots :f_{r}(\mathbf {x} )])\in \mathbb {A} ^{n}\times \mathbb {P} ^{r}\colon \mathbf {x} \in X\setminus Z(I)\}.$$

It follows from our discussion in Remark 2.1 that they must vanish on the whole of $\operatorname {Bl} _{(f_{0},\ldots ,f_{r})}X$$\operatorname {Bl} _{(f_{0},\ldots ,f_{r})}X$. Thus we can't have $\left\{\sum k_{0m}(\mathbf {x} )X_{m},\ldots ,\sum k_{sm}(\mathbf {x} )X_{m}\right\}$$\left\{\sum k_{0m}(\mathbf {x} )X_{m},\ldots ,\sum k_{sm}(\mathbf {x} )X_{m}\right\}$ all vanishing at a point in $\operatorname {Bl} _{(f_{0},\ldots ,f_{r})}X$$\operatorname {Bl} _{(f_{0},\ldots ,f_{r})}X$, since then vanishing of the $\psi _{i}$$\psi_i$ will imply $X_{i}=0\;\forall i$$X_{i}=0\;\forall i$, a contradiction.

So $F$$F$ is well-defined and it is clearly invertible with an analogously constructed inverse morphism. Hence $\operatorname {Bl} _{(f_{0},\ldots ,f_{r})}X\cong \operatorname {Bl} _{(g_{0},\ldots ,g_{s})}X$$\operatorname {Bl} _{(f_{0},\ldots ,f_{r})}X\cong \operatorname {Bl} _{(g_{0},\ldots ,g_{s})}X$.

 

We now define one more notion, regarding subvarieties of the one we wish to blow up.

Definition 2.3

Let $Y\subseteq X$$Y\subseteq X$ be a closed subvariety different from $Z$$Z$. We define the strict transform of $Y$$Y$ in $\pi \colon \operatorname {Bl} _{I}X\to X$$\pi \colon \operatorname {Bl} _{I}X\to X$ to be ${}{\overline {\pi ^{-1}(Y\setminus Z(I))}}$${}{\overline {\pi ^{-1}(Y\setminus Z(I))}}$.

 

Remark 2.2

Note that with this definition it is clear that the strict transform of $Y$$Y$ is in fact $\operatorname {Bl} _{I\cap \mathbb {I} (Y)}Y$$\operatorname {Bl} _{I\cap \mathbb {I} (Y)}Y$. This is the crucial functoriality property of blow ups -- we can compute the blow up of a variety by first embedding it into a larger one, blowing that up and taking strict transform. The price we'll pay for writing down a general, coordinate free definition of blowing up is that this functoriality property will not be obvious (though, of course, still true).

 

Using the theory we've just developed, we will now revisit the example from the beginning.

Example 2.1

The blow up of $\mathbb {C} ^{2}=Spec\,$$\mathbb {C} ^{2}=Spec\,$ at the (reduced) origin. The ideal we wish to blow up in is then $(x_{1},x_{2})\triangleleft \mathbb {C}$$(x_{1},x_{2})\triangleleft \mathbb {C}$ and we will write $\operatorname {Bl} _{\mathbf {0} }\mathbb {C} ^{2}$$\operatorname {Bl} _{\mathbf {0} }\mathbb {C} ^{2}$ as a shorthand for $\operatorname {Bl} _{(x_{1},x_{2})}\mathbb {C} ^{2}$$\operatorname {Bl} _{(x_{1},x_{2})}\mathbb {C} ^{2}$ . So we have

$${\begin{aligned}\operatorname {Bl} _{\textbf {0}}\mathbb {C} ^{2}&={}{\overline {\{((x_{1},x_{2}),[x_{1}:x_{2}])\in \mathbb {C} ^{2}\times \mathbb {P} ^{1}\colon (x_{1},x_{2})\neq (0,0)\}}}\\&=\{((x_{1},x_{2}),[X_{1}:X_{2}])\in \mathbb {C} ^{2}\times \mathbb {P} ^{1}\colon x_{1}X_{2}-X_{1}x_{2}=0\}\\&={\mathcal {O}}_{\mathbb {P} ^{1}}(-1)\end{aligned}}$$

$${\begin{aligned}\operatorname {Bl} _{\textbf {0}}\mathbb {C} ^{2}&={}{\overline {\{((x_{1},x_{2}),[x_{1}:x_{2}])\in \mathbb {C} ^{2}\times \mathbb {P} ^{1}\colon (x_{1},x_{2})\neq (0,0)\}}}\\&=\{((x_{1},x_{2}),[X_{1}:X_{2}])\in \mathbb {C} ^{2}\times \mathbb {P} ^{1}\colon x_{1}X_{2}-X_{1}x_{2}=0\}\\&={\mathcal {O}}_{\mathbb {P} ^{1}}(-1)\end{aligned}}$$

 

Exercise 2.1

If we are working over $\mathbb {R}$$\mathbb{R}$, show that $\operatorname {Bl} _{\mathbf {0} }\mathbb {R} ^{2}$$\operatorname {Bl} _{\mathbf {0} }\mathbb {R} ^{2}$ is topologically $\mathbb {R} ^{2}\,\#\,\mathbb {R} P^{2}$$\mathbb {R} ^{2}\,\#\,\mathbb {R} P^{2}$ (remove a small disc around the origin and identify antipodal points on the resulting boundary). Then we see that in fact ${\mathcal {O}}_{\mathbb {R} P^{1}}(-1)$${\mathcal {O}}_{\mathbb {R} P^{1}}(-1)$ is the Mbius bundle and this justifies the common depiction of the blow up we give in Figure (MISSING)

 

Exercise 2.2

Similarly, show that $\operatorname {Bl} _{\mathbf {0} }\mathbb {C} ^{2}$$\operatorname {Bl} _{\mathbf {0} }\mathbb {C} ^{2}$ is topologically $\mathbb {C} ^{2}\,\#\,{\overline {\mathbb {P} ^{2}}}$$\mathbb {C} ^{2}\,\#\,{\overline {\mathbb {P} ^{2}}}$, where the bar denotes opposite orientation. To do this, show that the blow up removes a small $D^{4}$$D^{4}$ centered at the origin and then identifies the Hopf fibres on the introduced boundary $S^{3}$$S^{3}$.

 

Exercise 2.3

Let $E$$E$ denote the exceptional divisor in $\operatorname {Bl} _{\mathbf {0} }\mathbb {C} ^{n}$$\operatorname {Bl} _{\mathbf {0} }\mathbb {C} ^{n}$. Show that the intersection product of $E$$E$ with itself equals $-1$$-1$.