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maybe this is an idiot question, but I have to ask it. Usually, in the classical background, one defines the blow up $Bl_Y(X)$ at a variety $Y$ in as the closure of the graph of the function $f: X/Z(f_1, f_2, ...f_m) \longrightarrow \mathbb{P}^{m-1}$ where $f$ evaluate the polynomials. However, when one blow up at a point, the definition seens kinda different, it's the subvariety of $X \times \mathbb{P}^{n - 1}$ determined by $x_i t_j = x_j t_i$ where $t_i$ is the homogeneous coordinate. The definitions seems to not match each other.

Thanks in advance.

user40276
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1 Answers1

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The two definitions match in the following sense:

Let $X\subset \mathbb{A}^n$ be an affine variety. Let $f_0,\cdots,f_r\in k[x_1,\cdots,x_n]$ where they don't vanish identically on $X.$ Then $U=X \setminus Z(f_0,\cdots,f_r)$ is a non-empty open subset of $X.$ Define the well-defined morphism

$$f:U \to \mathbb{P}^r \; \text{given by}\; p \mapsto [f_0(p):\cdots,f_r(p)]$$

Now the closure of the graph of $f$ inside $X \times \mathbb{P}^r$ is the blow up of $X$ in $(f_0,\cdots,f_r).$

In particular, let $X=\mathbb{A}^2$ with coordinates $x,y$ and take $f_0=x,f_1=y.$ The blow up of $\mathbb{A}^2$ in $(x,y)$ or at zero is then the following. The morphism

$$f:U \to \mathbb{P}^1 \; \text{given by} \; (x,y) \mapsto [x:y]$$

is well defined, where $U=\mathbb{A}^2 \setminus \{(0,0)\}.$ The graph of $f$ on $U$ is

$$\{((x,y),[s:t])|xt=sy \} \subset U \times \mathbb{P}^1$$

and its closure is simply $\{((x,y),[s:t])|xt=sy\} \subset \mathbb{A}^2 \times \mathbb{P}^1.$ The reason for the existence of the relation $xt=sy$ is to make sure that $[s:t]$ is in the image of $f.$

Ehsan M. Kermani
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  • Oh!Thanks, now I got it. But why don't use the same definition of a a blow up at a subvariety (dealing a point as a subvariety)? – user40276 Nov 08 '13 at 22:04
  • You're welcome. What do you mean by the same notation? if you mean to write the graph as ${(p,f(p))} \subset \mathbb{A}^2 \times \mathbb{P}^1,$ note that it's the same subset, but for computational matters, it's better to write the relations as well. – Ehsan M. Kermani Nov 08 '13 at 22:11
  • Yes, you're right I didn't noticed that the blow up is at $0$. – user40276 Nov 08 '13 at 22:13
  • There is a very similar question here: http://math.stackexchange.com/questions/240301/blowing-up-at-a-subvariety/875204#875204 and the answer there may be useful for you. – Qiao Jul 22 '14 at 22:05