Blow Up of a Variety

Question

Let $Y$ be an affine variety of $\mathbb{A}^n$ and it contains $0$. Think of $\mathbb{A}^n \times \mathbb{P}^{n-1}$ as a quasi-projective variety. Define a closed subset of $\mathbb{A}^n \times \mathbb{P}^{n-1}$, call it $X$, by the zero locus of the polynomial $f=\sum_{i,j}(x_i y_j - x_jy_i)$. We define the morphism $\varphi: X\to \mathbb{A}^n$ given by projection onto its first component. The blow-up of $Y$ (at $0$) is given by $B=\overline{\varphi^{-1}(Y\setminus \{0\})}$ (closure taken in either $X$ or $\mathbb{A}^n$, it does not mater because $X$ is closed).

We claim that $B$ is birational to $Y$. Because $K(B)\simeq K(B \setminus \varphi^{-1}(0))$ and $K(Y) \simeq K(Y\setminus \{0\})$. However, $\varphi$ induces an isomorphism between $B\setminus \varphi^{-1}(0)$ and $Y\setminus \{0\}$, in particular $B$ is birational to $Y$.

I am trying to understand why $\varphi$ induces an isomorphism between those two open sets. I forgot what the reason was, just remember it was something pretty simple.

Answer 1

The two sets are birational because away from $0$, the point in $\mathbb{A}^n$ uniquely determines the point in $\mathbb{P}^{n-1}$, and so the projection $\mathbb{A}^n\times\mathbb{P}^{n-1}\to \mathbb{A}^n$ is an isomorphism between $\mathbb{A}^n\setminus 0$ and $X\setminus \varphi^{-1}(0)$. To see that the point in $\mathbb{P}^{n-1}$ is uniquely determined, suppose $(x_1,\cdots,x_n)$ is a point on $Y$ not equal to $0$. Then there exists a coordinate $x_i$ such that $x_i\neq 0$. It is clear that the point $[\frac{x_0}{x_i}:\cdots:\frac{x_n}{x_i}]$ is the unique point on $\mathbb{P}^{n-1}$ satisfying the equations $x_jy_k=x_ky_j$, and so the preimage of $\mathbb{A}^n\setminus 0$ under the projection is exactly $X\setminus \varphi^{-1}(0)$.