# Projection Morphism of Blowup

I'm currently reading the article A Short Course on Geometric Motivic Integration, by Manuel Blickle. In his proof of Theorem 3.3, the author considers the following.

Let $$X'=\operatorname{Bl}_{0}\mathbb{A}^3_\mathbb{C}$$, equipped with the usual projection $$\pi:X'\longrightarrow\mathbb{A}^3_\mathbb{C}.$$ Suppose $$\mathbb{A}_{\mathbb{C}}^3$$ has coordinates $$x_0,x_1,x_2$$. Note that $$X'=\operatorname{Proj}\left(\frac{\mathbb{C}[x_0,x_1,x_2,z_0,z_1,z_2]}{I}\right),$$ where $$I=(x_0z_1-x_1z_0,x_0z_2-x_2z_0,x_1z_2-x_2z_1)$$.

In the proof, the author considers the affine patch of $$X'$$ given by $$z_0=1$$. It is claimed that $$\pi$$, restricted to this patch, corresponds to the canonical ring homomorphism $$f:\mathbb{C}[x_0,x_0z_1,x_0z_2]\longrightarrow \mathbb{C}[x_0,z_1,x_2].$$

I'm having trouble seeing why $$\pi$$ corresponds to this form. When we consider the patch given by $$z_0=1$$, aren't we considering the open subscheme $$D_+(z_0)$$ of $$X'$$? If this is the case, shouldn't we have $$D_+(z_0)=\operatorname{Spec}\left(\frac{\mathbb{C}[x_0,x_1,x_2,z_1/z_0,z_2/z_0]}{J}\right),$$ where $$J=\left(\frac{x_0z_1}{z_0}-x_1,\frac{x_0z_2}{z_0}-x_2,\frac{x_1z_2}{z_0}-\frac{x_2z_1}{z_0}\right)?$$

I'm not able to see how this construction leads to the aforementioned homomorphism $$f$$. Could someone let me know where I'm going wrong? Thank you for any help.

• The idea is that we can use $J$ to eliminate variables. Write $Z_j = z_j/z_0$. First, note that the third generator is redundant since both monomials are equal to $x_0Z_1Z_2$ in the quotient. Taking the first two generators into account we get an isomorphism to $\mathbb{C}[x_0, x_0Z_1, x_0Z_2, Z_1, Z_2] = \mathbb{C}[x_0,Z_1, Z_2]$. Describing the morphism is then just a matter of tracking these identifications down. Commented Jun 30 at 23:52
• (If you track it down, the map should send $x_0 \mapsto x_0$, $x_1 \mapsto x_0Z_1$, and $x_2 \mapsto x_0Z_2$) Commented Jun 30 at 23:54
• @Daniel: Thank you! I just have one question. When you write $\mathbb{C}[x_0, x_0 Z_1, x_0 Z_2, Z_1, Z_2]=\mathbb{C}[x_0, Z_1, Z_2]$, how are you eliminating the terms $x_0 Z_1, x_0 Z_2$? Commented Jul 1 at 0:50

Let me elaborate on my comment, since writing polynomial rings like this can be a little sloppy. Let’s write $$R$$ for the coordinate ring of $$D_+(z_0)$$.
Since $$x_1 - x_0Z_1 \in J$$, for instance, we can replace the $$x_1$$ variable with $$x_0Z_1$$. That way you can remove that generator and describe the ring more simply.
More concretely, since $$x_1 = x_0Z_1$$ and $$x_2 = x_0Z_2$$ in $$R$$, we see it is generated as a $$\mathbb{C}$$-algebra by $$x_0, Z_1,$$ and $$Z_2$$. As such there is a surjection $$\mathbb{C}[x_1, Z_1, Z_2] \to R$$. The kernel of this map is then $$\mathbb{C}[x_1, Z_1, Z_2] \cap J = \{0\}$$, where we view these as subsets of $$\mathbb{C}[x_1, x_2, x_3, Z_1, Z_2].$$ Hence this is an isomorphism, as we needed,