# Projection from Blowup is Isomorphism Away from Exceptional Set

I'm following Eisenbud's description of blowing up: let $$X$$ be an affine algebraic variety, $$R$$ the coordinate ring of $$X$$, and let $$a_1,\ldots,a_r$$ generate $$R$$ as a $$k$$-algebra. Let $$Y\subseteq X$$ be an affine subvariety corresponding to an ideal $$I\subseteq R$$ generated by $$g_0,\ldots,g_s$$.

Let $$S=k[x_1,\ldots,x_r,y_0,\ldots,y_s]$$ and define $$\varphi:S\to R[t]$$ by sending $$x_i\mapsto a_i$$ and $$y_i\mapsto g_it$$. Then $$\ker(\varphi)$$ is an ideal homogeneous in the $$y_i$$, and so corresponds to an algebraic subset $$Z\subseteq\mathbb{A}^r\times\mathbb{P}^s$$ (the blowup of $$Y$$ in $$X$$).

Let $$\pi: \mathbb{A}^r\times\mathbb{P}^s\to\mathbb{A}^r$$ be projection, and let $$E=Z\cap\pi^{-1}(Y)$$ (the exceptional set of the blowup).

I am trying to prove the following:

The restriction $$\pi|_{Z\setminus E}:Z\setminus E\to X\setminus Y$$ is an isomorphism.

I can show that $$\pi(Z)=X$$, and so the map is surjective, but I am struggling to show injectivity.

Eisenbud also states:

$$E$$ corresponds to $$R[It]/IR[It]$$ as a subvariety of $$Z$$.

However I can't seem to show this either, or use this to help me prove the isomorphism. Since $$S/\ker(\varphi)\cong\text{im}(\varphi)=R[It]$$, would we have that $$E$$ corresponds to the ideal $$(y_0,\ldots,y_s)\cap\ker(\varphi)$$ in $$S$$?

Any help would be much appreciated.

Claim: The projection $$\pi \from Z \subseteq X \times \IP^s \to X$$ induces an isomorphism $$Z\setminus \pi^{-1}(Y) \iso X\setminus Y$$.
Proof: We construct the inverse map by sending $$x \in X \setminus Y$$ to $$(x,(g_0(x):\ldots:g_s(x)) \in Z.$$ To see that this is well-defined, note first that because $$x \notin Y$$, at least one of the $$g_j(x)$$ does not vanish, so we have an element of $$X \times \IP^s$$. To see that this is indeed an element of $$Z$$, let $$F \in J$$ be of degree $$d$$ as above. But then $$F(x,g_0(x), \dots, g_s(x)) = \bigl(\sum f_{i_0,\dots, i_s} g_0^{i_0} \dots g_s^{i_s}\bigr)(x) = 0.$$ It is clear that $$\pi(x,(g_0(x) : \ldots : g_s(x)) = x$$. Conversely, suppose that $$(x, (b_0 : \ldots : b_s)) \in Z \setminus \pi^{-1}(Y)$$. Note that we have $$g_j Y_{j'} - g_{j'} Y_j \in J\qquad \text{for all }j, j'.$$ Hence, these functions vanish identically on $$Z$$. Plugging in $$(x,(b_0:\ldots:b_s))$$ we find that $$g_j(x)b_{j'} = g_{j'}(x) b_j \qquad \text{ for all } j, j'.$$ This shows that $$(b_0 : \ldots : b_s) = (g_0(x) : \ldots : g_s(x)) \in \IP^s$$ and we see that the function is indeed inverse to $$\pi$$.
• Many thanks for such a clear explanation, it has been very helpful for me. I was wondering if you would mind explaining Eisenbud's claim I also mentioned that $\pi^{-1}(Y)$ corresponds to $B_IR/IB_IR$? No problem if not or if you would rather not answer a comment, I'm happy to post another question. Thanks again – Dave May 13 at 21:47
• @Dave I'm glad my answer was helpful to you. Regarding you follow-up question: Let us write $\pi^\ast \colon R \to B_IR$ for the natural map into degree $0$. The ideal $IB_IR$ is generated by $\pi^\ast(g_0), \dots, \pi^\ast(g_s)$. Given $(x,b) \in Z$ you have $(x,b) \in \pi^{-1}(Y) \iff x \in Y \iff g_0(x) = \dots = g_s(x) = 0 \iff \pi^\ast(g_0)(x,b) = \dots = \pi^\ast(g_s)(x,b) = 0$, so $\pi^{-1}(Y)$ is the zero set of functions generating $IB_IR$. In much generality, pulling back closed subsets amounts to pushing forward their ideals. – Trauerschwein May 14 at 10:04