Showing equivalence of two definitions of the Blow-Up of a Variety

Question

Let $X=\mathop{\mathrm{Spec}}(A)$ be an affine variety over some algebraically closed field $\Bbbk$ and $I\subseteq A$ an ideal of $A$. There are two ways to define the blow-up $\tilde X$ of $X$ along $I$, namely

1. Set $\tilde X := \mathop{\mathrm{Proj}}(A[IT])$ where $A[IT]$ is the graded ring $\bigoplus_{d\ge 0} I^dT^d \subseteq A[T]$ and $I^0:= A$.
2. Let $I=(f_0,\ldots,f_r)$ be a set of generators for $I$ and define a rational map $\varphi: X \to \mathbb{P}_\Bbbk^r$ by $\varphi(P):=[f_0(P):\ldots:f_r(P)]$. It is defined over $U:=X\setminus Z(I)$. Then, we define $\tilde X := \Gamma_\varphi$ to be the graph of $\varphi$, which is the closure of the graph of $\varphi|_U$.

I would like to show that both definitions are equivalent; let me give you my approach (which is basically just a more general variant of Example II.7.12.1 in Hartshorne):

Define a map $\pi: A[y_0,\ldots,y_r] \to A[IT]$ by $y_i\mapsto f_iT$. It induces an embedding of $\tilde X$ into $\mathbb{P}_\Bbbk^r \times X$ whose image should be $\Gamma_\varphi$. On any open subset $D(f_iT)$, we should now be able to prove that the kernel of the induced map

$A\left[\frac{y_0}{y_i},\ldots,\frac{y_r}{y_i}\right]\to \left(A[IT]_{f_iT}\right)_0$

is equal to $\left(y_kf_j-y_jf_k\,\vert\,0\le k<j\le r\right)$. However, I don't seem to be able to verify this. If anyone could show me how to proceed from here or even give a completely different approach, I would be very grateful.

Thanks in advance!

Answer 1

You try to show that the kernel of $\pi:A[y_0, \ldots, y_r] \rightarrow A[IT]$ by $y_i \mapsto f_iT$ is generated by the $y_if_j - y_jf_i$ for $1\le i < j \le r$. It is not a surprise that you don't succeed, because this is not true for general ideals $I$. Consider for example $$I = (x^2,xy,y^2)$$ in $k[x,y]$. The kernel of $\pi: k[x,y,a,b,c] \rightarrow k[x,y][IT]$ under $a \mapsto x^2T$, $b\mapsto xyT$ and $c\mapsto y^2T$ contains $ac-b^2$ which does not lie in the ideal $(axy - bx^2, ay^2-cx^2,by^2-cxy)$. The statement holds if (iff?) $f_0,\ldots, f_r$ is a regular sequence according to Fulton's "Introduction to Intersection Theory in Algebraic Geometry", if I remember correctly (I don't have a copy at hand).

However, it holds the following: $$\ker \pi = (y_1 - f_1T, \ldots, y_r-f_rT) \cap A[y_1, \ldots, y_r],$$ where $(y_1 - f_1T, \ldots, y_r-f_rT)$ is an ideal in $A[y_0,\ldots,y_r,T]$. So I think one could build a proof like this:

1) $\mathrm{Proj}(A[IT]) = \mathrm{Proj}(A[y_0, \ldots,y_r]/\ker \pi)$ should be clear.

2) Furthermore, $\mathrm{Proj}(A[y_0, \ldots,y_r]/\ker \pi)$ is the closure of the projection of $$\mathrm{Proj}(A[y_0,\ldots,y_r,T]/(y_1 - f_1T, \ldots, y_r-f_rT) )$$ to the hyperplane $\{t=0\}$. This is the geometrical interpretation of elimination according to e.g. Cox, Little, O'Shea "Ideals, Varieties and Algorithms". Remember that $\ker \pi$ is obtained by elimination of $T$ from $(y_1 - f_1T, \ldots, y_r-f_rT)$. The closure of the projection should be the closure of $(y_0 : \ldots:y_r)=(f_0:\ldots:f_r)$, hence the closure of the graph of $\varphi|_U$, i.e. $\Gamma_\varphi$.

Please ask me to clarify if something is unclear.

Answer 2

I think my comments have, by now, answered the question: Since $X$ is irreducible, so is $U$. Let $U' := \mathbb{P}_\Bbbk^r \times U$. Then, $U'\cap\Gamma_\varphi \cong U$ is irreducible. Since $\tilde X \cap U'$ is a closed subset of $U'\cap\Gamma_\varphi$ and of the same dimension, we must have $U'\cap\Gamma_\varphi = U'\cap\tilde X$. By passing to the closure, we obtain the desired result.