Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
By the way, it is obvious that $f_ky_j-f_jy_k\in\ker(\pi)$ for all $j$ and $k$. Hence, the inclusion $\tilde X \subseteq \Gamma_\varphi$ is obvious. – Jesko Hüttenhain Jun 2 '11 at 12:44
Also, since $\dim(A[IT])>\dim(A)$, we can immediately see that $\dim(\tilde X)\ge\dim(\Gamma_\varphi)$. Hence by my previous comment, $\dim(\tilde X)=\dim(\Gamma_\varphi)$. This is me trying to upgrade the inclusion to an equality by some topological argument, but I am not really there yet. – Jesko Hüttenhain Jun 2 '11 at 21:04

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.

share|cite|improve this answer
up vote 0 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.