# Blow-up, strict transform and tangent cone (Gathmann Notes, Exercise 9.22)

I'm studying Gatmann's Notes (version of 2014) https://www.mathematik.uni-kl.de/~gathmann/de/alggeom.php

I'm currently reading the Chapter 9. Birational Maps and Blowing Up. I'm trying to do exercise 9.22 which appears to be important.

Exercise 9.22 (Computation of tangent cones). Let $$I\trianglelefteq K[x_1,\dots,x_n]$$ be an ideal, and assume that the corresponding affine variety $$X=V(I)\subseteq \mathbb{A}^n$$ contains the origin. Consider the blow-up $$\tilde{X}\subseteq \widetilde{\mathbb{A}^n}\subseteq \mathbb{A}^n\times \mathbb{P}^{n-1}$$ at $$x_1,\dots,x_n$$, and denote the homogeneous coordinates of $$\mathbb{P}^{n-1}$$ by $$y_1,\dots,y_n$$.

(a) By example 9.15 we know that $$\widetilde{\mathbb{A}^n}$$ can be covered by affine spaces, with one coordinate patch being \begin{align} \mathbb{A}^n&\to \widetilde{\mathbb{A}^n}\subseteq \mathbb{A}^n\times \mathbb{P}^{n-1}\\ (x_1,y_2,\dots,y_n)&\mapsto((x_1,x_1y_2,\dots,x_1y_n),(1:y_2:\dots:y_n)). \end{align} Prove that on this coordinate patch the blow-up $$\tilde X$$ is given as the zero locus of the polynomials $$$$\frac{f(x_1,x_1y_2,\dots,x_1y_n)}{x_1^{\min\deg f}}$$$$ for all non-zero $$f\in I$$, where $$\min\deg f$$ denotes the smallest degree of a monomial in $$f$$.

(b) Prove that the exceptional hypersurface of $$\tilde X$$ is $$$$V_p(f^{in}:f\in I)\subseteq \{0\}\times \mathbb{P}^{n-1}$$$$ where $$f^{in}$$ is the initial term of $$f$$, i.e. the sum of all monomials in $$f$$ of smallest degree. Consequently, the tangent cone of $$X$$ at the origin is $$$$C_0X=V_a(f^{in}:f\in I)\subseteq \mathbb{A}^n.$$$$

(c) If $$I=(f)$$ is a principal ideal prove that $$C_0X=V_a(f^{in})$$. However, for a general ideal $$I$$, show that $$C_0X$$ is in general not the zero locus of the initial terms of a set of generators for $$I$$.

I'm stuck at (a), which I think it's related to (b) and (c). I have done the following but it might be wrong:

First, I'm gonna state two Lemmas that I think are right

Lemma 1: Let $$X=X_1\cup\dots\cup X_r$$ be the decomposition of a Noetherian space into irreducible subspaces. If $$A$$ is a closed subset of $$X$$ such that for each $$i=1,\dots,n$$, $$X_i\not\subseteq A$$, then $$X\setminus A$$ is dense in $$X$$.

Lemma 2: If $$f\neq 0$$, then $$$$\frac{f(x_1,x_1y_2,\dots,x_1y_n)}{x_1^{\min\deg f}}\notin (x_1).$$$$

Let's call $$\phi:\mathbb{A}^n\to \widetilde{\mathbb{A}^n}$$ the morphism defined in (a). If $$\pi:\widetilde{\mathbb{A}^n}\to \mathbb{A}^n$$ is the map associated to the blow-up, I believe I can prove the following equality

$$$$\phi(V(\frac{f(x_1,x_1y_2,\dots,x_1y_n)}{x_1^{\min\deg f}}:f\in I\setminus\{0\})\setminus V(x_1))=\pi^{-1}(X\setminus\{0\})\cap \phi(\mathbb{A}^n).$$$$

From there, If I could prove that $$$$\overline{V(\frac{f(x_1,x_1y_2,\dots,x_1y_n)}{x_1^{\min\deg f}}:f\in I\setminus\{0\})\setminus V(x_1)}=V(\frac{f(x_1,x_1y_2,\dots,x_1y_n)}{x_1^{\min\deg f}}:f\in I\setminus\{0\}),$$$$ then the exercise would be done just by taking closures. I think Lemma 1 and 2 come into play here. The problem is that $$V(\frac{f(x_1,x_1y_2,\dots,x_1y_n)}{x_1^{\min\deg f}}:f\in I\setminus\{0\})\setminus V(x_1)$$ might not be dense in $$V(\frac{f(x_1,x_1y_2,\dots,x_1y_n)}{x_1^{\min\deg f}}:f\in I\setminus\{0\})$$ because it may happen that $$X_i\subseteq V(x_1)$$ for some irreducible component $$X_i$$ of $$V(\frac{f(x_1,x_1y_2,\dots,x_1y_n)}{x_1^{\min\deg f}}:f\in I\setminus\{0\})$$.

Again, I might have made a mistake, so please read critically. I am mainly interested in (a), but a complete answer is also welcome.

• This is an excellent question, in that you have clearly tried very hard to answer it yourself and have provided lots of context. There's just one problem: what exactly is the question that you want answered? Or, perhaps better: what are you looking for in an answer? Make the answerer's life easier by asking a specific question. Mar 23, 2021 at 20:48
• @Will R I want a full answer to the exercise, or at least an answer to a) which is the troublesome part for me.
– Zero
Mar 23, 2021 at 21:06

A preliminary lemma:

Lemma: Let $$I\subset k[x_1,\cdots,x_n]$$ be an ideal and let $$f\in k[x_1,\cdots,x_n]$$ be a nonzero element. The closure of $$V(I)\cap D(f)$$ is given by the vanishing locus of the ideal $$J=\left(\frac{e}{f^{\deg_f e}}\mid e\in I\right),$$ where $$\deg_f e$$ is the number of times $$f$$ divides $$e$$.

Proof: The morphism of varieties $$V(I)\cap D(f) \to \Bbb A^n$$ corresponds to the ring map $$k[x_1,\cdots,x_n]\to k[x_1,\cdots,x_n]_f/I_f$$, and the closure of $$V(I)\cap D(f)$$ is the variety cut out by the kernel of this ring morphism. As $$I_f=\left\{ \frac{e}{f^m}\mid e\in I, m\in\Bbb Z_{\geq0} \right\}$$, if $$e\in I$$ then if $$f^m\mid e$$, we have $$\frac{e}{f^m}$$ is an element in $$k[x_1,\cdots,x_n]$$ which maps to zero. As every such element is $$f^l\cdot \frac{e}{f^{\deg_f e}}$$ for some $$l\geq 0$$, we have the result. $$\blacksquare$$

On to the exercise.

By remark 9.11, we know that $$\widetilde{X}$$ is the closure of $$\pi^{-1}(X\setminus\{0\})$$ in $$\widetilde{\Bbb A^n}$$. On $$U_1$$, we have that $$\pi^{-1}(X)$$ is the variety cut out by the ideal $$(f(x_1,x_1y_2,\cdots,x_1y_n)\mid f\in I)$$, while $$\pi^{-1}(0)=V(x_1)$$. As $$\pi^{-1}(X\setminus \{0\})$$ is $$\pi^{-1}(X)$$ without $$\pi^{-1}(0)$$, we see that on $$U_1$$, the closure of $$\pi^{-1}(X\setminus \{0\})$$ is exactly the desired ideal by our lemma. This proves (a).

For (b), we use the computation from (a). The exceptional divisor is the intersection of $$\widetilde{X}$$ with the $$\Bbb P^{n-1}$$ living over the origin in $$\widetilde{A^n}$$, so we can find this on $$U_1$$ by setting $$x_1=0$$. As evaluating $$\frac{f(x_1,x_1y_2,\cdots,x_1y_n)}{x_1^{\deg_{x_1} f}}$$ at $$x_1=0$$ gives the initial term of $$f(y_1,\cdots,y_n)$$ and then sets $$y_1=1$$, we see that the exceptional set is cut out by the ideal generated by $$f^{in}(y_1,\cdots,y_n)$$ as $$f$$ runs over the polynomials in $$I$$.

In order to attack (c), expand $$f=f_r+f_{r+1}+\cdots$$ and $$g=g_s+g_{s+1}+\cdots$$ in to homogeneous parts: then $$fg=f_rg_s+(f_rg_{s+1}+f_{r+1}g_s)+\cdots$$, and thus $$(fg)^{in}=f^{in}\cdot g^{in}$$. If $$I=(f)$$, this shows that the ideal we constructed in (b) is just $$(f^{in})$$. The counterexample is a standard sort of trick: consider the ideal $$(x,x-y^2)\subset k[x,y,z]$$. Then the initial ideal of this is $$(x,y^2)$$, but using the set of generators $$\{x,x-y^2\}$$, the ideal which is generated by the initial terms is just $$(x)$$.

• It looks like this is it. However I have two things to say: 1. Why is $m\in \mathbb{Z}$ in the definition of $I_f$, shouldn't it be $m\in \mathbb{N}$? 2. Could you explain the sentence: On $U_1$, we have that $\pi^{-1}(X)$ is the variety cut out by the ideal $(f(x_1,x_1y_2,\dots,x_1y_n)\mid f\in I_X)$
– Zero
Mar 24, 2021 at 4:37
• 1. I usually write $\Bbb Z_{\geq 0}$ instead of $\Bbb N$ and left off the $_{\geq 0}$ part (I fixed it in an edit). Though if you interpret $e/f^{-1}=fe$, I guess it was still correct as written. 2. The composite map $U_1\to\widetilde{\Bbb A^n}\stackrel{\pi}{\to}\Bbb A^n$ pulls back functions by $x_1\mapsto x_1$ and $x_i\mapsto x_1y_i$ (this is example 9.14). As the inverse image of $V(I)$ is the vanishing locus of the image of $I$ along the pullback ring map, the conclusion follows. Mar 24, 2021 at 4:53
• I'm not sure if you should write $I_X$ instead of $I$ alone, because by Hilbert's Nullstelensatz $I_X=\sqrt{I}$.
– Zero
Mar 24, 2021 at 5:09
• I don't think that's hugely consequential, but if one stray reference to $I_X$ instead of $I$ is troublesome, I can remove it. Mar 24, 2021 at 5:12
• Is there a problem with $I$ alone? I thought your solution was totally correct now. Or I'm missing something?
– Zero
Mar 24, 2021 at 5:17