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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 \begin{equation} \frac{f(x_1,x_1y_2,\dots,x_1y_n)}{x_1^{\min\deg f}} \end{equation} 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 \begin{equation} V_p(f^{in}:f\in I)\subseteq \{0\}\times \mathbb{P}^{n-1} \end{equation} 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 \begin{equation} C_0X=V_a(f^{in}:f\in I)\subseteq \mathbb{A}^n. \end{equation}

(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 \begin{equation} \frac{f(x_1,x_1y_2,\dots,x_1y_n)}{x_1^{\min\deg f}}\notin (x_1). \end{equation}

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

\begin{equation} \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). \end{equation}

From there, If I could prove that \begin{equation} \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\}), \end{equation} 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.

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  • $\begingroup$ 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. $\endgroup$
    – Will R
    Mar 23, 2021 at 20:48
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    $\begingroup$ @Will R I want a full answer to the exercise, or at least an answer to a) which is the troublesome part for me. $\endgroup$
    – Zero
    Mar 23, 2021 at 21:06

1 Answer 1

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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 $(x)$.

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  • $\begingroup$ 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 $(f(x_1,x_1y_2,\dots,x_1y_n)\mid f\in I_X)$ $\endgroup$
    – Zero
    Mar 24, 2021 at 4:37
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    $\begingroup$ 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. $\endgroup$
    – KReiser
    Mar 24, 2021 at 4:53
  • $\begingroup$ I'm not sure if you should write $I_X$ instead of $I$ alone, because by Hilbert's Nullstelensatz $I_X=\sqrt{I}$. $\endgroup$
    – Zero
    Mar 24, 2021 at 5:09
  • $\begingroup$ 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. $\endgroup$
    – KReiser
    Mar 24, 2021 at 5:12
  • $\begingroup$ Is there a problem with $I$ alone? I thought your solution was totally correct now. Or I'm missing something? $\endgroup$
    – Zero
    Mar 24, 2021 at 5:17

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