Take the 2-minute tour ×
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.

Fulton in his book "Intersection theory" uses local description of this deformation that I can't understand. I quote paragraph from page 87 and insert my questions.

Assume $Y=\operatorname{Spec}(A)$, and $X$ is defined by the ideal $I$ in $A$. To study blow-up of $Y \times \mathbb{P}^1$ near $\infty$ identify $\mathbb{P}^1-\{0\}$ with $\mathbb{A}^1=\operatorname{Spec} K[T]$, where $K$ is the ground field. The blow-up of $Y \times \mathbb{A}^1$ along $X \times \{0\}$ is $\operatorname{Proj}(S^{\bullet})$, with $$S^n=(I, T)^n=I^n+I^{n-1}T+\ldots+ AT^n+AT^{n+1}+\ldots.$$

First thing that I don't understand is why sum don't stop after $AT^n$, i.e. I'd suppose that $$S^n=(I, T)^n=I^n+I^{n-1}T+\ldots+ AT^n.$$

$\operatorname{Proj}(S^{\bullet})$ is covered by affine open sets $\operatorname{Spec}(S^{\bullet}_{(a)})$, where $S^{\bullet}_{(a)}$ is the ring of fractions $$S^{\bullet}_{(a)}=\{s/a^n | s \in S^n\},$$ and $a$ runs through a set of generators for the ideal $(I, T)$ in $A[T].$

Next claim that I don't understand is: for $a \in I,$ the exceptional divisor $P(C \otimes 1)$ is defined in $\operatorname{Spec}(S^{\bullet}_{(a)})$ by the equation $a/1$, $a \in S^0$. How is that possible first invert element $a$ and than consider equation $a=0$?

Any help would be much appreciated.

Update: graded algebra that correspond to the exceptional locus is $R=\sum_{n } (I, T)^n/(I,T)^{n+1}$, then we have short exact sequence $$0 \to (I, T)S^{\bullet} \to S^{\bullet} \to R^{\bullet} \to 0.$$

So $((I,T)S^{\bullet})_{(a)}=\{s/a^n | s \in S^{n+1}\}=\{a \frac{s}{a^{n+1}}| s \in S^{n+1}\}=aS^{\bullet}_{(a)}.$ After localization we get $$0 \to aS^{\bullet}_{(a)} \to (S^{\bullet})_{(a)} \to (R^{\bullet})_{(a)} \to 0.$$

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

For your first question, $S^n$ is an ideal in $A[T]$, so you're looking at $A[T]$-linear combinations of $(I,T)^n$. Since $T^n \in S^n$, so is $A[T] \cdot T^n$, which includes $A T^k$ for all $k \geq n$.

For your second question, keep in mind that when $a \in I$, $1/a$ is not an element of $S_{(a)}^{\bullet}$ because $1 \notin S^1$. Thus, you have not actually inverted the element $a \in I$ in your construction. (I interpret Fulton's notation of $a / 1$, $a \in S^0$ to mean to view $a$ as an element of $S^0$, not $S^1$, since $a / 1$ is not a well-defined expression of an element in $S^1_{(a)}$. Note that $a = a/1$ is also clearly not inveritble in $S^0 = A[T]$.)

share|improve this answer
    
Thanks a lot. Could you also explain why $a=0$, is an equation for exceptional divisor? –  Alex Oct 29 '12 at 14:43
    
It's essentially a straightforward (though complicated) unpacking of the definition of the exceptional divisor of a blowup. Have you read the relevant material in Appendix B of Fulton's book? He explains the general blow-up construction including a description of the ideal sheaf of the exceptional divisor there. –  Michael Joyce Oct 29 '12 at 16:25
    
I added an update about this unpacking, is that correct? –  Alex Oct 29 '12 at 18:05
    
At quick glance, it looks good. I'll have a more careful look at it tonight when I have my copy of Fulton handy. –  Michael Joyce Oct 29 '12 at 19:16
add comment

Your Answer

 
discard

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.