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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 $P(C \oplus 1)$ 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 $$0 \to aS^{\bullet}_{(a)} \to (S^{\bullet})_{(a)} \to (R^{\bullet})_{(a)} \to 0.$$

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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 $S^0 = A[T]$.)

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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

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