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