I know this is a stupid question, but I can't find a reference, or the result stated in in full generality online. I was hoping somebody knew:


Let $X = \textrm{Spec }A$ a noetherian scheme, $Z\hookrightarrow Y \hookrightarrow X$ a sequence of closed immersions with $Z,Y$ corresponding to the ideals $I,J$ respectively. Then we have a canonical inclusion (closed immersion) of blowups $\tilde{Y} = \textrm{Bl}\_Z Y \hookrightarrow \tilde{X} = \textrm{Bl}\_Z X$.


The scheme $\tilde{X}$ has an affine open cover with $U(a) = \textrm{Spec } B\_a$ and $B\_a$ the degree zero piece of the localization of the Rees algebra at $a \in I$.


$\textbf{Question}$: What is the ideal defining $\tilde{Y}\vert\_{U(a)}$ in $U(a)$.


$\textbf{Question}$: At the level of Rees algebras, does the canonical map $\tilde{Y} \hookrightarrow \tilde{X}$ correspond to the natural surjection of graded rings $$ A \oplus I\oplus I^2 \cdots \twoheadrightarrow A/J \oplus \bar{I} \oplus \bar{I}^2 \oplus \cdots $$ where $\bar{I}$ denotes the image of $I$ in $A/J$. Certainly this is "natural".



Ideal of strict transform
Let me fix some notation, let's set $\pi : \widetilde{X} \to X$ be the blowup and set $\bar{I}$ to be the ideal sheaf $(\pi^{-1} I) \cdot O\_{\widetilde{X}}$ (note this is an invertible sheaf) and set $\bar{J} := (\pi^{-1} J) \cdot O\_{\widetilde{X}}$ (this is probably not invertible). Finally, for clarity, let's set $B = [B\_a]\_0$, the degree zero piece of $B$ localized at $a$. Note there is a natural map $R \to B$ since $R$ maps to the degree zero piece of the Rees algebra.


In terms of your two questions:


Question 1
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The ideal sheaf $\widetilde{J}$ defining the strict transform $\widetilde{Y}$ on $\widetilde{X}$ is defined as follows.  

$$\widetilde{J} = \bigcup\_{n = 1}^{\infty} (\bar{J} : \bar{I}^n) =: (\bar{J} : \bar{I}^{\infty})$$
where the colon is taken over $O\_{\widetilde{X}}$ (and the infinite power is a formal notation). In particular, as you can see this is a pain to compute. In terms of local coordinates in the notation you wrote, this is just:
$$
\bigcup\_{n = 1}^{\infty} ((J \cdot B) :\_{B} \langle a^n \rangle\_{B}).
$$
You can find more about this for example in papers on resolution of singularities, I think I first learned this in section 7 of this [PAPER](http://arxiv.org/pdf/math/0206244.pdf) by Bravo, Encinas and Villamayor.




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Example
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Let's do an example. Consider $X = \text{Spec } k[x,y]$ and let $Z = V(x,y)$ be the origin. Let's let $Y = V(x^3-y^4)$, some sort of particularly nasty cusp, so $J = (x^3-y^4)$. 


There are two affine charts on the blowup. $B = k[x,y/x]$ and $B' = k[x/y,y]$. We first extend $J$ to these two charts. We get
$$
J \cdot B = (x^3-y^4) \cdot B = (x^3 - (y/x)^4 x^4) \cdot B = = x^3(1 - (y/x)^4 x) \cdot B.
$$
and
$$
J \cdot B' = (x^3 - y^4) \cdot B' = ( (x/y)^3 y^3 - y^4 ) \cdot B' = y^3( (x/y)^3 - y) \cdot B'
$$
The ideal sheaf $\bar{J}$ just corresponds to $J \cdot B$ and $J \cdot B'$. The ideal sheaf corresponding to the strict transform corresponds to $(1 - (y/x)^4 x) \cdot B$ and $(1 - (y/x)^4 x) \cdot B$. In other words, strip away the $x^3$ and $y^3$ (respectively) which simply vanish on the exceptional divisor.




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Question 2
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I believe this is right. Think about what the kernel of that map is on the $B\_a$, certainly you have $J$, but you also have things that are knocked in there by powers of $a$ (due to the localization).


