Ideal of strict transform 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".
 A: 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
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 by Bravo, Encinas and Villamayor.

Example
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.

Question 2
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).
