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".


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.


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).

  • $\begingroup$ Hi Karl, Thanks for your answer! I'm having a little bit of difficulty understanding your $\tilde{J}$. It seems like your $J$ is the same as my $J$, and that $J\cdot B_a$ refers to the intersection with the degree zero part of the localization of the Rees algebra at $a$. I very recently found an article of Brian Conrad which says that this $J\cdot B_a$ is already the ideal of the strict transform. Did I understand your notation correctly? (The article is here: line 23, p5, math.stanford.edu/~conrad/papers/nagatafinal.pdf) $\endgroup$
    – 6672
    Nov 1 '12 at 14:54
  • $\begingroup$ For me, $\widetilde{J}$ is the ideal sheaf of the strict transform. $\bar{J}$ is definitely not the strict transform. Brian may be being a little loose with his usage of pull-back or his map might be a small map on $X$ or something (I didn't read the surrounding text). $\endgroup$ Nov 1 '12 at 15:14
  • $\begingroup$ I've changed notation a little. In particular, I'm not using $B_a$ for the degree zero part any more since that seems to be causing confusion. $\endgroup$ Nov 1 '12 at 15:26
  • $\begingroup$ Karl, I worked it out and the two statements (Brian's and yours) are in fact the same, and the difference is purely notational. However, the statement I made above is wrong, and is not what appears in Brian's article. To get a correct statement I should replace $J\cdot B_a$ with (in terms of your new notation) $J_a \cdot B$ where the intersection happens inside not $B_a$, but the localization $C$ of $B$ at $a \in A \subset B$ (so not just the degree $1$ a of the Rees alg) and $J_a$ denotes the extension of the ideal $J$ to $C$. If you would like to see it, I can include a careful proof below. $\endgroup$
    – 6672
    Nov 2 '12 at 23:55

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