# Blowing up intersections of two divisors make them disjoint?

Let $D,E \subset \mathbb{C}^3$ be prime divisors where $D$ is smooth and $E$ is not necessarily smooth. Assume that $D \cap E$ has SNC support and let

$D \cap E = \bigcup \Gamma_i$ be a decomposition into irreducible components.

(Added) Let $f_1 \colon X_1 \rightarrow \mathbb{C}^3$ be a blow-up along a smooth curve $\Gamma_i$ and $\tilde{D}_1 \subset X_1$ be the strict transform of $D$. $f_1$ induces an isomorphism $g_1 \colon \tilde{D}_1 \rightarrow D$.

There is a SNC curve $\bigcup \Gamma^1_i \subset \tilde{D}_1$. We next blow-up one of $\Gamma^1_i$.

Question Is it possible to make the strict transforms of $D$ and $E$ disjoint by blowing-up smooth curves $\Gamma_i$ or its strict transforms on the strict transforms of $D$ several times?

I know that the blow-up of $\mathbb{C}^3$ along the ideal sheaf $\mathcal{I}_D + \mathcal{I}_E$ makes $D$ and $E$ disjoint. However I want the smooth centre blow-ups.

(My thought which uses Karl Schwede's answer comment)

Let $\pi:Y \rightarrow X$ be a principalization of $I_D + I_E$ as in Karl Schwede's answer.

Let $Z \subset X$ be a finite set which is the union of the images of the smooth points centers. I think that $\pi^{-1}(X \setminus Z) \rightarrow X \setminus Z$ is the blow-ups of smooth curves $\Gamma_i \setminus Z$ or their strict transform curves on the strict transforms of $D$.

Let $\pi': Y' \rightarrow X$ be a composite of blow-ups of the smooth curves $\Gamma_i$ or its strict transform inside strict transforms of $D$ which are in same order as $\pi$. That is, we can define $\pi'$ by forgetting smooth points centers and curve centers which are contained in the inverse image of smooth point centers. $\pi'$ is same as $\pi$ outside $Z$.

Let $\tilde{D}', \tilde{E}' \subset Y'$ be the strict transforms of the original divisors $D, E \subset X$. If $\tilde{D}' \cap \tilde{E}' \neq \emptyset$, then $\tilde{D}' \cap \tilde{E}' \subset \pi'^{-1}(Z) \cap \bigcup \tilde{\Gamma}'_i$ and it is finite points. This is a contradiction since $\tilde{D}', \tilde{E}'$ are Cartier divisors on a smooth 3-fold $Y$. Hence $\tilde{D}' \cap \tilde{E} = \emptyset$.

Sorry for the long sentence, but I think that the idea is simple. I just ignored centers which is concerned with 0-dimensional centers. Are there gaps in this argument?

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What do you mean by "blowing up $\Gamma_i$ several times"? After the first blowup the curve is replaced by a divisor, so what are you going to blowup next time? –  Sasha Apr 9 '12 at 3:17
Blowing up one of $\Gamma_i$ doesn't change $D$, that is, it induces an isomorphism $\tilde{D} \rightarrow D$ where $\tilde{D}$ is the strict transform. So I blow-up the inverse image of one of $\Gamma_i$ on $\tilde{D}$ next time. I will edit the question. –  tarosano Apr 9 '12 at 8:19
Dear tarosano, do we know that $I_D + I_E$ is a radical ideal? –  Karl Schwede Apr 9 '12 at 11:39
I think I can't assume $I_D + I_E$ is radical ideal since the typical case is $I_D = (x), I_E= (x +y^3)$ and $D$ and $E$ does not intersect transversally. –  tarosano Apr 9 '12 at 11:57
I added explanations in my question. Please let me know if there is a gap in the tentative proof. –  tarosano Apr 9 '12 at 12:07

I'm not exactly sure what you want (see Sasha's question above) but algorithms for embedded resolution of singularities might do what you want (for example, see the recent book of Kollár, various papers by Villamayor and co-authors, or Wlodarczyk)

Indeed, do a log resolution (ie, principalization) of the ideal $I_D + I_E$. What this will produce is a proper birational map $$\pi : Y \to X = \mathbb{C}^3$$ such that

1. $(I_D + I_E) \cdot O_Y = O_Y(-F)$ for some divisor $F$. In particular, this implies that $\pi$ factors through the blowup of $I_D + I_E$ by the universal property of blowing up and thus separates $D$ and $E$.
2. $\pi$ is an isomorphism outside of $D \cap E = V(I_D + I_E)$.
3. $\pi$ is obtained by a sequence of blowups of smooth centers.

You don't need the SNC support at all, although this will make things much easier.

Of course, I don't quite know what you mean by blowing up the support of the $\Gamma_i$ several times.

In the SNC case: What the resolution algorithms would start with is blowing up the strata of the $D \cap E$, in other words, it first will blow up point where the curves of $D \cap E$ intersect. After doing that, it will next blow up the the strict transforms of irreducible components of $D \cap E$ (ie, the curves). EDIT: It might blow up some more points again at various steps. It might also blow up other curves lying over, but themselves are not the $\Gamma_i$ at a particular step. The reason I say this is because blowing up the $I_D + I_E$ doesn't actually know about the divisors $D$ and $E$ and won't keep track of their strict transforms...

Is this ok for what you have in mind? Or do you want to avoid blowing up the points too?

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Thanks for the comment. Actually, I want to avoid a smooth point blow-up. –  tarosano Apr 9 '12 at 8:36
I see, let me think about this then. Do you get to assume that $I_D + I_E$ is a radical ideal? I guess we know its unmixed which is good. –  Karl Schwede Apr 9 '12 at 11:42
Thanks for editing. Anyway, is it true that there exists a finite set $Z$ such that $\pi^{-1}(X \setminus Z) \rightarrow X \setminus Z$ is a composite of blow-ups of $\Gamma_i$ or their ST on the ST of $D$? (ST = strict transform) –  tarosano Apr 9 '12 at 15:17
Dear tarosano, I don't see why this should be the case, but I'd have to think about it. One may as well assume then there is only one $\Gamma_i$ to start with, but it is non-reduced. After one blows up the the support of that $\Gamma_i$, I don't see why the support of $D \cap E$ (or rather their strict transforms) still need intersect in something with non-singular support (maybe this follows from unmixedness). What else can we assume about $D$ and $E$? Maybe this is all fine though, it just needs to be gone through carefully. –  Karl Schwede Apr 9 '12 at 16:18
$f_1:X_1 \rightarrow \mathbb{C}^3$ which is the blow-up of a smooth curve $\Gamma_1$ induces an isomorphism $f_{D_1}: D_1 \rightarrow D$ and a birational morphism $f_{E_1}:E_1 \rightarrow E$ where $D_1, E_1$ are strict transforms. We can see that $D_1 \cap E_1 \subset f_{D_1}^{-1}(D \cap E) \simeq D \cap E = \bigcup \Gamma_i$ and $D_1 \cap E_1 \setminus \Gamma_1 \simeq f_{D_1}^{-1}(D \cap E \setminus \Gamma_1)$. Hence $D_1 \cap E_1$ is isomorphic to $\bigcup \Gamma_i$ or $\bigcup_{i \neq 1} \Gamma_i$ and the intersection stays SNC. Is there a problem? –  tarosano Apr 9 '12 at 18:14