Let $X$ a variety. Say $X=\operatorname{Spec} A$.


Consider two ideals of $A$, say $I$ and $J$, with equal radical ; and consider the blow-ups of X with centre $I$ and $J$, say $Y\_I$ and $Y\_J$. **How can I decide if the blow-up map $Y\_I \to X$ factors through $Y\_J$ ?**


For example, if $X = \operatorname{Spec}k[x,y]$ is the affine plane. Then we can consider the blow-ups with centre $I\_1 = (x,y)$, $I\_2 = (x,y^2)$, $I\_3=(x^2,y^2)$ and $I\_4=(x^2,xy,y^2)=I\_1^2$. We have $I\_4\subset I\_3\subset I\_2\subset I\_1$.
Let $Y\_i$ denote the corresponding blow-ups.


It is well-known that $Y\_1$ and $Y\_4$ are isomorphic. However $Y\_2$ and $Y\_3$ are two other different varieties.
Using the universal property of the blow-up we see that $Y\_1\to X$ factors through $Y\_3$ but not through $Y\_2$.


**What kind of property should I look at to predict this kind of factorization, without computing the actual blow-up ?**



Comparing blow-ups with comparable centres
There are *two* conditions I can think of.


1. Perhaps $I = J \cdot J'$ for some other ideal $J'$. Indeed, blowing up a product of ideals is the same as blowing up one ideal (say $J$) and then blowing up the total transform of the other (say $J'$).
2. The other is requiring that $I = \overline{J}$, here this is integral closure of ideals (more generally $I$ and $J$ have the same integral closure and $I \subseteq J$). Indeed, two ideals have the same integral closure if and only if they have the same normalized blow-up and if their total transforms on the common normalized blow-up agree.


Let me give an alternate explanation of 2., suppose for example that $I = \overline{J}$. Then the blow-up of $I$ and the blow-up of $J$ have the same normalization *and* $Y\_I$ is just a partial normalization of $Y\_J$. 


So this explains some of your examples. Indeed, the integral closure of $I\_3$ is just $I\_4$. On the other hand, $I\_4 = I\_1^2$. So by 1., $I\_1$ and $I\_4$ have the same blowup but by 2., the blow up of $I\_4$ factors through the blowup of $I\_3$. However, if you consider $I\_1 \cdot I\_2$ or $I\_3 \cdot I\_2$, you will find that their blowups factor through $I\_2$.


For smooth surfaces, there is a theory of factorization of complete (integrally closed) ideals due to Zariski. This should allow you to answer these sorts of questions even more precisely.


