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Let $X$ be a regular scheme and $Z \subset X$ a close subscheme of codimension two. Suppose $Z$ has two components, say $Z = Z_1 \cup Z_2$. Let $f: Y \to X$ be the blowup of $X$ with the center $Z$. It is well-known that the inverse image of $Z$ in $Y$ is a Cartier divisor. The question is, is the inverse image of $Z_1$ (or $Z_2$) a Cartier divisor as well? Or under what condition is this the case?

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It depends on what scheme structure you give $Z$.

If $I_1$ defines $Z_1$ and $I_2$ defines $Z_2$ then the blowup of $Z = V(I_1 \cdot I_2)$ does turn $Z_1$ and $Z_2$ into Cartier divisors. This is actually a pretty straightforward exercise from several perspectives (either working out charts or using universal properties), so I'll let you do it. (You don't need $X$ to be regular for this).

On the other hand, if you set $Z = V(I_1 \cap I_2)$ then you are out of luck. Let me give you an example. Set $X = \mathbb{A}^2 = \text{Spec}[x,y]$, $Z_1 = V(x, y^2)$ and $Z_2 = V(x^2, y)$. Fix $Z = V( (x, y^2) \cap (x^2, y) ) = V(x^2, xy, y^2)$. The blowup of $Z$, $Z_1$ and $Z_2$ all have one exceptional divisor, but they correspond to different valuations. It follows that the inverse image of $Z_1, Z_2$ are not Cartier divisors in the blowup of $Z$ (if this isn't clear, write down affine charts).

It's not so hard to find reduced examples that behave the same way, but you'd need to go to dimension $\geq 3$.

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  • $\begingroup$ Thank you very much. If I understand correctly, everything should work in a nice way if $Z_1$ and $Z_2$ are both integral. Am I right? $\endgroup$ – marker Jun 12 '13 at 14:18
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    $\begingroup$ Unfortunately, that is not the case. The product and intersection of ideals can be quite different even in that case. For instance, consider $k[x,y,z]$ and $I_1 = (x,y), I_2 = (y,z)$. Then $$ I_1 \cdot I_2 = (xy, xz, yz, y^2) $$ but $$ I_1 \cap I_2 = (xy, xz, yz, y). $$ The blowups are different too. $\endgroup$ – Karl Schwede Jun 12 '13 at 15:08

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