Total transform of a Cartier divisor

Assume the Axiom of Choice (The Axiom of Choice) as inherited from the blowup construction, which is a relative Proj. Let $\pi \text{\hspace{0.17em}⁣}:{\mathrm{Bl}}_{\mathcal{I}}X\to X$ be the blowup of a scheme $X$ along a quasi-coherent ideal sheaf $\mathcal{I}$ of finite type (Blowup of a scheme along an ideal sheaf), and let $D$ be a Cartier divisor on $X$ (Cartier divisor) whose pullback ${\pi }^{\ast }D$ along $\pi$ is defined (Pullback of a Cartier divisor). The total transform of $D$ under $\pi$ is the pullback Cartier divisor

${\pi }^{\ast }D\mathrm{.}$

Its associated invertible sheaf is the pullback of the invertible sheaf of $D$: there is a canonical isomorphism ${\mathcal{O}}_{\mathrm{Bl}}({\pi }^{\ast }D)\cong {\pi }^{\ast }{\mathcal{O}}_X(D)$ of ${\mathcal{O}}_{\mathrm{Bl}}$-modules (Pullback of a Cartier divisor computes the pullback of its line bundle).

The pullback is defined for every effective Cartier divisor. Indeed, on a standard chart $A[I\mathrm{/}a]\subseteq A_a$, a nonzerodivisor $f\in A$ remains a nonzerodivisor after localization and on this subalgebra. Thus every effective local equation pulls back to a regular equation, without a dominance assumption. For integral $X$ and nonzero $\mathcal{I}$, the chart embeddings in the function field similarly pull back nonzero rational local equations, so every Cartier divisor has a pullback. If $\mathcal{I}=0$, the blowup is empty and these assertions hold vacuously.

For a reduced curve $D$ on a regular surface, blowing up a closed point with two-dimensional regular local ring gives the formula ${\pi }^{\ast }D=D^{\mathrm{\prime }}+mE,$ where $m$ is the order of its local equation at that point. This formula is proved in Total transform equals strict transform plus multiplicity times the exceptional divisor ↗; it is not part of the definition for arbitrary centers. For example, on ${\mathbb{A}}_k^2$, the blowup of $(x^2)$ is the identity, its exceptional Cartier divisor is $2V(x)$, and the strict transform of $V(x)$ is empty. No integer $m$ expresses $V(x)$ as $m\cdot 2V(x)$.