Lemma 31.34.6. Let $X$ be a locally Noetherian scheme. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Let $s$ be a regular meromorphic section of $\mathcal{L}$. Let $U \subset X$ be the maximal open subscheme such that $s$ corresponds to a section of $\mathcal{L}$ over $U$. The blowup $b : X' \to X$ in the ideal of denominators of $s$ is $U$-admissible. There exists an effective Cartier divisor $D \subset X'$ and an isomorphism

$b^*\mathcal{L} = \mathcal{O}_{X'}(D - E),$

where $E \subset X'$ is the exceptional divisor such that the meromorphic section $b^*s$ corresponds, via the isomorphism, to the meromorphic section $1_ D \otimes (1_ E)^{-1}$.

Proof. From the definition of the ideal of denominators in Definition 31.23.10 we immediately see that $b$ is a $U$-admissible blowup. For the notation $1_ D$, $1_ E$, and $\mathcal{O}_{X'}(D - E)$ please see Definition 31.14.1. The pullback $b^*s$ is defined by Lemmas 31.32.11 and 31.23.8. Thus the statement of the lemma makes sense. We can reinterpret the final assertion as saying that $b^*s$ is a global regular section of $b^*\mathcal{L}(E)$ whose zero scheme is $D$. This uniquely defines $D$ hence to prove the lemma we may work affine locally on $X$ and $X'$. Assume $X = \mathop{\mathrm{Spec}}(A)$ is affine and $\mathcal{L} = \mathcal{O}_ X$. Then $s$ is a regular meromorphic function and shrinking further we may assume $s = a'/a$ with $a', a \in A$ nonzerodivisors. Then the ideal of denominators of $s$ corresponds to the ideal $I = \{ x \in A \mid xa' \in aA\}$. Recall that $X'$ is covered by spectra of affine blowup algebras $A' = A[\frac{I}{x}]$ with $x \in I$ (Lemma 31.32.2). Fix $x \in I$ and write $xa' = a a''$ for some $a'' \in A$. The divisor $E \subset X'$ is cut out by $x \in A'$ over the spectrum of $A'$ and hence $1/x$ is a generator of $\mathcal{O}_{X'}(E)$ over $\mathop{\mathrm{Spec}}(A')$. Finally, in the total quotient ring of $A'$ we have $a'/a = a''/x$. Hence $b^*s = a'/a$ restricts to a regular section of $\mathcal{O}_{X'}(E)$ which is over $\mathop{\mathrm{Spec}}(A')$ given by $a''/x$. This finishes the proof. (The divisor $D \cap \mathop{\mathrm{Spec}}(A')$ is cut out by the image of $a''$ in $A'$.) $\square$

Comment #5570 by Andres Fernandez Herrero on

In the last sentence perhaps $ab/x$ should be replaced by $ax/b$, which is in $A$ because $x \in I$.

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