Lemma 31.13.13 (02OO)—The Stacks project

Table of contents
Part 2: Schemes
Chapter 31: Divisors
Section 31.13: Effective Cartier divisors
Lemma 31.13.13 (cite)

Lemma 31.13.13. Let $f : X \to Y$ be a morphism of schemes. Let $D \subset Y$ be an effective Cartier divisor. The pullback of $D$ by $f$ is defined in each of the following cases:

$f(x) \not\in D$ for any weakly associated point $x$ of $X$,
$X$, $Y$ integral and $f$ dominant,
$X$ reduced and $f(\xi ) \not\in D$ for any generic point $\xi $ of any irreducible component of $X$,
$X$ is locally Noetherian and $f(x) \not\in D$ for any associated point $x$ of $X$,
$X$ is locally Noetherian, has no embedded points, and $f(\xi ) \not\in D$ for any generic point $\xi $ of an irreducible component of $X$,
$f$ is flat, and
add more here as needed.

Proof. The question is local on $X$, and hence we reduce to the case where $X = \mathop{\mathrm{Spec}}(A)$, $Y = \mathop{\mathrm{Spec}}(R)$, $f$ is given by $\varphi : R \to A$ and $D = \mathop{\mathrm{Spec}}(R/(t))$ where $t \in R$ is a nonzerodivisor. The goal in each case is to show that $\varphi (t) \in A$ is a nonzerodivisor.

In case (1) this follows from Algebra, Lemma 10.66.7. Case (4) is a special case of (1) by Lemma 31.5.8. Case (5) follows from (4) and the definitions. Case (3) is a special case of (1) by Lemma 31.5.12. Case (2) is a special case of (3). If $R \to A$ is flat, then $t : R \to R$ being injective shows that $t : A \to A$ is injective. This proves (6). $\square$

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