Lemma 31.23.5. Let $f : X \to Y$ be a morphism of schemes. In each of the following cases pullbacks of meromorphic functions are defined.

1. every weakly associated point of $X$ maps to a generic point of an irreducible component of $Y$,

2. $X$, $Y$ are integral and $f$ is dominant,

3. $X$ is integral and the generic point of $X$ maps to a generic point of an irreducible component of $Y$,

4. $X$ is reduced and every generic point of every irreducible component of $X$ maps to the generic point of an irreducible component of $Y$,

5. $X$ is locally Noetherian, and any associated point of $X$ maps to a generic point of an irreducible component of $Y$,

6. $X$ is locally Noetherian, has no embedded points and any generic point of an irreducible component of $X$ maps to the generic point of an irreducible component of $Y$, and

7. $f$ is flat.

Proof. The question is local on $X$ and $Y$. Hence we reduce to the case where $X = \mathop{\mathrm{Spec}}(A)$, $Y = \mathop{\mathrm{Spec}}(R)$ and $f$ is given by a ring map $\varphi : R \to A$. By the characterization of regular sections of the structure sheaf in Lemma 31.14.7 we have to show that $R \to A$ maps nonzerodivisors to nonzerodivisors. Let $t \in R$ be a nonzerodivisor.

If $R \to A$ is flat, then $t : R \to R$ being injective shows that $t : A \to A$ is injective. This proves (7).

In the other cases we note that $t$ is not contained in any of the minimal primes of $R$ (because every element of a minimal prime in a ring is a zerodivisor). Hence in case (1) we see that $\varphi (t)$ is not contained in any weakly associated prime of $A$. Thus this case follows from Algebra, Lemma 10.66.7. Case (5) is a special case of (1) by Lemma 31.5.8. Case (6) follows from (5) and the definitions. Case (4) is a special case of (1) by Lemma 31.5.12. Cases (2) and (3) are special cases of (4). $\square$

Comment #3770 by Laurent Moret-Bailly on

Beginning of statement: it should be "pullbacks of meromorphic functions", not "sections".

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