Lemma 31.32.10 (0BFM)—The Stacks project

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Table of contents
Part 2: Schemes
Chapter 31: Divisors
Section 31.32: Blowing up
Lemma 31.32.10 (cite)

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Lemma 31.32.10. Let $X$ be a scheme. Let $Z \subset X$ be a closed subscheme. Let $b : X' \to X$ be the blowing up of $X$ along $Z$. Then $b$ induces an bijective map from the set of generic points of irreducible components of $X'$ to the set of generic points of irreducible components of $X$ which are not in $Z$.

Proof. The exceptional divisor $E \subset X'$ is an effective Cartier divisor and $X' \setminus E \to X \setminus Z$ is an isomorphism, see Lemma 31.32.4. Thus it suffices to show the following: given an effective Cartier divisor $D \subset S$ of a scheme $S$ none of the generic points of irreducible components of $S$ are contained in $D$. To see this, we may replace $S$ by the members of an affine open covering. Hence by Lemma 31.13.2 we may assume $S = \mathop{\mathrm{Spec}}(A)$ and $D = V(f)$ where $f \in A$ is a nonzerodivisor. Then we have to show $f$ is not contained in any minimal prime ideal $\mathfrak p \subset A$. If so, then $f$ would map to a nonzerodivisor contained in the maximal ideal of $R_\mathfrak p$ which is a contradiction with Algebra, Lemma 10.25.1. $\square$

Comments (3)

Comment #6334 by Adi Caplan-Bricker on July 10, 2021 at 15:49

(Sorry I botched that mathjax, let me try again....)

In general nowhere dense sets will often contain generic points of irreducible components, so this argumentation doesn't work.

For example, if $A$ is an infinite product of fields, then every non-principal prime ideal $\mathfrak{p}$ is such that $\{\mathfrak{p}\}$ is a nowhere dense subset of $\operatorname{Spec}(A)$ as well as an irreducible component.

That said, we can just appeal directly to [Lemma 01WS]. The generic point of an irreducible component corresponds to a minimal prime in every affine open neighborhood.
If an effective cartier divisor $E$ contained such a point $x$ , then in any affine open $U \cong \operatorname{Spec}(A)$ of $x$ we would find a non-zero divisor $f \in A$ contained in the minimal prime $\mathfrak{p}_x$ of $A$ .
This is absurd because minimal primes consist of zero-divisors by [Lemma 00EU].

Comment #6335 by Johan on July 10, 2021 at 16:10

Great, thanks! Will make the corresponding edits soonish.

Comment #6437 by Johan on July 20, 2021 at 12:33

Thanks and fixed here.

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