The Stacks project

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

(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 is an infinite product of fields, then every non-principal prime ideal is such that is a nowhere dense subset of 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 contained such a point , then in any affine open of we would find a non-zero divisor contained in the minimal prime of .
This is absurd because minimal primes consist of zero-divisors by [Lemma 00EU].

Comment #6335 by on

Great, thanks! Will make the corresponding edits soonish.

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