The Stacks project

Proof. We will use Lemma 35.15.2. First we note that “being locally Noetherian” is local in the Zariski topology. This is clear from the definition, see Properties, Definition 28.5.1. Next, we show that if $S' \to S$ is a flat, finitely presented morphism of affines and $S$ is locally Noetherian, then $S'$ is locally Noetherian. This is Morphisms, Lemma 29.15.6. Finally, we have to show that if $S' \to S$ is a surjective flat, finitely presented morphism of affines and $S'$ is locally Noetherian, then $S$ is locally Noetherian. This follows from Algebra, Lemma 10.164.1. Thus (1), (2) and (3) of Lemma 35.15.2 hold and we win. $\square$

Proof. We will use Lemma 35.15.2. First we note that “being Jacobson” is local in the Zariski topology. This is Properties, Lemma 28.6.3. Next, we show that if $S' \to S$ is a flat, finitely presented morphism of affines and $S$ is Jacobson, then $S'$ is Jacobson. This is Morphisms, Lemma 29.16.9. Finally, we have to show that if $f : S' \to S$ is a surjective flat, finitely presented morphism of affines and $S'$ is Jacobson, then $S$ is Jacobson. Say $S = \mathop{\mathrm{Spec}}(A)$ and $S' = \mathop{\mathrm{Spec}}(B)$ and $S' \to S$ given by $A \to B$. Then $A \to B$ is finitely presented and faithfully flat. Moreover, the ring $B$ is Jacobson, see Properties, Lemma 28.6.3.

By Algebra, Lemma 10.168.10 there exists a diagram

with $A \to B'$ finitely presented, faithfully flat and quasi-finite. In particular, $B \to B'$ is finite type, and we see from Algebra, Proposition 10.35.19 that $B'$ is Jacobson. Hence we may assume that $A \to B$ is quasi-finite as well as faithfully flat and of finite presentation.

Assume $A$ is not Jacobson to get a contradiction. According to Algebra, Lemma 10.35.5 there exists a nonmaximal prime $\mathfrak p \subset A$ and an element $f \in A$, $f \not\in \mathfrak p$ such that $V(\mathfrak p) \cap D(f) = \{ \mathfrak p\} $.

This leads to a contradiction as follows. First let $\mathfrak p \subset \mathfrak m$ be a maximal ideal of $A$. Pick a prime $\mathfrak m' \subset B$ lying over $\mathfrak m$ (exists because $A \to B$ is faithfully flat, see Algebra, Lemma 10.39.16). As $A \to B$ is flat, by going down see Algebra, Lemma 10.39.19, we can find a prime $\mathfrak q \subset \mathfrak m'$ lying over $\mathfrak p$. In particular we see that $\mathfrak q$ is not maximal. Hence according to Algebra, Lemma 10.35.5 again the set $V(\mathfrak q) \cap D(f)$ is infinite (here we finally use that $B$ is Jacobson). All points of $V(\mathfrak q) \cap D(f)$ map to $V(\mathfrak p) \cap D(f) = \{ \mathfrak p\} $. Hence the fibre over $\mathfrak p$ is infinite. This contradicts the fact that $A \to B$ is quasi-finite (see Algebra, Lemma 10.122.4 or more explicitly Morphisms, Lemma 29.20.10). Thus the lemma is proved. $\square$

Proof. We will use Lemma 35.15.2. First we note that $\mathcal{P}$ is local in the Zariski topology. Next, we show that if $T \to S$ is a flat, finitely presented morphism of affines and $S$ has a finite number of irreducible components, then so does $T$. Namely, since $T \to S$ is flat, the generic points of $T$ map to the generic points of $S$, see Morphisms, Lemma 29.25.9. Hence it suffices to show that for $s \in S$ the fibre $T_ s$ has a finite number of generic points. Note that $T_ s$ is an affine scheme of finite type over $\kappa (s)$, see Morphisms, Lemma 29.15.4. Hence $T_ s$ is Noetherian and has a finite number of irreducible components (Morphisms, Lemma 29.15.6 and Properties, Lemma 28.5.7). Finally, we have to show that if $T \to S$ is a surjective flat, finitely presented morphism of affines and $T$ has a finite number of irreducible components, then so does $S$. This follows from Topology, Lemma 5.8.5. Thus (1), (2) and (3) of Lemma 35.15.2 hold and we win. $\square$