Lemma 35.16.2. The property $\mathcal{P}(S) =$“$S$ is Jacobson” is local in the fppf topology.

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

$\xymatrix{ B \ar[rr] & & B' \\ & A \ar[ru] \ar[lu] & }$

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. $\square$

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