Lemma 35.13.2. The property $\mathcal{P}(S) =$“$S$ is Jacobson” is local in the fppf topology.

**Proof.**
We will use Lemma 35.12.2. 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. $\square$

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