The Stacks project

Proof. Suppose that $\mathfrak p \subset \mathfrak p' \subset R$ and $\mathfrak q' \subset S$ lies over $\mathfrak p'$. As $\varphi $ is open, for every $g \in S$, $g \not\in \mathfrak q'$ we see that $\mathfrak p$ is in the image of $D(g) \subset \mathop{\mathrm{Spec}}(S)$. In other words $S_ g \otimes _ R \kappa (\mathfrak p)$ is not zero. Since $S_{\mathfrak q'}$ is the directed colimit of these $S_ g$ this implies that $S_{\mathfrak q'} \otimes _ R \kappa (\mathfrak p)$ is not zero, see Lemmas 10.9.9 and 10.12.9. Hence $\mathfrak p$ is in the image of $\mathop{\mathrm{Spec}}(S_{\mathfrak q'}) \to \mathop{\mathrm{Spec}}(R)$ as desired. $\square$

Proof. Omitted. $\square$

Proof. According to Lemma 10.41.3 this follows from Topology, Lemma 5.19.5 $\square$

Proof. We give two proofs.

First proof. Let $\mathfrak p \subset R$ be a prime ideal such that the corresponding point of $\mathop{\mathrm{Spec}}(R)$ is in the closure of $T$. This means that for every $f \in R$, $f \not\in \mathfrak p$ we have $D(f) \cap T \not= \emptyset $. Note that $D(f) \cap T$ is the image of $\mathop{\mathrm{Spec}}(S_ f)$ in $\mathop{\mathrm{Spec}}(R)$. Hence we conclude that $S_ f \not= 0$. In other words, $1 \not= 0$ in the ring $S_ f$. Since $S_{\mathfrak p}$ is the directed colimit of the rings $S_ f$ we conclude that $1 \not= 0$ in $S_{\mathfrak p}$. In other words, $S_{\mathfrak p} \not= 0$ and considering the image of $\mathop{\mathrm{Spec}}(S_{\mathfrak p}) \to \mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ we see there exists a $\mathfrak p' \in T$ with $\mathfrak p' \subset \mathfrak p$. As we assumed $T$ closed under specialization we conclude $\mathfrak p$ is a point of $T$ as desired.

Second proof. Let $I = \mathop{\mathrm{Ker}}(R \to S)$. We may replace $R$ by $R/I$. In this case the ring map $R \to S$ is injective. By Lemma 10.30.5 all the minimal primes of $R$ are contained in the image $T$. Hence if $T$ is stable under specialization then it contains all primes. $\square$

Proof. It is a general fact that specializations lift along a closed map of topological spaces, see Topology, Lemma 5.19.7. Hence the second condition implies the first.

Assume that going up holds for $R \to S$. Let $V(I) \subset \mathop{\mathrm{Spec}}(S)$ be a closed set. We want to show that the image of $V(I)$ in $\mathop{\mathrm{Spec}}(R)$ is closed. The ring map $S \to S/I$ obviously satisfies going up. Hence $R \to S \to S/I$ satisfies going up, by Lemma 10.41.4. Replacing $S$ by $S/I$ it suffices to show the image $T$ of $\mathop{\mathrm{Spec}}(S)$ in $\mathop{\mathrm{Spec}}(R)$ is closed. By Topology, Lemmas 5.19.2 and 5.19.6 this image is stable under specialization. Thus the result follows from Lemma 10.41.5. $\square$

Proof. First proof. The first assertion follows from Lemma 10.41.5 combined with Lemma 10.29.4. The second follows because the complement of a constructible set is constructible (see Topology, Lemma 5.15.2), the first part of the lemma and Topology, Lemma 5.19.2.

Second proof. Since $\mathop{\mathrm{Spec}}(R)$ is a spectral space by Lemma 10.26.2 this is a special case of Topology, Lemma 5.23.6. $\square$

Proof. If $R \to S$ is flat, then $R \to S$ satisfies going down by Lemma 10.39.19. Thus to prove the lemma we may assume that $R \to S$ has finite presentation and satisfies going down.

Since the standard opens $D(g) \subset \mathop{\mathrm{Spec}}(S)$, $g \in S$ form a basis for the topology, it suffices to prove that the image of $D(g)$ is open. Recall that $\mathop{\mathrm{Spec}}(S_ g) \to \mathop{\mathrm{Spec}}(S)$ is a homeomorphism of $\mathop{\mathrm{Spec}}(S_ g)$ onto $D(g)$ (Lemma 10.17.6). Since $S \to S_ g$ satisfies going down (see above), we see that $R \to S_ g$ satisfies going down by Lemma 10.41.4. Thus after replacing $S$ by $S_ g$ we see it suffices to prove the image is open. By Chevalley's theorem (Theorem 10.29.10) the image is a constructible set $E$. And $E$ is stable under generalization because $R \to S$ satisfies going down, see Topology, Lemmas 5.19.2 and 5.19.6. Hence $E$ is open by Lemma 10.41.7. $\square$

Proof. Let $\mathfrak p \subset R$ be in the image of the south-west arrow. This means (Lemma 10.18.6) that

is not the zero ring, i.e., $S' \otimes _ k \kappa (\mathfrak p)$ is not the zero ring and the image of $f$ in it is not nilpotent. The ring map $S' \otimes _ k \kappa (\mathfrak p) \to S \otimes _ k \kappa (\mathfrak p)$ is injective. Hence also $S \otimes _ k \kappa (\mathfrak p)$ is not the zero ring and the image of $f$ in it is not nilpotent. Hence $(S \otimes _ k R)_ f \otimes _ R \kappa (\mathfrak p)$ is not the zero ring. Thus (Lemma 10.18.6) we see that $\mathfrak p$ is in the image of the south-east arrow as desired. $\square$

Proof. Let $f \in S \otimes _ k R$. It suffices to prove that the image of the standard open $D(f)$ is open. Let $S' \subset S$ be a finite type $k$-subalgebra such that $f \in S' \otimes _ k R$. The map $R \to S' \otimes _ k R$ is flat and of finite presentation, hence the image $U$ of $\mathop{\mathrm{Spec}}((S' \otimes _ k R)_ f) \to \mathop{\mathrm{Spec}}(R)$ is open by Proposition 10.41.8. By Lemma 10.41.9 this is also the image of $D(f)$ and we win. $\square$

Proof. Consider any prime $\mathfrak q' \subset S$ which corresponds to a point of $\mathop{\mathrm{Spec}}(S_{\mathfrak p})$. This means that $\mathfrak p' = R \cap \mathfrak q'$ is contained in $\mathfrak p$. Here is a picture

Assume (1) and (2)(a). By going up there exists a prime $\mathfrak q'' \subset S$ with $\mathfrak q' \subset \mathfrak q''$ and $\mathfrak q''$ lying over $\mathfrak p$. By the uniqueness of $\mathfrak q$ we conclude that $\mathfrak q'' = \mathfrak q$. In other words $\mathfrak q'$ defines a point of $\mathop{\mathrm{Spec}}(S_{\mathfrak q})$.

Assume (1) and (2)(b). By going down there exists a prime $\mathfrak q'' \subset \mathfrak q$ lying over $\mathfrak p'$. By the uniqueness of primes lying over $\mathfrak p'$ we see that $\mathfrak q' = \mathfrak q''$. In other words $\mathfrak q'$ defines a point of $\mathop{\mathrm{Spec}}(S_{\mathfrak q})$.

In both cases we conclude that the map $\mathop{\mathrm{Spec}}(S_{\mathfrak q}) \to \mathop{\mathrm{Spec}}(S_{\mathfrak p})$ is bijective. Clearly this means all the elements of $S - \mathfrak q$ are all invertible in $S_{\mathfrak p}$, in other words $S_{\mathfrak p} = S_{\mathfrak q}$. $\square$

Proof. The meaning of the statement is as follows. Let $\mathfrak p \subset \mathfrak p' \subset R$ be primes. Let $\mathfrak q' \subset S$ be a prime $\mathfrak q' \in \text{Supp}(N)$ Then there exists a prime $\mathfrak q \subset \mathfrak q'$, $\mathfrak q \in \text{Supp}(N)$ lying over $\mathfrak p$. As $N$ is flat over $R$ we see that $N_{\mathfrak q'}$ is flat over $R_{\mathfrak p'}$, see Lemma 10.39.18. As $N_{\mathfrak q'}$ is finite over $S_{\mathfrak q'}$ and not zero since $\mathfrak q' \in \text{Supp}(N)$ we see that $N_{\mathfrak q'} \otimes _{S_{\mathfrak q'}} \kappa (\mathfrak q')$ is nonzero by Nakayama's Lemma 10.20.1. Thus $N_{\mathfrak q'} \otimes _{R_{\mathfrak p'}} \kappa (\mathfrak p')$ is also not zero. We conclude from Lemma 10.39.15 that $N_{\mathfrak q'} \otimes _{R_{\mathfrak p'}} \kappa (\mathfrak p)$ is nonzero. Let $J \subset S_{\mathfrak q'} \otimes _{R_{\mathfrak p'}} \kappa (\mathfrak p)$ be the annihilator of the finite nonzero module $N_{\mathfrak q'} \otimes _{R_{\mathfrak p'}} \kappa (\mathfrak p)$. Since $J$ is a proper ideal we can choose a prime $\mathfrak q \subset S$ which corresponds to a prime of $S_{\mathfrak q'} \otimes _{R_{\mathfrak p'}} \kappa (\mathfrak p)/J$. This prime is in the support of $N$, lies over $\mathfrak p$, and is contained in $\mathfrak q'$ as desired. $\square$