10.41 Going up and going down
Suppose \mathfrak p, \mathfrak p' are primes of the ring R. Let X = \mathop{\mathrm{Spec}}(R) with the Zariski topology. Denote x \in X the point corresponding to \mathfrak p and x' \in X the point corresponding to \mathfrak p'. Then we have:
x' \leadsto x \Leftrightarrow \mathfrak p' \subset \mathfrak p.
In words: x is a specialization of x' if and only if \mathfrak p' \subset \mathfrak p. See Topology, Section 5.19 for terminology and notation.
Definition 10.41.1. Let \varphi : R \to S be a ring map.
We say a \varphi : R \to S satisfies going up if given primes \mathfrak p \subset \mathfrak p' in R and a prime \mathfrak q in S lying over \mathfrak p there exists a prime \mathfrak q' of S such that (a) \mathfrak q \subset \mathfrak q', and (b) \mathfrak q' lies over \mathfrak p'.
We say a \varphi : R \to S satisfies going down if given primes \mathfrak p \subset \mathfrak p' in R and a prime \mathfrak q' in S lying over \mathfrak p' there exists a prime \mathfrak q of S such that (a) \mathfrak q \subset \mathfrak q', and (b) \mathfrak q lies over \mathfrak p.
So far we have see the following cases of this:
An integral ring map satisfies going up, see Lemma 10.36.22.
As a special case finite ring maps satisfy going up.
As a special case quotient maps R \to R/I satisfy going up.
A flat ring map satisfies going down, see Lemma 10.39.19
As a special case any localization satisfies going down.
An extension R \subset S of domains, with R normal and S integral over R satisfies going down, see Proposition 10.38.7.
Here is another case where going down holds.
Lemma 10.41.2. Let R \to S be a ring map. If the induced map \varphi : \mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R) is open, then R \to S satisfies going down.
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
Lemma 10.41.3. Let R \to S be a ring map.
R \to S satisfies going down if and only if generalizations lift along the map \mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R), see Topology, Definition 5.19.4.
R \to S satisfies going up if and only if specializations lift along the map \mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R), see Topology, Definition 5.19.4.
Proof.
Omitted.
\square
Lemma 10.41.4. Suppose R \to S and S \to T are ring maps satisfying going down. Then so does R \to T. Similarly for going up.
Proof.
According to Lemma 10.41.3 this follows from Topology, Lemma 5.19.5
\square
Lemma 10.41.5. Let R \to S be a ring map. Let T \subset \mathop{\mathrm{Spec}}(R) be the image of \mathop{\mathrm{Spec}}(S). If T is stable under specialization, then T is closed.
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
Lemma 10.41.6. Let R \to S be a ring map. The following are equivalent:
Going up holds for R \to S, and
the map \mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R) is closed.
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
Lemma 10.41.7. Let R be a ring. Let E \subset \mathop{\mathrm{Spec}}(R) be a constructible subset.
If E is stable under specialization, then E is closed.
If E is stable under generalization, then E is open.
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
Proposition 10.41.8. Let R \to S be flat and of finite presentation. Then \mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R) is open. More generally this holds for any ring map R \to S of finite presentation which satisfies going down.
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
Lemma 10.41.9. Let k be a field, and let R, S be k-algebras. Let S' \subset S be a sub k-algebra, and let f \in S' \otimes _ k R. In the commutative diagram
\xymatrix{ \mathop{\mathrm{Spec}}((S \otimes _ k R)_ f) \ar[rd] \ar[rr] & & \mathop{\mathrm{Spec}}((S' \otimes _ k R)_ f) \ar[ld] \\ & \mathop{\mathrm{Spec}}(R) & }
the images of the diagonal arrows are the same.
Proof.
Let \mathfrak p \subset R be in the image of the south-west arrow. This means (Lemma 10.18.6) that
(S' \otimes _ k R)_ f \otimes _ R \kappa (\mathfrak p) = (S' \otimes _ k \kappa (\mathfrak p))_ f
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
Lemma 10.41.10. Let k be a field. Let R and S be k-algebras. The map \mathop{\mathrm{Spec}}(S \otimes _ k R) \to \mathop{\mathrm{Spec}}(R) is open.
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
Here is a tricky lemma that is sometimes useful.
Lemma 10.41.11. Let R \to S be a ring map. Let \mathfrak p \subset R be a prime. Assume that
there exists a unique prime \mathfrak q \subset S lying over \mathfrak p, and
either
going up holds for R \to S, or
going down holds for R \to S and there is at most one prime of S above every prime of R.
Then S_{\mathfrak p} = S_{\mathfrak q}.
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
\xymatrix{ \mathfrak q' \ar@{-}[d] \ar@{-}[r] & ? \ar@{-}[r] \ar@{-}[d] & S \ar@{-}[d] \\ \mathfrak p' \ar@{-}[r] & \mathfrak p \ar@{-}[r] & R }
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
The following lemma is a generalization of going down for flat ring maps.
Lemma 10.41.12. Let R \to S be a ring map. Let N be a finite S-module flat over R. Endow \text{Supp}(N) \subset \mathop{\mathrm{Spec}}(S) with the induced topology. Then generalizations lift along \text{Supp}(N) \to \mathop{\mathrm{Spec}}(R).
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
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