## Tag `00HU`

## 10.40. Going up and going down

Suppose $\mathfrak p$, $\mathfrak p'$ are primes of the ring $R$. Let $X = \mathop{\rm Spec}(R)$ 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 10.40.1. Let $\varphi : R \to S$ be a ring map.

- We say a $\varphi : R \to S$ satisfies
going upif 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 downif 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 10.35.22.
- 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 10.38.18
- 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 10.37.7.
Here is another case where going down holds.

Lemma 10.40.2. Let $R \to S$ be a ring map. If the induced map $\varphi : \mathop{\rm Spec}(S) \to \mathop{\rm Spec}(R)$ 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 $D(g) \subset \mathop{\rm Spec}(S)$. 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 10.11.9. Hence $\mathfrak p$ is in the image of $\mathop{\rm Spec}(S_{\mathfrak q'}) \to \mathop{\rm Spec}(R)$ as desired. $\square$Lemma 10.40.3. Let $R \to S$ be a ring map.

- $R \to S$ satisfies going down if and only if generalizations lift along the map $\mathop{\rm Spec}(S) \to \mathop{\rm Spec}(R)$, see Topology, Definition 5.19.3.
- $R \to S$ satisfies going up if and only if specializations lift along the map $\mathop{\rm Spec}(S) \to \mathop{\rm Spec}(R)$, see Topology, Definition 5.19.3.

Proof.Omitted. $\square$Lemma 10.40.4. 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 10.40.3 this follows from Topology, Lemma 5.19.4 $\square$Lemma 10.40.5. Let $R \to S$ be a ring map. Let $T \subset \mathop{\rm Spec}(R)$ be the image of $\mathop{\rm Spec}(S)$. 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 $\mathop{\rm Spec}(R)$ 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 $\mathop{\rm Spec}(S_f)$ in $\mathop{\rm Spec}(R)$. 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 limit 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 $\mathop{\rm Spec}(S_{\mathfrak p}) \to \mathop{\rm Spec}(S) \to \mathop{\rm Spec}(R)$ 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 $I = \mathop{\rm Ker}(R \to S)$. We may replace $R$ by $R/I$. In this case the ring map $R \to S$ is injective. By Lemma 10.29.5 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 10.40.6. Let $R \to S$ be a ring map. The following are equivalent:

- Going up holds for $R \to S$, and
- the map $\mathop{\rm Spec}(S) \to \mathop{\rm Spec}(R)$ is closed.

Proof.It is a general fact that specializations lift along a closed map of topological spaces, see Topology, Lemma 5.19.6. Hence the second condition implies the first.Assume that going up holds for $R \to S$. Let $V(I) \subset \mathop{\rm Spec}(S)$ be a closed set. We want to show that the image of $V(I)$ in $\mathop{\rm Spec}(R)$ 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 10.40.4. Replacing $S$ by $S/I$ it suffices to show the image $T$ of $\mathop{\rm Spec}(S)$ in $\mathop{\rm Spec}(R)$ is closed. By Topology, Lemmas 5.19.2 and 5.19.5 this image is stable under specialization. Thus the result follows from Lemma 10.40.5. $\square$

Lemma 10.40.7. Let $R$ be a ring. Let $E \subset \mathop{\rm Spec}(R)$ 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 10.40.5 combined with Lemma 10.28.3. 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 $\mathop{\rm Spec}(R)$ is a spectral space by Lemma 10.25.2 this is a special case of Topology, Lemma 5.23.5. $\square$

Proposition 10.40.8. Let $R \to S$ be flat and of finite presentation. Then $\mathop{\rm Spec}(S) \to \mathop{\rm Spec}(R)$ is open. More generally this holds for any ring map $R \to S$ of finite presentation which satisfies going down.

Proof.Assume that $R \to S$ has finite presentation and satisfies going down. It suffices to prove that the image of a standard open $D(f)$ is open. Since $S \to S_f$ satisfies going down as well, we see that $R \to S_f$ satisfies going down. Thus after replacing $S$ by $S_f$ we see it suffices to prove the image is open. By Chevalley's theorem (Theorem 10.28.9) 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 5.19.5. Hence $E$ is open by Lemma 10.40.7. $\square$Lemma 10.40.9. 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 $$ \xymatrix{ \mathop{\rm Spec}((S \otimes_k R)_f) \ar[rd] \ar[rr] & & \mathop{\rm Spec}((S' \otimes_k R)_f) \ar[ld] \\ & \mathop{\rm Spec}(R) & } $$ 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 10.16.9) 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 10.16.9) we see that $\mathfrak p$ is in the image of the south-east arrow as desired. $\square$Lemma 10.40.10. Let $k$ be a field. Let $R$ and $S$ be $k$-algebras. The map $\mathop{\rm Spec}(S \otimes_k R) \to \mathop{\rm Spec}(R)$ is open.

Proof.Let $f \in R \otimes_k S$. 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 $\mathop{\rm Spec}((S' \otimes_k R)_f) \to \mathop{\rm Spec}(R)$ is open by Proposition 10.40.8. By Lemma 10.40.9 this is also the image of $D(f)$ and we win. $\square$Here is a tricky lemma that is sometimes useful.

Lemma 10.40.11. 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 $\mathop{\rm Spec}(S_{\mathfrak p})$. 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 $\mathop{\rm Spec}(S_{\mathfrak q})$.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 $\mathop{\rm Spec}(S_{\mathfrak q})$.

In both cases we conclude that the map $\mathop{\rm Spec}(S_{\mathfrak q}) \to \mathop{\rm Spec}(S_{\mathfrak p})$ 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 10.40.12. Let $R \to S$ be a ring map. Let $N$ be a finite $S$-module flat over $R$. Endow $\text{Supp}(N) \subset \mathop{\rm Spec}(S)$ with the induced topology. Then generalizations lift along $\text{Supp}(N) \to \mathop{\rm Spec}(R)$.

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 10.38.19. 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 10.19.1. Thus $N_{\mathfrak q'} \otimes_{R_{\mathfrak p'}} \kappa(\mathfrak p')$ is also not zero. We conclude from Lemma 10.38.15 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. $\square$

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 9144–9528 (see updates for more information).

```
\section{Going up and going down}
\label{section-going-up}
\noindent
Suppose $\mathfrak p$, $\mathfrak p'$ are primes
of the ring $R$. Let $X = \Spec(R)$ 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 \ref{topology-section-specialization}
for terminology and notation.
\begin{definition}
\label{definition-going-up-down}
Let $\varphi : R \to S$ be a ring map.
\begin{enumerate}
\item We say a $\varphi : R \to S$ satisfies {\it 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 $\mathfrak p'$.
\item We say a $\varphi : R \to S$ satisfies {\it 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 $\mathfrak p$.
\end{enumerate}
\end{definition}
\noindent
So far we have see the following cases of this:
\begin{enumerate}
\item An integral ring map satisfies going up, see
Lemma \ref{lemma-integral-going-up}.
\item As a special case finite ring maps satisfy going up.
\item As a special case quotient maps $R \to R/I$ satisfy going up.
\item A flat ring map satisfies going down, see
Lemma \ref{lemma-flat-going-down}
\item As a special case any localization satisfies going down.
\item An extension $R \subset S$ of domains, with $R$ normal
and $S$ integral over $R$ satisfies going down, see
Proposition \ref{proposition-going-down-normal-integral}.
\end{enumerate}
Here is another case where going down holds.
\begin{lemma}
\label{lemma-open-going-down}
Let $R \to S$ be a ring map. If the induced map
$\varphi : \Spec(S) \to \Spec(R)$ is open, then
$R \to S$ satisfies going down.
\end{lemma}
\begin{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 $D(g) \subset \Spec(S)$. 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 \ref{lemma-localization-colimit} and
\ref{lemma-tensor-products-commute-with-limits}.
Hence $\mathfrak p$ is in the image of
$\Spec(S_{\mathfrak q'}) \to \Spec(R)$ as desired.
\end{proof}
\begin{lemma}
\label{lemma-going-up-down-specialization}
Let $R \to S$ be a ring map.
\begin{enumerate}
\item $R \to S$ satisfies going down if and only if
generalizations lift along the map $\Spec(S) \to \Spec(R)$,
see Topology, Definition \ref{topology-definition-lift-specializations}.
\item $R \to S$ satisfies going up if and only if
specializations lift along the map $\Spec(S) \to \Spec(R)$,
see Topology, Definition \ref{topology-definition-lift-specializations}.
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-going-up-down-composition}
Suppose $R \to S$ and $S \to T$ are ring maps satisfying
going down. Then so does $R \to T$. Similarly for going up.
\end{lemma}
\begin{proof}
According to Lemma \ref{lemma-going-up-down-specialization}
this follows from
Topology, Lemma \ref{topology-lemma-lift-specialization-composition}
\end{proof}
\begin{lemma}
\label{lemma-image-stable-specialization-closed}
Let $R \to S$ be a ring map. Let $T \subset \Spec(R)$
be the image of $\Spec(S)$. If $T$ is stable under specialization,
then $T$ is closed.
\end{lemma}
\begin{proof}
We give two proofs.
\medskip\noindent
First proof. Let $\mathfrak p \subset R$ be a prime ideal such that
the corresponding point of $\Spec(R)$ 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 $\Spec(S_f)$ in $\Spec(R)$. 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 limit
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 $\Spec(S_{\mathfrak p})
\to \Spec(S) \to \Spec(R)$ 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.
\medskip\noindent
Second proof. Let $I = \Ker(R \to S)$. We may replace $R$ by $R/I$.
In this case the ring map $R \to S$ is injective.
By Lemma \ref{lemma-injective-minimal-primes-in-image}
all the minimal primes of $R$ are contained in the image $T$. Hence
if $T$ is stable under specialization then it contains all primes.
\end{proof}
\begin{lemma}
\label{lemma-going-up-closed}
Let $R \to S$ be a ring map. The following are equivalent:
\begin{enumerate}
\item Going up holds for $R \to S$, and
\item the map $\Spec(S) \to \Spec(R)$ is closed.
\end{enumerate}
\end{lemma}
\begin{proof}
It is a general fact that specializations lift along a
closed map of topological spaces, see
Topology, Lemma \ref{topology-lemma-closed-open-map-specialization}.
Hence the second condition implies the first.
\medskip\noindent
Assume that going up holds for $R \to S$.
Let $V(I) \subset \Spec(S)$ be a closed set.
We want to show that the image of $V(I)$ in $\Spec(R)$ 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 \ref{lemma-going-up-down-composition}.
Replacing $S$ by $S/I$ it suffices to show the image $T$
of $\Spec(S)$ in $\Spec(R)$ is closed.
By Topology, Lemmas \ref{topology-lemma-open-closed-specialization}
and \ref{topology-lemma-lift-specializations-images} this
image is stable under specialization. Thus the result follows
from Lemma \ref{lemma-image-stable-specialization-closed}.
\end{proof}
\begin{lemma}
\label{lemma-constructible-stable-specialization-closed}
Let $R$ be a ring. Let $E \subset \Spec(R)$ be a constructible subset.
\begin{enumerate}
\item If $E$ is stable under specialization, then $E$ is closed.
\item If $E$ is stable under generalization, then $E$ is open.
\end{enumerate}
\end{lemma}
\begin{proof}
First proof. The first assertion
follows from Lemma \ref{lemma-image-stable-specialization-closed}
combined with Lemma \ref{lemma-constructible-is-image}.
The second follows because the complement of a constructible
set is constructible
(see Topology, Lemma \ref{topology-lemma-constructible}),
the first part of the lemma and Topology,
Lemma \ref{topology-lemma-open-closed-specialization}.
\medskip\noindent
Second proof. Since $\Spec(R)$ is a spectral space by
Lemma \ref{lemma-spec-spectral} this is a special case of
Topology, Lemma
\ref{topology-lemma-constructible-stable-specialization-closed}.
\end{proof}
\begin{proposition}
\label{proposition-fppf-open}
Let $R \to S$ be flat and of finite presentation.
Then $\Spec(S) \to \Spec(R)$ is open.
More generally this holds for any ring map $R \to S$ of
finite presentation which satisfies going down.
\end{proposition}
\begin{proof}
Assume that $R \to S$ has finite presentation and satisfies
going down.
It suffices to prove that the image of a standard open $D(f)$ is open.
Since $S \to S_f$ satisfies going down as well, we see that
$R \to S_f$ satisfies going down. Thus after replacing
$S$ by $S_f$ we see it suffices to prove the image is
open. By Chevalley's theorem
(Theorem \ref{theorem-chevalley})
the image is a constructible set $E$. And $E$ is stable
under generalization because $R \to S$ satisfies going down,
see Topology, Lemmas \ref{topology-lemma-open-closed-specialization}
and \ref{topology-lemma-lift-specializations-images}.
Hence $E$ is open by
Lemma \ref{lemma-constructible-stable-specialization-closed}.
\end{proof}
\begin{lemma}
\label{lemma-same-image}
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
$$
\xymatrix{
\Spec((S \otimes_k R)_f) \ar[rd] \ar[rr] & &
\Spec((S' \otimes_k R)_f) \ar[ld] \\
& \Spec(R) &
}
$$
the images of the diagonal arrows are the same.
\end{lemma}
\begin{proof}
Let $\mathfrak p \subset R$ be in the image of the south-west
arrow. This means (Lemma \ref{lemma-in-image}) 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 \ref{lemma-in-image})
we see that $\mathfrak p$ is in the image of the south-east arrow
as desired.
\end{proof}
\begin{lemma}
\label{lemma-map-into-tensor-algebra-open}
Let $k$ be a field.
Let $R$ and $S$ be $k$-algebras.
The map $\Spec(S \otimes_k R) \to \Spec(R)$
is open.
\end{lemma}
\begin{proof}
Let $f \in R \otimes_k S$.
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
$\Spec((S' \otimes_k R)_f) \to \Spec(R)$ is open
by Proposition \ref{proposition-fppf-open}.
By Lemma \ref{lemma-same-image} this is also the image of $D(f)$ and we win.
\end{proof}
\noindent
Here is a tricky lemma that is sometimes useful.
\begin{lemma}
\label{lemma-unique-prime-over-localize-below}
Let $R \to S$ be a ring map.
Let $\mathfrak p \subset R$ be a prime.
Assume that
\begin{enumerate}
\item there exists a unique prime $\mathfrak q \subset S$ lying over
$\mathfrak p$, and
\item either
\begin{enumerate}
\item going up holds for $R \to S$, or
\item going down holds for $R \to S$ and there is at most one prime
of $S$ above every prime of $R$.
\end{enumerate}
\end{enumerate}
Then $S_{\mathfrak p} = S_{\mathfrak q}$.
\end{lemma}
\begin{proof}
Consider any prime $\mathfrak q' \subset S$ which corresponds to
a point of $\Spec(S_{\mathfrak p})$. 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 $\Spec(S_{\mathfrak q})$.
\medskip\noindent
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 $\Spec(S_{\mathfrak q})$.
\medskip\noindent
In both cases we conclude that the map
$\Spec(S_{\mathfrak q}) \to \Spec(S_{\mathfrak p})$
is bijective. Clearly this means all the elements of $S - \mathfrak q$
are all invertible in $S_{\mathfrak p}$, in other words
$S_{\mathfrak p} = S_{\mathfrak q}$.
\end{proof}
\noindent
The following lemma is a generalization of going down for
flat ring maps.
\begin{lemma}
\label{lemma-going-down-flat-module}
Let $R \to S$ be a ring map. Let $N$ be a finite $S$-module flat over $R$.
Endow $\text{Supp}(N) \subset \Spec(S)$ with the induced topology.
Then generalizations lift along $\text{Supp}(N) \to \Spec(R)$.
\end{lemma}
\begin{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 \ref{lemma-flat-localization}.
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 \ref{lemma-NAK}.
Thus $N_{\mathfrak q'} \otimes_{R_{\mathfrak p'}} \kappa(\mathfrak p')$
is also not zero. We conclude from Lemma \ref{lemma-ff}
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.
\end{proof}
```

## Comments (0)

## Add a comment on tag `00HU`

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner).

All contributions are licensed under the GNU Free Documentation License.

There are no comments yet for this tag.