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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.

  1. 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 $\mathfrak p'$.
  2. 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:

  1. An integral ring map satisfies going up, see Lemma 10.35.22.
  2. As a special case finite ring maps satisfy going up.
  3. As a special case quotient maps $R \to R/I$ satisfy going up.
  4. A flat ring map satisfies going down, see Lemma 10.38.18
  5. As a special case any localization satisfies going down.
  6. 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.

  1. $R \to S$ 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.
  2. $R \to S$ 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:

  1. Going up holds for $R \to S$, and
  2. 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.

  1. If $E$ is stable under specialization, then $E$ is closed.
  2. 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

  1. there exists a unique prime $\mathfrak q \subset S$ lying over $\mathfrak p$, and
  2. either
    1. going up holds for $R \to S$, or
    2. 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 9143–9527 (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}

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