Exercise 3.21

Question

\begin{enumerate}[label=oman*)]
  \item Let $A$ be a ring, $S$ a multiplicatively closed subset of $A$, and $\phi : A 	o S^{-1}A$ the canonical homomorphism. Show that $\phi^* : \operatorname{Spec}(S^{-1}A) 	o \operatorname{Spec}(A)$ is a homeomorphism of $\operatorname{Spec}(S^{-1}A)$ onto its image in $X = \operatorname{Spec}(A)$. Let this image be denoted by $S^{-1}X$.
    \par In particular, if $f \in A$, the image of $\operatorname{Spec}(A_f)$ in $X$ is the basic open set $X_f$ (Chapter $1$, Exercise $17$).
  \item Let $f : A 	o B$ be a ring homomorphism. Let $X = \operatorname{Spec}(A)$ and $Y = \operatorname{Spec}(B)$, and let $f^* : Y 	o X$ be the mapping associated with $f$. Identifying $\operatorname{Spec}(S^{-1}A)$ with its canonical image $S^{-1}X$ in $X$, and $\operatorname{Spec}(S^{-1}B) (=\operatorname{Spec}(f(S)^{-1}B))$ with its canonical image $S^{-1}Y$ in $Y$, show that $S^{-1}f^* : \operatorname{Spec}(S^{-1}B) 	o \operatorname{Spec}(S^{-1}A)$ is the restriction of $f^*$ to $S^{-1}Y$, and that $S^{-1}Y = f^{*-1}(S^{-1}X)$.
  \item Let $\mathfrak{a}$ be an ideal of $A$ and let $\mathfrak{b} = \mathfrak{a}^e$ be its extension in $B$. Let $\overline{f} : A/\mathfrak{a} 	o B/\mathfrak{b}$ be the homomorphism induced by $f$. If $\operatorname{Spec}(A/\mathfrak{a})$ is identified with its canonical image $V(\mathfrak{a})$ in $X$, and $\operatorname{Spec}(B/\mathfrak{b})$ with its image $V(\mathfrak{b})$ in $Y$, show that $\overline{f}^*$ is the restriction of $f^*$ to $V(\mathfrak{b})$.
  \item Let $\mathfrak{p}$ be a prime ideal of $A$. Take $S = A - \mathfrak{p}$ in $ii)$ and then reduce mod $S^{-1}\mathfrak{p}$ as in $iii)$. Deduce that the subspace $f^{*-1}(\mathfrak{p})$ of $Y$ is naturally homeomorphic to $\operatorname{Spec}(B_{\mathfrak{p}}/\mathfrak{p}B_{\mathfrak{p}}) = \operatorname{Spec}(k(\mathfrak{p})\otimes_A B)$, where $k(\mathfrak{p})$ is the residue field of the local ring $A_{\mathfrak{p}}$.
    \par $\operatorname{Spec}(k(\mathfrak{p}) \otimes_A B)$ is called the \emph{fiber} of $f^*$ over $\mathfrak{p}$.
  \end{enumerate}

Amit Awasthi · Accepted Answer

\begin{proof}[Proof of $i)$]
  $\phi^*$ is continuous by AM Exercise $1.21i)$. Since every prime ideal of $S^{-1}A$ is an extended ideal by AM Proposition $3.11i)$, we see that $\phi^*$ is injective by AM Exercise $3.20ii)$, and therefore bijective onto its image. We consider $S^{-1}X = \operatorname{im}(\phi^*)$; this is the set of prime ideals that do not meet $S$ by Proposition $3.11iv)$. It, therefore, remains to show that $\phi^*$ is closed. Since any arbitrary ideal of $S^{-1}A$ is an extended ideal by AM Proposition $3.11i)$, we only have to consider basis elements of the form $\mathbb{V}(\mathfrak{a}^e) \subseteq \operatorname{Spec}(S^{-1}A)$ for $\mathfrak{a} \in A$. We claim $\phi^*(\mathbb{V}(\mathfrak{a}^e)) = S^{-1}X \cap \mathbb{V}(\mathfrak{a}^{ec})$ (note the latter is closed in $S^{-1}X$ since $\mathbb{V}(\mathfrak{a}^{ec})$ is closed in $X$). If $\mathfrak{p} \in \phi^*(\mathbb{V}(\mathfrak{a}^e))$, then $\mathfrak{p} \cap S = \emptyset$ and $\mathfrak{a}^e \subseteq \mathfrak{p}^e \implies \mathfrak{a}^{ec} \subseteq \mathfrak{p}^{ec} = \mathfrak{p} \implies \mathfrak{p} \in S^{-1}X \cap \mathbb{V}(\mathfrak{a}^{ec})$. On the other hand if $\mathfrak{p} \in S^{-1}X \cap \mathbb{V}(\mathfrak{a}^{ec})$, then $\mathfrak{p} \cap S = \emptyset$ and $\mathfrak{a}^{ec} \subseteq \mathfrak{p} \implies \mathfrak{a}^e \subseteq \mathfrak{p}^e$. Thus $\phi$ is a homeomorphism onto its image.
  \par In particular, if $f \in A$, $\phi^*(\operatorname{Spec}(A_f)) = X_f$ by having $S = \braket{f}$.
\end{proof}
\begin{proof}[Proof of $ii)$]
  We first claim the diagram, with $\varphi_A : A \hookrightarrow S^{-1}A, \varphi_B : B \hookrightarrow S^{-1}B$ the natural embedding maps,
  \begin{center}
    \begin{tikzcd}
      A \arrow{r}{f}\arrow{d}[swap]{\varphi_A} & B \arrow{d}{\varphi_B}\
      S^{-1}A \arrow{r}{S^{-1}f} & S^{-1}B
    \end{tikzcd}
  \end{center}
  commutes. But this is clear since $S^{-1}B \simeq f(S)^{-1}B$ by AM Exercise $3.4$, and since $S^{-1}f(a/s) = f(a)/f(s)$. Moreover, calculating explicitly, if $a \in A$, $(\varphi_B \circ f) (a) = \varphi_B (f(a)) = f(a)/1$, while $(S^{-1}f \circ \varphi_A)(a) = S^{-1}f (a/1) = f(a)/1$. Since AM Exercise $1.21$ implies $f^* \circ \varphi_B^* = (\varphi_B \circ f)^* = (S^{-1}f \circ \varphi_A)^* = \varphi_A^* \circ (S^{-1}f)^*$, we have the commutative diagram
  \begin{center}
    \begin{tikzcd}
      \operatorname{Spec}(S^{-1}B) = S^{-1}Y \arrow{r}{(S^{-1}f)^*}\arrow{d}[swap]{\varphi_B^*} & \operatorname{Spec}(S^{-1}A) = S^{-1}X \arrow{d}{\varphi_A^*}\
      \operatorname{Spec}(B) = Y \arrow{r}{f^*} & \operatorname{Spec}(A) = X
    \end{tikzcd}
  \end{center}
  where the identification is by $i)$, which shows the compatibility of $(S^{-1}f)^*$ and $f^*$.
  \par We now show $S^{-1}Y = f^{*-1}(S^{-1}X)$. The diagram above shows that $f^*(S^{-1}Y) \subseteq S^{-1}X$, and so $S^{-1}Y \subseteq f^{*-1}(S^{-1}X)$. On the other hand, suppose $\mathfrak{p} \in f^{*-1}(S^{-1}X)$. Then, $\mathfrak{p}^c = f^*(\mathfrak{p}) \in S^{-1}X$, so $\mathfrak{p}^c \cap S = \emptyset$ by AM Proposition $3.11iv)$. To show $\mathfrak{p} \in S^{-1}Y$, it suffices to show $\mathfrak{p} \cap f(S) = \emptyset$. So, suppose $x \in \mathfrak{p} \cap f(S)$; then, $x = f(s)$ for some $s \in S \cap \mathfrak{p}^c = \emptyset$, which is a contradiction. Thus, $S^{-1}Y \supseteq f^{*-1}(S^{-1}X)$, and $S^{-1}Y = f^{*-1}(S^{-1}X)$.
\end{proof}
\begin{proof}[Proof of $iii)$]
  Letting $\pi_A : A 	woheadrightarrow A/\mathfrak{a},\pi_B : B 	woheadrightarrow B/\mathfrak{b}$ be the natural quotient maps, we have the commutative diagram
  \begin{center}
    \begin{tikzcd}
      A \arrow{r}{f}\arrow{d}[swap]{\pi_A} & B \arrow{d}{\pi_B}\
      A/\mathfrak{a} \arrow{r}{\overline{f}} & B/\mathfrak{b}
    \end{tikzcd}
  \end{center}
  and by the same argument as in $iv)$, we get
  \begin{center}
    \begin{tikzcd}
      \operatorname{Spec}(B/\mathfrak{b}) = \mathbb{V}(\mathfrak{b}) \arrow{r}{\overline{f}^*}\arrow{d}[swap]{\pi_B^*} & \operatorname{Spec}(A/\mathfrak{a}) = \mathbb{V}(\mathfrak{a}) \arrow{d}{\pi_A^*}\
      \operatorname{Spec}(B) = Y \arrow{r}{f^*} & \operatorname{Spec}(A) = X
    \end{tikzcd}
  \end{center}
  The identification comes from AM Exercise $1.21iv)$, which says $\pi_B^*$ is a homeomorphism $\operatorname{Spec}(B/\mathfrak{b}) 	o \mathbb{V}(\ker(\pi_B)) = \mathbb{V}(\mathfrak{b})$ and $\pi_A^*$ is a homeomorphism $\operatorname{Spec}(A/\mathfrak{a}) 	o \mathbb{V}(\ker(\pi_A)) = \mathbb{V}(\mathfrak{a})$. Thus, $\overline{f}^*$ and $f^*$ are compatible.
  %\par Much like in $ii)$, we also show $\mathbb{V}(\mathfrak{b}) = f^{*-1}(\mathbb{V}(\mathfrak{a}))$. The diagram above shows that $f^*(\mathbb{V}(\mathfrak{b})) \subseteq \mathbb{V}(\mathfrak{a})$, and so $\mathbb{V}(\mathfrak{b}) \subseteq f^{*-1}(\mathbb{V}(\mathfrak{a}))$. On the other hand, suppose $\mathfrak{p} \in f^{*-1}(\mathbb{V}(\mathfrak{a}))$. Then, $\mathfrak{p}^c = f^{*-1}(\mathfrak{p}) \in \mathbb{V}(\mathfrak{a})$, and so $\mathfrak{p}^c \supseteq \mathfrak{a}$. Taking the extension of both, we get $\mathfrak{p} = \mathfrak{p}^{ce} \supseteq \mathfrak{a}^e = \mathfrak{b}$, and so $\mathfrak{p}^c \in \mathbb{V}(\mathfrak{b})$. Thus, $f^{*-1}(\mathbb{V}(\mathfrak{a})) \subseteq \mathbb{V}(\mathfrak{b})$, and $\mathbb{V}(\mathfrak{b}) = f^{*-1}(\mathbb{V}(\mathfrak{a}))$.
\end{proof}
\begin{proof}[Proof of $iv)$]
  Following the steps given, we get the commutative diagram
  \begin{center}
    \begin{tikzcd}
      \operatorname{Spec}(B_{\mathfrak{p}}/\mathfrak{p}B_{\mathfrak{p}}) \arrow{r}{\overline{f_{\mathfrak{p}}}^*}\arrow{d}[swap]{\pi_B^*} & \operatorname{Spec}(A_{\mathfrak{p}}/\mathfrak{p}_{\mathfrak{p}})\arrow{d}{\pi_A^*}\
      \operatorname{Spec}(B_{\mathfrak{p}}) \arrow{r}{f_{\mathfrak{p}}^*}\arrow{d}[swap]{\varphi_B^*} & \operatorname{Spec}(A_{\mathfrak{p}}) \arrow{d}{\varphi_A^*}\
      \operatorname{Spec}(B) \arrow{r}{f^*} & \operatorname{Spec}(A)
    \end{tikzcd}
  \end{center}
  By $iii)$, $\operatorname{Spec}(B_{\mathfrak{p}}/\mathfrak{p}B_{\mathfrak{p}})$ is homeomorphic to $\mathbb{V}(\mathfrak{p}B_{\mathfrak{p}})$. By $ii)$, $\mathbb{V}(\mathfrak{p}B_{\mathfrak{p}})$ is homeomorphic to $\varphi_B^*(\mathbb{V}(\mathfrak{p}B_{\mathfrak{p}}))$. We now claim that $\varphi_B^*(\mathbb{V}(\mathfrak{p}B_{\mathfrak{p}})) = f^{*-1}(\mathfrak{p})$. Suppose $\mathfrak{q} \in f^{*-1}(\mathfrak{p})$, which gives $\mathfrak{p} \in \operatorname{im}(\varphi_A^*) \implies \mathfrak{q} \in \operatorname{im}(\varphi_B^*)$. Then, since $\mathfrak{p} = f^{-1}(\mathfrak{q})$, i.e., $f(\mathfrak{p}) \subseteq \mathfrak{q}$, i.e., $\mathfrak{p}B \subseteq \mathfrak{q}$. Thus, $\mathfrak{q}_{\mathfrak{p}}$ is prime in $B_{\mathfrak{p}}$ and contains $\mathfrak{p}B_{\mathfrak{p}}$, i.e., $f^{*-1}(\mathfrak{p}) \subseteq \varphi_B^*(\mathbb{V}(\mathfrak{p}B_{\mathfrak{p}}))$. In the other direction, suppose $\mathfrak{q} \in \varphi_B^*(\mathbb{V}(\mathfrak{p}B_{\mathfrak{p}}))$. Then, $\mathfrak{p}B_{\mathfrak{p}} \subseteq \mathfrak{q}_{\mathfrak{p}}$ so $\mathfrak{p}B \subseteq \mathfrak{q}_{\mathfrak{p}}^c = \mathfrak{q}$, i.e., $f(\mathfrak{p}) \subseteq \mathfrak{q}$. So, $\mathfrak{p} \subseteq f^{-1}(\mathfrak{q})$. On the other hand, $f^{-1}(\mathfrak{q}) \subseteq \mathfrak{p}$ since $\mathfrak{q} \cap f(A \setminus \mathfrak{p}) = \emptyset$ by choice of $\mathfrak{q}$. Thus, $\mathfrak{p} = f^{-1}(\mathfrak{q})$, and so $\mathfrak{q} \in f^{*-1}(\mathfrak{p})$, i.e., $\varphi_B^*(\mathbb{V}(\mathfrak{p}B_{\mathfrak{p}})) = f^{*-1}(\mathfrak{p})$.
  \par We now want to show the isomorphism given. We have $B_{\mathfrak{p}}/\mathfrak{p}B_{\mathfrak{p}} \simeq B_{\mathfrak{p}}/(\mathfrak{p}B)_{\mathfrak{p}} \simeq (B/\mathfrak{p}B)_{\mathfrak{p}}$ by AM Proposition $3.11v)$. Then, $(B/\mathfrak{p}B)_{\mathfrak{p}} \simeq A_{\mathfrak{p}} \otimes_A B/\mathfrak{p}B$ by Proposition $3.5$. AM Exercise $1.2$ gives $A_{\mathfrak{p}} \otimes_A B/\mathfrak{p}B \simeq A_{\mathfrak{p}} \otimes_A (A/\mathfrak{p} \otimes_A B)$. The associativity of the tensor product from AM Proposition $2.14ii)$ then gives $A_{\mathfrak{p}} \otimes_A (A/\mathfrak{p} \otimes_A B) \simeq (A/\mathfrak{p} \otimes_A A_{\mathfrak{p}}) \otimes_A B$. Applying AM Exercise $1.2$ again yields $(A/\mathfrak{p} \otimes_A A_{\mathfrak{p}}) \otimes_A B \simeq A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}} \otimes_A B$. But then $A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}} = A_{\mathfrak{p}}/\mathfrak{p}_{\mathfrak{p}} = k(\mathfrak{p})$, since $A_{\mathfrak{p}}$ is local and therefore $\mathfrak{p}_{\mathfrak{p}}$ can only be the maximal ideal, and so we get the isomorphism $B_{\mathfrak{p}}/\mathfrak{p}B_{\mathfrak{p}} \simeq k(\mathfrak{p}) \otimes_A B$. We note that this also preserves ring structure since this is just the map $b/f(x) + \mathfrak{p}B_{\mathfrak{p}} \leftrightarrow (1/x + \mathfrak{p}_{\mathfrak{p}}) \otimes b$.
\end{proof}

Exercise 3.21

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Exercise 3.21

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