Exercise 1.21

Question

Let $\varphi\colon A 	o B$ be a ring homomorphism. Let $X = \operatorname{Spec}(A)$ and $Y = \operatorname{Spec}(B)$. If $\mathfrak{q} \in Y$, then $\varphi^{-1}(\mathfrak{q})$ is a prime ideal of $A$, i.e., a point of $X$. Hence $\varphi$ induces a mapping $\varphi^*\colon Y 	o X$. Show that
  \begin{enumerate}[label=oman*)]
    \item If $f \in A$ then $\varphi^{*-1}(X_f) = Y_{\varphi(f)}$, hence that $\varphi^*$ is continuous.
    \item If $\mathfrak{a}$ is an ideal of $A$, then $\varphi^{*-1}(V(\mathfrak{a})) = V(\mathfrak{a}^e)$.
    \item If $\mathfrak{b}$ is an ideal of $B$, then $\overline{\varphi^*( V(\mathfrak{b}))} = V(\mathfrak{b}^c)$.
    \item If $\varphi$ is surjective, then $\varphi^*$ is a homeomorphism of $Y$ onto the closed subset $V( \operatorname{Ker}(\varphi) )$ of $X$. (In particular, $\operatorname{Spec}(A)$ and $\operatorname{Spec}(A/\mathfrak{N})$ (where $\mathfrak{N}$ is the nilradical of $A$) are naturally homeomorphic.)
    \item If $\varphi$ is injective, then $\varphi^*(Y)$ is dense in $X$. More precisely, $\varphi^*(Y)$ is dense in $X \Leftrightarrow \operatorname{Ker}(\varphi) \subseteq \mathfrak{N}$.
    \item Let $\psi\colon B 	o C$ be another ring homomorphism. Then $(\psi \circ \varphi)^* = \varphi^* \circ \psi^*$.
    \item Let $A$ be an integral domain with just one non-zero prime ideal $\mathfrak{p}$, and let $K$ be the field of fractions of $A$. Let $B = (A/\mathfrak{p}) 	imes K$. Define $\varphi\colon A 	o B$ by $\varphi(x) = (\overline{x},x)$, where $\overline{x}$ is the image of $x$ in $A/\mathfrak{p}$. Show that $\varphi^*$ is bijective but not a homeomorphism.
  \end{enumerate}

Amit Awasthi · Accepted Answer

\begin{proof}[Proof of $i)$]
  $\varphi^{*-1}(X_f) = \{\mathfrak{q} \in Y \mid f 
otin \mathfrak{q}^c\} = \{\mathfrak{q} \in Y \mid \varphi(f) 
otin \mathfrak{q}\} = Y_{\varphi(f)}$, so $\varphi^*$ is continuous since the inverse images of basis elements are basis elements.
\end{proof}
\begin{proof}[Proof of $ii)$]
  Using $i)$ and Exercise $15iii)$, we have
  \begin{align*}
    \varphi^{*-1}\left(V(\mathfrak{a})ight) &= \varphi^{*-1}\left( V\left( \bigcup_{x \in \mathfrak{a}} \{x\} ight) ight) = \varphi^{*-1}\left( \bigcap_{x \in \mathfrak{a}}V(x) ight)\
    &= \bigcap_{x \in \mathfrak{a}}\varphi^{*-1}\left( V(x) ight) = \bigcap_{x \in \mathfrak{a}}\varphi^{*-1}\left( X \setminus X_x ight) = \bigcap_{x \in \mathfrak{a}}\left(Y \setminus \varphi^{*-1}\left(X_xight)ight)\
    &= \bigcap_{x \in \mathfrak{a}}\left(Y \setminus Y_{\varphi(x)}ight) = \bigcap_{x \in \mathfrak{a}} V(\varphi(x)) = V(\mathfrak{a}^e).\qedhere
  \end{align*}
\end{proof}
\begin{proof}[Proof of $iii)$]
  First, we have $\overline{\varphi^*\left( V(\mathfrak{b}) ight)} \subseteq V(\mathfrak{b}^c)$ since
  \begin{equation*}
    \varphi^*( V(\mathfrak{b})) = \{\mathfrak{q}^c \in X \mid \mathfrak{q} \in Y,\ \mathfrak{b} \subseteq \mathfrak{q}\} \subseteq \{\mathfrak{p} \in X \mid \mathfrak{b}^c \subseteq \mathfrak{p}\} = V(\mathfrak{b}^c).
  \end{equation*}
  Conversely, suppose $\mathfrak{p} \in V(\mathfrak{b}^c)$; to show $\mathfrak{p} \in \overline{\varphi^*\left( V(\mathfrak{b}) ight)}$, it suffices to show every $X_f 
i \mathfrak{p}$ intersects $\varphi^*\left( V(\mathfrak{b}) ight)$. Now $X_f 
i \mathfrak{p}$ implies $f 
otin \mathfrak{p}$, hence $f 
otin r(\mathfrak{b}^c) = r(\mathfrak{b})^c$ by Exercise $1.18$ in the text. Thus, $\varphi(f) 
otin r(\mathfrak{b})$, and so there exists $\mathfrak{q} \in V(\mathfrak{b})$ such that $\varphi(f) 
otin \mathfrak{q}$ by Prop.~1.14. Then, $f 
otin \mathfrak{q}^c$, and so $\mathfrak{q}^c \in X_f$, i.e., $\varphi^*(V(\mathfrak{b})) \cap X_f 
e \emptyset$.
\end{proof}
\begin{proof}[Proof of $iv)$]
  If $\varphi$ is surjective, we have a bijection between primes containing $\operatorname{Ker}\varphi$ in $A$ and primes in $B$ by Prop.~1.1. We claim this induces a homeomorphism $\varphi^*\colon Y 	o V(\operatorname{Ker}\varphi)$; since it is continuous by $i)$, it remains to show it is closed. By $iii)$, we know $\varphi^*(V(\mathfrak{b})) \subseteq V(\mathfrak{b}^c)$ for any ideal $\mathfrak{b} \subseteq B$. If $\mathfrak{p} \in V(\mathfrak{b}^c)$ then $\varphi(\mathfrak{p}) \supseteq \varphi(\mathfrak{b}^c) = \mathfrak{b}$ by surjectivity of $\varphi$, and so $\varphi(\mathfrak{p}) \in V(\mathfrak{b})$. But then $\mathfrak{p} = \varphi^*(\varphi(\mathfrak{p})) \in \varphi^*(V(\mathfrak{b}))$.
  \par Finally, $A 	o A/\mathfrak{N}$ is surjective, hence $\operatorname{Spec}(A/\mathfrak{N})$ is homeomorphic to $V(\mathfrak{N})$. Since $V(\mathfrak{N})=\operatorname{Spec}(A)$ by Prop.~1.1 and Prop.~1.8, we have that $\operatorname{Spec}(A)$ and $\operatorname{Spec}(A/\mathfrak{N})$ are homeomorphic.
\end{proof}
\begin{proof}[Proof of $v)$]
  We see by $iii)$ that $X \supseteq \overline{\varphi^*(Y)} = \overline{\varphi^*(V(0))} = V(0^c) = V(\operatorname{Ker}(\varphi))$, so $\varphi^*(Y)$ dense in $X$ if and only if $V(\operatorname{Ker}(\varphi)) = X$ if and only if $\operatorname{Ker}(\varphi) \subseteq \mathfrak{N}$ by Prop.~1.8. In particular, if $\varphi$ injective, $\operatorname{Ker}(\varphi) = 0 \subseteq \mathfrak{N}$, so $\varphi^*(Y)$ is dense in $X$.
\end{proof}
\begin{proof}[Proof of $vi)$]
  $(\psi \circ \varphi)^*(\mathfrak{r}) = (\psi \circ \varphi)^{-1}(\mathfrak{r}) = \varphi^{-1}(\psi^{-1}(\mathfrak{r})) = \varphi^*(\psi^*(\mathfrak{r}))$ for every $\mathfrak{r} \in \operatorname{Spec}(C)$, hence $(\psi \circ \varphi)^* = \varphi^* \circ \psi^*$.
\end{proof}
\begin{proof}[Proof of $vii)$]
  $A/\mathfrak{p}$ is a field since $\mathfrak{p}$ is maximal in $A$. The ideals $\mathfrak{q}_1 = \{(\overline{x},0) \in B\}$ and $\mathfrak{q}_2 = \{(0,x) \in B\}$ are maximal, since $B/\mathfrak{q}_1 \cong K$ and $B/\mathfrak{q}_2 \cong A/\mathfrak{p}$. If $\mathfrak{q}$ is another prime of $B$, then $\mathfrak{q}_1\mathfrak{q}_2 = 0 \subseteq \mathfrak{q}$, hence $\mathfrak{q}_1 \subseteq \mathfrak{q}$ or $\mathfrak{q}_2 \subseteq \mathfrak{q}$, which implies $\mathfrak{q} = \mathfrak{q}_1$ or $\mathfrak{q}_2$ by maximality, and so $\operatorname{Spec}(B) = \{\mathfrak{q}_1,\mathfrak{q}_2\}$. Since $\varphi^*(\mathfrak{q}_1) = 0$ and $\varphi^*(\mathfrak{q}_2) = \mathfrak{p}$, $\varphi^*$ is bijective. But $\varphi^*$ is not a homeomorphism since $\operatorname{Spec}(A)$ has a non-closed point while $\operatorname{Spec}(B)$ does not. 
\end{proof}

Exercise 1.21

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Exercise 1.21

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