Exercise 1.18

Question

For psychological reasons it is sometimes convenient to denote a prime ideal of $A$ by a letter such as $x$ or $y$ when thinking of it as a point of $X = \operatorname{Spec}(A)$. When thinking of $x$ as a prime ideal of $A$, we denote it by $\mathfrak{p}_x$ (logically, of course, it is the same thing). Show that
  \begin{enumerate}[label=\roman*)]
    \item the set $\{x\}$ is closed (we say that $x$ is a ``closed point'') in $\operatorname{Spec}(A) \Leftrightarrow \mathfrak{p}_x$ is maximal;
    \item $\overline{\{x\}} = V(\mathfrak{p}_x)$;
    \item $y \in \overline{\{x\}} \Leftrightarrow \mathfrak{p}_x \subseteq \mathfrak{p}_y$;
    \item $X$ is a $T_0$-space (this means that if $x,y$ are distinct points of $X$, then either there is a neighborhood of $x$ which does not contain $y$, or else there is a neighborhood of $y$ which does not contain $x$).
  \end{enumerate}

Amit Awasthi · Accepted Answer

\begin{proof}[Proof of $i)$]
  $\Leftarrow$. If $\mathfrak{p}_x$ is maximal, $V(\mathfrak{p}_x) = \{x\}$, and so $\{x\}$ is closed.
  \par $\Rightarrow$. If $\{x\}$ is closed, $\{x\} = V(\mathfrak{a})$ for some ideal $\mathfrak{a} \subsetneq A$ by Exercise $15i)$. By Cor.~1.4, there exists a maximal ideal $\mathfrak{m}$ containing $\mathfrak{a}$, and so $\mathfrak{m} \in V(\mathfrak{a}) = \{x\}$, and so $\mathfrak{p}_x = \mathfrak{m}$ is maximal.
\end{proof}
\begin{proof}[Proof of $ii)$]
  $\overline{\{x\}} \subseteq V(\mathfrak{p}_x)$ since $V(\mathfrak{p}_x) 
i \mathfrak{p}_x$. Conversely let $\mathfrak{p}_y \in V(\mathfrak{p}_x)$; we want to show that every $X_f 
i y$ intersects $\{x\}$. Suppose not; then there exists a basis element $X_f$ containing $y$ that does not contain $x$, and so $X \setminus X_f = V(f) 
i x$ but $V(f) 
ot
i y$. In particular, $\mathfrak{p}_x 
ot\subseteq \mathfrak{p}_y$, and so $\mathfrak{p}_y 
otin V(\mathfrak{p}_x)$, a contradiction.
\end{proof}
\begin{proof}[Proof of $iii)$]
  By $ii)$, $y \in \overline{\{x\}}$ if and only if $y \in V(\mathfrak{p}_x)$ if and only if $\mathfrak{p}_x \subseteq \mathfrak{p}_y$.
\end{proof}
\begin{proof}[Proof of $iv)$]
  Let $x,y\in X$ be distinct; then, say, $\mathfrak{p}_x 
ot\subseteq \mathfrak{p}_y$. So there exists $f \in \mathfrak{p}_x \setminus \mathfrak{p}_y$, and $\mathfrak{p}_y \in X_f$ but $\mathfrak{p}_x 
otin X_f$.
\end{proof}

Exercise 1.18

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Exercise 1.18

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