Exercise 1.15

Question

Let $A$ be a ring and let $X$ be the set of all prime ideals of $A$. For each subset $E$ of $A$, let $V(E)$ denote the set of all prime ideals of $A$ which contain $E$. Prove that
  \begin{enumerate}[label=\roman*)]
    \item if $\mathfrak{a}$ is the ideal generated by $E$, then $V(E) = V(\mathfrak{a}) = V(r(\mathfrak{a}))$.
    \item $V(0) = X$, $V(1) = \emptyset$.
    \item if $(E_i)_{i \in I}$ is any family of subsets of $A$, then
      \begin{equation*}
        V\left( \bigcup_{i\in I} E_i \right) = \bigcap_{i \in I} V(E_i).
      \end{equation*}
    \item $V(\mathfrak{a} \cap \mathfrak{b}) = V(\mathfrak{a}\mathfrak{b}) = V(\mathfrak{a}) \cup V(\mathfrak{b})$ for any ideals $\mathfrak{a},\mathfrak{b}$ of $A$.
  \end{enumerate}
  \par These results show that the sets $V(E)$ satisfy the axioms for closed sets in a topological space. The resulting topology is called the \emph{Zariski topology}. The topological space $X$ is called the \emph{prime spectrum} of $A$, and is written $\operatorname{Spec}(A)$.

Amit Awasthi · Accepted Answer

\begin{proof}[Proof of $i)$]
  We have $V(E) \supseteq V(\mathfrak{a}) \supseteq V(r(\mathfrak{a}))$, so it suffices to show $V(E) \subseteq V(r(\mathfrak{a}))$ to get equalities throughout. Suppose $\mathfrak{p} \in V(E)$, so $E \subseteq \mathfrak{p}$. Then, $\mathfrak{a} = AE \subseteq A\mathfrak{p} = \mathfrak{p}$, and $r(\mathfrak{a}) \subseteq r(\mathfrak{p}) = \mathfrak{p}$ by Prop.~1.14, so $V(E) \subseteq V(r(\mathfrak{a}))$.
\end{proof}
\begin{proof}[Proof of $ii)$]
  $0$ is contained in every prime ideal of $A$, so $V(0) = X$. No prime ideal can contain $1$, so $V(1) = \emptyset$.
\end{proof}
\begin{proof}[Proof of $iii)$]
  A prime ideal contains $\bigcup_{i \in I} E_i$ if and only if it contains each $E_i$.
\end{proof}
\begin{proof}[Proof of $iv)$]
  By $i)$ and Exercise $1.13iii)$ in the text, we have $V(\mathfrak{a} \cap \mathfrak{b}) = V(r(\mathfrak{a} \cap \mathfrak{b})) = V(r(\mathfrak{a}\mathfrak{b})) = V(\mathfrak{a}\mathfrak{b})$. Now clearly $V(\mathfrak{a}\mathfrak{b}) \subseteq V(\mathfrak{a}) \cup V(\mathfrak{b})$ since if $\mathfrak{a}$ or $\mathfrak{b}$ is contained in $\mathfrak{p}$, then $\mathfrak{a} \cap \mathfrak{b} \subseteq \mathfrak{p}$. The converse holds by Prop.~$1.11ii)$.
\end{proof}

Exercise 1.15

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Exercise 1.15

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