Exercise 1.26

Question

Let $A$ be a ring. The subspace of $\operatorname{Spec}(A)$ consisting of the \emph{maximal} ideals of $A$, with the induced topology, is called the \emph{maximal spectrum} of $A$ and is denoted by $\operatorname{Max}(A)$. For arbitrary commutative rings it does not have the nice functorial properties of $\operatorname{Spec}(A)$ (see Exercise 21), because the inverse image of a maximal ideal under a ring homomorphism need not be maximal.
  \par Let $X$ be a compact Hausdorff space and let $C(X)$ denote the ring of all real-valued continuous functions on $X$ (add and multiply functions by adding and multiplying their values). For each $x \in X$, let $\mathfrak{m}_x$ be the set of all $f \in C(X)$ such that $f(x) = 0$. The ideal $\mathfrak{m}_x$ is maxmal, because it is the kernel of the (surjective) homomorphism $C(X) 	o \mathbb{R}$ which takes $f$ to $f(x)$. If $	ilde{X}$ denotes $\operatorname{Max}(C(X))$, we have therefore defined a mapping $\mu\colon X 	o 	ilde{X}$, namely $x \mapsto \mathfrak{m}_x$.
  \par We shall show that $\mu$ is a homeomorphism of $X$ onto $	ilde{X}$.
  \begin{enumerate}[label=oman*)]
    \item Let $\mathfrak{m}$ be any maximal ideal of $C(X)$, and let $V = V(\mathfrak{m})$ be the set of common zeros of the functions in $\mathfrak{m}$: that is,
      \begin{equation*}
        V = \{x \in X : f(x) = 0~	ext{for all}~f \in \mathfrak{m}\}.
      \end{equation*}
      Suppose that $V$ is empty. Then for each $x \in X$ there exists $f_x \in \mathfrak{m}$ such that $f_x(x) 
e 0$. Since $f_x$ is continuous, there is an open neighborhood $U_x$ of $x$ in $X$ on which $f_x$ does not vanish. By compactness a finite number of neighborhoods, say $U_{x_1},\ldots,U_{x_n}$, cover $X$. Let
      \begin{equation*}
        f = f_{x_1}^2 + \cdots + f_{x_n}^2.
      \end{equation*}
      Then $f$ does not vanish at any point of $X$, hence is a unit in $C(X)$. But this contradicts $f \in \mathfrak{m}$, hence $V$ is not empty.
      \par \hspace{1.25em} Let $x$ be a point of $V$. Then $\mathfrak{m} \subseteq \mathfrak{m}_x$, hence $\mathfrak{m} = \mathfrak{m}_x$ because $\mathfrak{m}$ is maximal. Hence $\mu$ is surjective.
    \item By Urysohn's lemma (this is the only non-trivial fact required in the argument) the continuous functions separate the points of $X$. Hence $x 
e y \Rightarrow \mathfrak{m}_x 
e \mathfrak{m}_y$, and therefore $\mu$ is injective.
    \item Let $f \in C(X)$; let
      \begin{equation*}
        U_f = \{x \in X : f(x) 
e 0\}
      \end{equation*}
      and let
      \begin{equation*}
        	ilde{U}_f = \{\mathfrak{m} \in 	ilde{X} : f 
otin \mathfrak{m}\}
      \end{equation*}
      Show that $\mu(U_f) = 	ilde{U}_f$. The open sets $U_f$ (resp.~$	ilde{U}_f$) form a basis of the topology of $X$ (resp.~$	ilde{X}$) and therefore $\mu$ is a homeomorphism.
      \par Thus $X$ can be reconstructed from the ring of functions $C(X)$.
  \end{enumerate}

Amit Awasthi · Accepted Answer

\begin{proof}[Proof of $i)$]
  Let $\mathfrak{m} \in 	ilde{X}$, and let $V = \{x \in X : f(x) = 0~	ext{for all}~f \in \mathfrak{m}\}$. If $V$ is empty, then for each $x$ there is $f_x \in \mathfrak{m}$ such that $f_x(x) 
e 0$; since each $f_x$ is continuous, there is an open neighborhood $U_x 
i x$ in $X$ such that $f_x 
e 0$ on $U_x$. Since $X$ is compact, finitely many $U_x$ say $U_{x_1},\ldots,U_{x_n}$ cover $X$. Letting $f = \sum f_{x_i}^2$, we see $f 
e 0$ anywhere on $X$, hence $f \in C(X)^	imes$ with inverse $1/f$. But this contradicts $f \in \mathfrak{m}$, hence $V 
e \emptyset$. Now if $x \in V$, then $\mathfrak{m} \subseteq \mathfrak{m}_x$, and this is moreover an equality since $\mathfrak{m}$ is maximal. Thus, $\mu(x) = \mathfrak{m}$ and so $\mu$ is surjective.
\end{proof}
\begin{proof}[Proof of $ii)$]
  Since $X$ is compact Hausdorff, it is normal \cite[Thm.~32.3]{Mun00}, and so by Urysohn's lemma \cite[Thm.~33.1]{Mun00}, if $x 
e y$ there exists $f_x$ such that $f_x(x) = 0$ while $f_x(y) = 1$; similarly, there exists $f_y$ such that $f_y(x) = 1$ while $f_y(y) = 0$. Then, $f_x \in \mathfrak{m}_x \setminus \mathfrak{m}_y$ and $f_y \in \mathfrak{m}_y \setminus \mathfrak{m}_x$, hence $\mathfrak{m}_x 
e \mathfrak{m}_y$, and so $\mu$ is injective.
\end{proof}
\begin{proof}[Proof of $iii)$]
  By $i)$, every $\mathfrak{m} \in 	ilde{X}$ is of the form $\mathfrak{m}_x$ for some $x \in X$, so
  \begin{equation*}
    \mu(U_f) = \{\mathfrak{m}_x \in 	ilde{X} : f(x) 
e 0\} = \{\mathfrak{m}_x \in 	ilde{X} : f 
otin \mathfrak{m}_x\} = 	ilde{U}_f.
  \end{equation*}
  $U_f$ is open since $f$ is continuous. Now if $x \in U \subseteq X$, then since $X$ is normal as in $ii)$, there is a neighborhood $V 
i x$ such that $\bar{V} \subseteq U$ \cite[Lem.~$31.1(b)$]{Mun00}. By Urysohn's lemma, there exists $f \in C(X)$ such that $f = 1$ on $\bar{V}$ and $f = 0$ on $X \setminus U$. So, $x \in U_f \subseteq U$, hence the $U_f$ are a basis for $X$.
  \par The $	ilde{U}_f$ are open since $	ilde{X}$ has the subspace topology induced from $\operatorname{Spec}(C(X))$, and $	ilde{U}_f = (\operatorname{Spec}(C(X)))_f \cap 	ilde{X}$ where $(\operatorname{Spec}(C(X)))_f$ is as in Exercise 1.17. Similarly, the $	ilde{U}_f$ form a basis for $	ilde{X}$ since the $(\operatorname{Spec}(C(X)))_f$ form a basis for $\operatorname{Spec}(C(X))$ by Exercise 1.17.
  \par Now since $\mu$ maps basis elements of $X$ to basis elements of $	ilde{X}$, and vice versa for $\mu^{-1}$, we have that $\mu$ is a homeomorphism $X 	o 	ilde{X}$, and so given $C(X)$, we can recover $X$ as $\operatorname{Max}(C(X))$.
\end{proof}

Exercise 1.26

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Exercise 1.26

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