Added basic facts on maximal ideals.

src/op/banach/multiplicative.tex

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+\section{Multiplicative Functionals}
+\label{section:multiplicative-functional}
+
+\begin{definition}[Multiplicative Functional]
+\label{definition:multiplicative-functional}
+    Let $A$ be a unital Banach algebra and $\phi \in A^*$, then $\phi$ is \textbf{multiplicative} if $\phi \ne 0$ and for each $x, y \in A$, $\phi(xy) = \phi(x)\phi(y)$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:multiplicative-unit}
+    Let $A$ be a unital Banach algebra and $\phi \in A^*$ be a multiplicative functional, then
+    \begin{enumerate}
+        \item For each $x \in A$, $|\phi(x)| \le [x]_{sp} \le \norm{x}_A$. 
+        \item $\norm{\phi}_{A^*} = 1$. 
+        \item $\phi(G(A)) \subset \complex \setminus \bracs{0}$. 
+    \end{enumerate}
+    
+\end{proposition}
+\begin{proof}
+    (3): For each $x \in G(A)$, $1 = \phi(xx^{-1}) = \phi(x)\phi(x^{-1}) \ne 0$. 
+    
+    (1): By (3), for every $\lambda \in \complex$ with $|\lambda| > [x]_{sp}$, $\lambda x \in G(A)$ and $\lambda - \phi(x) \ne 0$. Therefore $\phi(x) \in \ol{B(0, [x]_{sp})}$. 
+
+    (2): For each $\lambda \in \complex$, $\phi(\lambda 1) = \lambda$, so $\norm{\phi}_{A^*} \le 1$. 
+\end{proof}
+
+\begin{definition}[Space of Multiplicative Linear Functionals]
+\label{definition:multiplicative-linear-functional-space}
+    Let $A$ be a unital Banach algebra, then $\Omega(A)$ is the \textbf{space of multiplicative linear functionals}, which is a compact Hausdorff space under the weak-* topology. 
+\end{definition}
+\begin{proof}
+    By the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:commutative-maximal-ideal-space}
+    Let $A$ be a commutative unital Banach algebra, then the mapping
+    \[
+    \Omega(A) \to \cm(A) \quad \phi \mapsto \ker(\phi)
+    \]
+
+    is a bijection. 
+\end{proposition}
+\begin{proof}
+    For each $\phi \in \cm(A)$, $\ker(\phi)$ is an ideal of codimension $1$, and must be maximal. 
+
+    On the other hand, for each $I \in \Omega(A)$, the quotient $A/I$ is a field, so by the \hyperref[Gelfand-Mazur Theorem]{theorem:gelfand-mazur}, it is isomorphic to $\complex$. Thus the canonical projection $A \to A/I \iso \complex$ induces a multiplicative functional on $A$. 
+\end{proof}
+
+