The Gelfand-Mazur Theorem
 Table of Contents

# The Gelfand-Mazur Theorem

 Definition: Let $\mathfrak{A}$ be an algebra with unit. Then $\mathfrak{A}$ is said to be a Division Algebra if $\mathrm{Sing}(\mathfrak{A}) = \{ 0 \}$.

The theorem below tells us that the only Banach algebra over $\mathbb{C}$ with unit that is a division algebra is $\mathbb{C}$ itself.

 Theorem 1 (The Gelfand-Mazur Theorem): Let $\mathfrak{A}$ be a Banach algebra over $\mathbb{C}$ with unit. If $\mathfrak{A}$ is also a division algebra then $\mathfrak{A}$ is isometrically isomorphic to $\mathbb{C}$.
• Proof: Let $\mathfrak{A}$ be a Banach algebra over $\mathbb{C}$ with unit that is also a division algebra. Let $1_{\mathfrak{A}}$ denote the unit in $\mathfrak{A}$.
• Recall from the theorem on The Spectrum of an Element in a Normed or Banach Algebra over C page that for each $x \in \mathfrak{A}$, $\mathrm{Sp}(\mathfrak{A}, x)$ is a nonempty compact subset of $\mathbb{C}$. For each $x \in \mathfrak{A}$ let $\lambda(x) \in \mathrm{Sp}(\mathfrak{A}, x)$. Then $x - \lambda(x)1_{\mathfrak{A}} \in \mathrm{Sing}(\mathfrak{A})$.
• Since $\mathfrak{A}$ is a division algebra, the only singular element in $\mathfrak{A}$ is $0$. Thus $x - \lambda(x)1_{\mathfrak{A}} = 0$. Let $T : \mathbb{C} \to \mathfrak{A}$ be defined for all $\lambda \in \mathbb{C}$ by:
(1)
\begin{align} \quad T(\lambda) = \lambda1_{\mathfrak{A}} \end{align}
• We first show that $T$ is linear homomorphism of $\mathbb{C}$ to $\mathfrak{A}$. Let $\lambda_1, \lambda_2 \in \mathbb{C}$ and let $\alpha \in \mathbb{C}$. Then:
(2)
\begin{align} \quad T(\lambda_1 + \lambda_2) = (\lambda_1 + \lambda_2)1_{\mathfrak{A}} = \lambda_1 1_{\mathfrak{A}} + \lambda_2 1_{\mathfrak{A}} = T(\lambda_1) + T(\lambda_2) \end{align}
(3)
\begin{align} \quad T(\alpha \lambda_1) = (\alpha \lambda_1)1_{\mathfrak{A}} = \alpha (\lambda_1 1_{\mathfrak{A}}) = \alpha T(\lambda_1) \end{align}
(4)
\begin{align} \quad T(\lambda_1\lambda_2) = \lambda_1\lambda_2 1_{\mathfrak{A}} = \lambda_1 1_{\mathfrak{A}} \lambda_2 1_{\mathfrak{A}} = T(\lambda_1) T(\lambda_2) \end{align}
• So indeed $T$ is a linear homomorphism of $\mathbb{C}$ to $\mathfrak{A}$. We now show that $T$ is bijective.
• First let $\lambda_1, \lambda_2 \in \mathbb{C}$ and suppose that $T(\lambda_1) = T(\lambda_2)$. Then $\lambda_1 1_{\mathfrak{A}} = \lambda_2 1_{\mathfrak{A}}$. So $(\lambda_1 - \lambda_2)1_{\mathfrak{A}} = 0$, which implies $\lambda_1 = \lambda_2$. Thus $T$ is injective.
• Now let $x \in \mathfrak{A}$. Again, $\mathrm{Sp}(\mathfrak{A}, x)$ is nonempty so there exists a $\lambda(x) \in \mathrm{Sp}(\mathfrak{A}, x)$. So $x - \lambda(x)1_{\mathfrak{A}} \in \mathrm{Sing}(\mathfrak{A})$. Since $\mathfrak{A}$ is a division algebra, this implies that $x - \lambda(x)1_{\mathfrak{A}} = 0$, so $x = \lambda(x)1_{\mathfrak{A}}$. So:
(5)
\begin{align} \quad T(\lambda(x)) = \lambda(x)1_{\mathfrak{A}} = x \end{align}
• Thus $T$ is surjective and hence we conclude that $T : \mathbb{C} \to \mathfrak{A}$ is an algebra isomorphism from $\mathbb{C}$ to $\mathfrak{A}$.
• Lastly, observe that for all $\lambda \in \mathbb{C}$ we have that:
(6)
\begin{align} \quad \| T(\lambda) \| = \| \lambda 1_{\mathfrak{A}} \| = |\lambda| \| 1_{\mathfrak{A}} \| = |\lambda| \end{align}
• So $T$ is an isometric isomorphism from $\mathbb{C}$ to $\mathfrak{A}$. $\blacksquare$
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