The Gelfand-Mazur Theorem

The Gelfand-Mazur Theorem

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

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 $X$ be a Banach algebra over $\mathbf{C}$ with unit. If $X$ is also a division algebra then $X$ is isometrically isomorphic to $\mathbb{C}$.
  • Proof: Let $X$ be a Banach algebra over $\mathbb{C}$ with unit that is also a division algebra. Let $1_X$ denote the unit in $X$.
  • Recall from the theorem on The Spectrum of an Element in a Normed or Banach Algebra page that for each $x \in X$, $\mathrm{Sp}(X, x)$ is a nonempty compact subset of $\mathbb{C}$. For each $x \in X$ let $\lambda(x) \in \mathrm{Sp}(X, x)$. Then $x - \lambda(x)1_X \in \mathrm{Sing}(X)$.
  • Since $X$ is a division algebra, the only singular element in $X$ is $0$. Thus $x - \lambda(x)1_X = 0$. Let $T : \mathbb{C} \to X$ be defined for all $\lambda \in \mathbb{C}$ by:
(1)
\begin{align} \quad T(\lambda) = \lambda1_X \end{align}
  • We first show that $T$ is linear homomorphism of $\mathbb{C}$ to $X$. 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_X = \lambda_1 1_X + \lambda_2 1_X = T(\lambda_1) + T(\lambda_2) \end{align}
(3)
\begin{align} \quad T(\alpha \lambda_1) = (\alpha \lambda_1)1_X = \alpha (\lambda_1 1_X) = \alpha T(\lambda_1) \end{align}
(4)
\begin{align} \quad T(\lambda_1\lambda_2) = \lambda_1\lambda_21_X = \lambda_1 1_X \lambda_2 1_X = T(\lambda_1) T(\lambda_2) \end{align}
  • So indeed $T$ is a linear homomorphism of $\mathbb{C}$ to $X$. 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 $\lambda_1 1_X = \lambda_2 1_X$. So $(\lambda_1 - \lambda_2)1_X = 0$, which implies $\lambda_1 = \lambda_2$. Thus $T$ is injective.
  • Now let $x \in X$. Again, $\mathrm{Sp}(X, x)$ is nonempty so there exists a $\lambda(x) \in \mathrm{Sp}(X, x)$. So $x - \lambda(x)1_X \in \mathrm{Sing}(X)$. Since $X$ is a division algebra, this implies that $x - \lambda(x)1_X = 0$, so $x = \lambda(x)1_X$. So:
(5)
\begin{align} \quad T(\lambda(x)) = \lambda(x)1_X = x \end{align}
  • Thus $T$ is surjective and hence we conclude that $T : \mathbb{C} \to X$ is an algebra isomorphism from $\mathbb{C}$ to $X$.
  • Lastly, observe that for all $\lambda \in \mathbb{C}$ we have that:
(6)
\begin{align} \quad \| T(\lambda) \| = \| \lambda 1_X \| = |\lambda| \| 1_X \| = |\lambda| \end{align}
  • So $T$ is an isometric isomorphism from $\mathbb{C}$ to $X$. $\blacksquare$
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