Gelfand-Mazur Theorem

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Theorem

Let $\struct {A, \norm {\, \cdot \,} }$ be a unital normed algebra over $\C$ where:

$\map G A = A \setminus \set { {\mathbf 0}_A}$

where $\map G A$ is the group of units of $A$.

Then $A$ is isometrically algebra isomorphic to $\C$.


Proof

Define $\theta : \C \to A$ by:

$\map \theta \lambda = \lambda {\mathbf 1}_A$

for each $\lambda \in \C$.

For $z, w, \lambda \in \C$ we have:

\(\ds \map \theta {z + \lambda w}\) \(=\) \(\ds \paren {z + \lambda w} {\mathbf 1}_A\)
\(\ds \) \(=\) \(\ds z {\mathbf 1}_A + \lambda \paren {w {\mathbf 1}_A}\)
\(\ds \) \(=\) \(\ds \map \theta z + \lambda \map \theta w\)

so $\theta$ is linear.

Further, for $\lambda, \mu \in \C$ we have:

$\map \theta {\lambda \mu} = \paren {\lambda \mu} {\mathbf 1}_A = \paren {\lambda {\mathbf 1}_A} \paren {\mu {\mathbf 1}_A} = \map \theta \lambda \map \theta \lambda$

Hence $\theta$ is an algebra homomorphism.

For $\lambda \in \C$, we have:

\(\ds \norm {\map \theta \lambda}\) \(=\) \(\ds \norm {\lambda {\mathbf 1}_A}\)
\(\ds \) \(=\) \(\ds \cmod \lambda \norm { {\mathbf 1}_A}\) Norm Axiom $\text N 2$: Positive Homogeneity
\(\ds \) \(=\) \(\ds \cmod \lambda\) Definition of Unital Normed Algebra

Hence $\theta$ is a isometry.

It remains to show that $\theta$ is surjective.

Let $x \in A$.

From Spectrum of Element of Banach Algebra is Non-Empty: Corollary, we have $\map {\sigma_A} x \ne \O$ where $\map {\sigma_A} x$ is the spectrum of $x$ in $A$.

Let $\lambda \in \map {\sigma_A} x$.

Then $\lambda {\mathbf 1}_A - x \in A \setminus \map G A = \set { {\mathbf 0}_A}$.

Hence we have $x = \lambda {\mathbf 1}_A = \map \theta \lambda$.

That is, $x \in \Img \theta$.

So $\theta$ is surjective.

So $\theta$ is an isometric isomorphism onto $\C$.

Hence $A$ is isometrically isomorphic to $\C$.

$\blacksquare$


Source of Name

This entry was named for Stanisław Mieczysław Mazur and Israel Moiseevich Gelfand.