# Gelfand-Mazur theorem

Theorem - Let $\mathcal{A}$ be a unital Banach algebra^{} over $\u2102$ that is also a division algebra^{} (i.e. every non-zero element is invertible). Then $\mathcal{A}$ is isometrically isomorphic to $\u2102$.

Proof : Let $e$ denote the unit of $\mathcal{A}$.

Let $x\in \mathcal{A}$ and $\sigma (x)$ be its spectrum. It is known that the spectrum is a non-empty set (http://planetmath.org/SpectrumIsANonEmptyCompactSet) in $\u2102$.

Let $\lambda \in \sigma (x)$. Since $x-\lambda e$ is not invertible and $\mathcal{A}$ is a division algebra, we must have $x-\lambda e=0$ and so $x=\lambda e$

Let $\varphi :\u2102\u27f6\mathcal{A}$ be defined by $\varphi (\lambda )=\lambda e$.

It is clear that $\varphi $ is an injective algebra homomorphism.

By the above discussion, $\varphi $ is also surjective.

It is isometric because $\parallel \lambda e\parallel =|\lambda |\parallel e\parallel =|\lambda |$

Therefore, $\mathcal{A}$ is isometrically isomorphic to $\u2102$. $\mathrm{\square}$

Title | Gelfand-Mazur theorem |
---|---|

Canonical name | GelfandMazurTheorem |

Date of creation | 2013-03-22 17:29:03 |

Last modified on | 2013-03-22 17:29:03 |

Owner | asteroid (17536) |

Last modified by | asteroid (17536) |

Numerical id | 7 |

Author | asteroid (17536) |

Entry type | Theorem |

Classification | msc 46H05 |