Kernel of the Gelfand transform is the radical

Assume the Axiom of Choice (The Axiom of Choice). Let $A$ be a nonzero commutative unital complex Banach algebra (Unital Banach algebra) with Gelfand transform $\mathrm{\Gamma }={\mathrm{\Gamma }}_A:A\to C(\mathrm{\Delta }(A))$ (Gelfand transform) and Jacobson radical $\mathrm{rad}(A)$ (Jacobson radical and semisimple commutative Banach algebra). Then

$\ker \mathrm{\Gamma }={\bigcap }_{\chi \in \mathrm{\Delta }(A)}\ker \chi =\mathrm{rad}(A)\mathrm{.}$

Consequently $\mathrm{\Gamma }$ is injective if and only if $A$ is semisimple. No claim about $\mathrm{\Gamma }$ being isometric or surjective is made; injectivity of $\mathrm{\Gamma }$ is a statement about the radical only.

• The two intersections in the statement are the same set for two different reasons. The first is the definition of $\mathrm{\Gamma }$ evaluated at zero, the second is the maximal ideal theorem; the theorem is the equality of the two descriptions.
• Semisimplicity still does not give isometry. By Gelfand transform is a contractive unital homomorphism one always has $\mathrm{\parallel }\widehat{a}{\mathrm{\parallel }}_{\mathrm{\infty }}=r(a)\le \mathrm{\parallel }a\mathrm{\parallel }$, and semisimplicity upgrades injectivity of $\mathrm{\Gamma }$, not the norm equality $\mathrm{\parallel }\widehat{a}{\mathrm{\parallel }}_{\mathrm{\infty }}=\mathrm{\parallel }a\mathrm{\parallel }$.