Consider the $C^{\ast}$-algebra $A = C_0(\mathbb R)$ and the ideal $J = \{f \in A\ |\ f(0) = 0 \}$ of $A.$ Show that $A/J \cong \mathbb C$ as $C^{\ast}$-algebras.
Here $J$ is clearly a closed maximal ideal of $A$ since it is the kernel of a non-zero multiplicative linear functional on $A$ namely the evaluation $f \mapsto f(0).$ So it is obvious that $A/J$ happens to be a field. But how do I make sure that it is the field of complex numbers $\mathbb C$ as a $C^{\ast}$-algebra? Any help in this regard would be warmly appreciated.
Thanks for your time.
Apply the first isomorphism theorem for $C^*$-algebras:
Theorem: Let $\varphi: A \to B$ be a $*$-homomorphism between $C^*$-algebras. Then the induced map $$\overline{\varphi}: A/\ker(\varphi) \to \varphi(B): a + \ker (\varphi)\mapsto \varphi(a)$$ is a well-defined $*$-isomorphism.
Apply this to the canonical $*$-homomorphism $$\operatorname{ev}_0: C_0(\mathbb{R})\to \mathbb{C}: f \mapsto f(0).$$