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Let $G = \mathbb Z^n$, the fi nite cyclic group of order $n$, and let $A = l ^1(G)$, a Banach algebra over $\mathbb C$ when the product is convolution, de fined for $f, g \in A$ by $f * g(x) =\sum _{y \in G}f(y)g(x -y)$. Can we determine the maximal ideal space of $A$, equivalently the space of algebra homomorphisms from $A$ onto $\mathbb C$?

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Is there something missing or unclear in the answer below? –  Martin Feb 15 '13 at 9:25
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The group $G$ embeds into $\ell^1(G)$ via the homomorphism $g \mapsto \delta_g$ since $\delta_g \ast \delta_h = \delta_{gh}$. Since $\lVert \delta_g \rVert_1 = 1$, a surjective homomorphism $\ell^1(G) \to \mathbb{C}$ restricts to a homomorphism $G \to \mathbb{C}^\times$ with bounded range. The maximal bounded subgroup of $\mathbb{C}^\times$ is $\mathbb{T} = \{z \in \mathbb{C} \mid |z| =1\}$, so we obtain a map from the maximal ideal space of $\ell^1(G)$ to the Pontryagin dual $\widehat{G} =\operatorname{Hom}(G,\mathbb T)$. Conversely, a homomorphism $G \to \mathbb{T}$ determines a surjective homomorphism $\ell^1(G) \to \mathbb{C}$ since the $\delta_g$ span a dense subspace of $\ell^1(G)$. From this one easily deduces an identification of the maximal ideal space of $\ell^1(G)$ with the Pontryagin dual of $G$ which is $\widehat{G} = \mathbb{T}^n$ or $\widehat{G} \cong \mathbb{Z}_n$ depending on whether you meant $G = \mathbb{Z}^n$ or $G = \mathbb{Z}_n$.

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