A commutative Banach algebra A without any minimal ideals.

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Let A = A(Δ)^{1} , J be a minimal ideal and, for n ≥ 0, I_{n} = {f ∈ A ; f(0) = f′(0) = … = f^{(n)}(0) = 0} ( recall f^{(0)} = f ). Then (I_{n})_{n≥0} is a strictly decreasing sequense of (primary) ideals. Assuming 0 ≠ f ∈ J, then 0 ≠ z^{n+1}f ∈ I_{n} ∩ J. So I_{n} ∩ J = J. Hence (⋂_{n=1}^{∞} I_{n}) ∩ J = J and so J = 0, since ⋂_{n=1}^{∞} I_{n} = {0}. Thus A has no minimal ideal.

1Let Δ denote the closed unit disc {z ∈ C, |z| ≤ 1}. Suppose that A(Δ) denoted the set of all elements of C(Δ) which are analytic on the interior of Δ. A(Δ) is a closed subalgebra of C(Δ)