Let A = A(D)¹ , J be a minimal ideal and, for n ³ 0, I_n = {f Î A ; f(0) = f¢(0) = ¼ = f⁽ⁿ⁾(0) = 0} ( recall f⁽⁰⁾ = f ). Then (I_n)_{n ³ 0} is a strictly decreasing sequense of (primary) ideals. Assuming 0 ¹ f Î J, then 0 ¹ zⁿ⁺¹f Î I_nÇJ. So I_n ÇJ = J. Hence (Ç_{n = 1}^¥ I_n)ÇJ = J and so J = 0, since Ç_{n = 1}^¥ I_n = {0}. Thus A has no minimal ideal.

Footnotes:

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