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How do I find the maximal ideals of the algebra of holomorphic functions in one variable?thanks.

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    i guess you should make your question more precise. Anyway, consider the kernel of the map $f \mapsto f(0)$2011-12-07

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I assume you're talking about the algebra $H(\Omega)$ of holomorphic functions in a domain $\Omega$ of the complex plane. The obvious maximal ideals are $J_p = \{f: f(p) = 0\}$ for $p \in \Omega$. However, these are not all. For example, consider a sequence $\{p_n\} \subset \Omega$ that has no limit point in $\Omega$.
The set $J$ of functions $f$ analytic in $\Omega$ such that $f(p_n) = 0$ for all sufficiently large $n$ is a proper ideal of $H(\Omega)$, and not contained in any $J_p$: in fact, by the Weierstrass factorization theorem, for each positive integer $N$ there exists a function $f$ analytic in $\Omega$ whose set of zeros is $\{p_n: n \ge N\}$. By Zorn's lemma $J$ is contained in a maximal ideal, which is not a $J_p$.

As far as I know there is no way to explicitly construct one of these "exotic" maximal ideals of $H(\Omega)$: they live out in Axiom-of-Choice Land.