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Let $X = \mathcal{C}([0,1],\mathbb{R})$, be the ring of all continuous valued functions $f:[0,1] \to \mathbb{R}$. For $x \in [0,1]$, let $M_{x} = \{ f \in M \ | \ f(x)=0\}$. One can show by using compactness of $[0,1]$ that every maximal ideal is of this form.

Extending the Question to Entire functions: Let $\mathsf{C}(z)$ be the ring of complex entire functions. For $ \lambda \in \mathsf{C}$ let $M_{\lambda}$ denote the set of all entire functions which have a zero at $\lambda$. Then is $M_{\lambda}$ a maximal ideal in $\mathsf{C}(z)$, and does every ideal happen to be of this form? I don't know to prove this!

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up vote 14 down vote accepted

No.

Let $f(z)=1/\Gamma(z)$ where $\Gamma$ is the gamma function. Then $f$ vanishes at $0$, $-1,-2,\ldots$ and nowhere else. For integers $m$ let $f_m(z)=f(z+m)$. Then $f_m$ vanishes at $-m,-m-1,\ldots$. Let $I$ be the ideal of the ring of the entire functions generated by the $f_m$. Then $I$ is proper, since otherwise finitely many of the $f_m$ would generate an ideal containing the constant function $1$. But any finite set of $f_m$ have a common zero, so this doesn't hold. So $I\subseteq J$ for some maximal ideal $J$ (using the standard Zorn's lemma argument). But $J\ne M_\lambda$ for any $\lambda$, as for each $\lambda$ there is some $m$ with $f_m(\lambda)\ne0$, and $f_m\in J$.

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Nice! Btw, when you say "$f$ is proper", you mean "$I$ is [a] proper [ideal]", don't you? –  a.r. Sep 4 '10 at 23:50
    
Corrected, thanks Agusti. –  Robin Chapman Sep 5 '10 at 6:31
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If I remember correctly from my student days, if D is a connected domain in the complex plane, then the ring O(D) of holomorphic functions on D has the property that every finitely generated ideal is actually principal, but there are ideas tat cannot be finitely generated (such as in Robin's answer above)

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