Maximal ideals in function spaces Hello friends of mathematics,
i have got a question. In the lecture we proved that the maximal ideals of $C(X)$ are the sets of functions which vanishes on a closed subset of $X$. But now i will look to $C^1[0,1]$ which is a Banach algebra. I will compute the maximal ideal space of this Banach algebra. If i know the maximal ideals i know the maximal ideal space and the other way around. I think it is something with evaluation functions and that the maximal ideal space is isomorphic to the uni interval $[0,1]$. But this is just a feeling from me. Can someone help me to underline this?!
Thank you.
 A: The proof of this fact is identical to the proof that every maximal ideal in $C[0,1]$ is determined by a point in $[0,1]$.
Suppose that $M\subseteq C^{1}[0,1]$ is a maximal ideal. Let $I_{x}=\{f\in C^{1}[0,1]|f(x)=0\}.$ Then clearly $I_{x}$ is a maximal ideal since it is the kernel of the homomorphism $C^{1}[0,1]\rightarrow\mathbb{R},f\mapsto f(x)$ and $\mathbb{R}$ is a field. I now claim that $M\subseteq I_{x}$ for some $x\in[0,1]$. Suppose to the contrary that $M\not\subseteq I_{x}$ for all $x\in[0,1]$. Then for all $x\in[0,1]$, there is a function $f_{x}\in M$ with $f_{x}(x)\neq 0$. Let $U_{x}=\{z|f(z)\neq 0\}$. Then $\{U_{x}|x\in[0,1]\}$ is an open cover of $[0,1]$, so by compactness there is a finite subcover $U_{x_{1}}\cup...\cup U_{x_{n}}=[0,1]$. In particular, we have $f_{x_{1}}^{2}+...+f_{x_{n}}^{2}\in M$ be an invertible function. This contradicts the fact that $M$ is a proper ideal. Therefore the maximal ideals of $C^{1}[0,1]$ are in a one-to-one correspondence with the points of $[0,1]$.
