Ideals In $\mathcal{C}[0,1]$

$\boxed{1}$ .- Let $x \in [0,1]$ a fixed point, we are going to prove that $M_x \subset \mathcal{C}[0,1]$ is a maximal ideal. Let $I$ to be a ideal strictly containig $M_x$ , then there exists $g \in I \subset \mathcal{C}[0,1]$ such that $g(x) = a \neq 0$ , then $\forall h \in \mathcal{C}[0,1]$ , $h(x) = b = k \cdot a$ for some $k \in \mathbb{R} \Rightarrow h - kg = f \in M_x \subset I \Rightarrow h = kg + f \in I \Rightarrow$ $I = \mathcal{C}[0,1]$ .

Other more elegant way: let $v_x : \mathcal{C}[0,1] \to \mathbb{R} \space / \space v_x(f) = f(x)$ then $v_x$ is an epimorphism between $\mathbb{R}$ - vectorial spaces , which implies that $\mathcal{C}[0,1] / Ker {v_x}$ is isomorphic to $\mathbb{R}$ which is a field, which implies that $Ker{v_x} = \{f \in \mathcal{C}[0,1] \space / \space f(x) = 0\} = M_x$ is a maximal ideal

$\boxed{2}$ .- Let $I$ be a proper maximal ideal of the ring $\mathcal{C}([0,1])$ of continuous functions on [0,1]. We prove that there exists $x \in [0,1]$ such that $f(x)=0 \space \forall f \in I$ .

Assume that for each point of $x \in [0,1]$ there exists a function $f \in I$ such that $f(x) \neq 0$ . Then there exists a neighborhood $N_x$ of x in [0,1] such that $f \neq 0$ on $N_x$ . Since [0,1] is compact and $\{N_x\}$ is an open cover, there exists a finite cover. That is, there are functions $f_1,⋯,f_n \in I$ such that for each $x \in [0,1]$ , we have $f_i (x) \neq 0$ for some i (depending on x). Then the function $f = f_1^{2} + ... +f_n^{2}$ is also a member of $I$ such that f > 0 on $I$ . Therefore $Id = f/f \in I$ and $I = \mathcal{C}([0,1])$ , a contradiction.