Let $X$ denote the closed unit interval, and let $R$ be the ring of continuous functions $X \to \mathbb{R}$.

Question

$\textbf{Problem}$

$\bullet~$ Let $X$ denote the closed unit interval, and let $R$ be the ring of continuous functions $X \to \mathbb{R}$.

$\textbf{(a)}~$ Let $f_{1}, f_{2}, \dots, f_{n}$ be functions with no common zero on $X$. Prove that the ideal generated by these functions is the unit ideal.

$%\boxed{\textit{Hint: }~ \text{ Consider the sum } \sum_{i = 1}^{n}f_{i}^{2}.}$$\textbf{(b)}~$ Establish a bijective correspondence between maximal ideals of $R$ and points on the interval.

Any sort of Idea?

Answer 1

Maximal ideals of $C(X) = R$ are given by $$I_{t_0} = \{f \in C(X) : f(t_0) = 0\}$$ for $t_0 \in X$. Namely, every such ideal is maximal since considering the map $C(X) \to \Bbb{R} : f\mapsto f(t_0)$ gives the isomorphism

$$C(X)/I_{t_0} \cong \Bbb{R}$$

since $I_{t_0}$ is the kernel of this map. This implies that $I_{t_0}$ is maximal since in general $m \subset R$ is a maximal ideal iff $R/m$ is a field.

Conversely, let $I$ be a maximal ideal in $C(X)$ and assume that $I \ne I_{t_0}$ for all $t_0$. Then for every $t \in X$ there is a function $f_t \in I$ such that $f_t(t) \ne 0$. By continuity there is an open neighbourhood $U_t$ of $t$ such that $|f_t| > 0$ on $U_t$. Since $X = [0,1]$ is compact and $(U_t)_{t\in X}$ is an open cover, there are $t_1, \ldots, t_n \in X$ such that $X \subseteq U_{t_1} \cup \cdots \cup U_{t_n}$. Now the function $$f = \sum_{i=1}^n f_{t_i}^2$$ is in $I$ since all of the $f_i$ are in $I$. Furthermore, $f$ is positive everywhere and hence invertible, which implies that $I = C(X)$. This contradicts maximality of $I$ so we conclude $I=I_{t_0}$ for some $t_0 \in X$.

Hence $t \mapsto I_t$ is a bijection between points of $X$ and maximal ideals of $C(X)$.

Answer 2

Hint for (a):

The function $\sum_i f_i^2$ has no zeros and is thus invertible.

Hint for (b):

Consider $t \mapsto \ker(ev_t)$ where $ev_t$ evaluates the function $f$ in $t\in [0,1]$.