## Characterization of maximal ideals in the ring of all continuous real-valued functions on [0,1]

Let $R$ be the ring of all continuous functions from the closed interval $[0,1]$ to $\mathbb{R}$ and for each $c \in [0,1]$, let $M_c$ be the maximal ideal of all functions $f$ such that $f(c) = 0$.

1. Prove that if $M \subseteq R$ is a maximal ideal, then there exists a number $c \in [0,1]$ such that $M = M_c$.
2. Prove that if $b \neq c$, then $M_b \neq M_c$.
3. Prove that $M_c$ is not equal to $(x-c)$.
4. Prove that $M_c$ is not finitely generated.

[With loads of help from Ravi Vakil’s notes.]

1. Let $M \subseteq R$ be a maximal ideal, and suppose $M \neq M_c$ for all $c \in [0,1]$. Then for all such $c$, there exists a function $f_c \in M$ such that $f_c(c) \neq 0$. Without loss of generality, we may assume that $f_c(c) > 0$. Since $f_c$ is continuous, there exists a positive real number $\epsilon_c > 0$ such that $f_c[(c-\epsilon_c,c+\epsilon_c)] = 0$. Clearly the set $\{ (c-\epsilon_c,c+\epsilon_c) \cap [0,1] \ |\ c \in [0,1] \}$ covers $[0,1]$. Since $[0,1]$ is compact, this cover has a finite subcover; say $K \subseteq [0,1]$ is finite and $\{(c-\epsilon_c,c+\epsilon_c) \cap [0,1] \ |\ c \in K \}$ covers $[0,1]$. Now for each $c \in K$, define $u_c(x) = 1 + (x-c)/\epsilon_c$ if $x \in (c-\epsilon_c,c) \cap [0,1]$, $1 + (c-x)/\epsilon_c$ if $x \in [c,c+\epsilon_c) \cap [0,1]$, and 0 otherwise. Evidently $u_c \in R$ and $u_c$ vanishes outside of $(c-\epsilon_c,c+\epsilon_c)$. Consider $g = \sum u_cf_c$. Since $f_c \in M$, we have $g \in M$. However, for all $x \in [0,1]$, $g(x)$ is positive since each $u_c(x)f_c(x)$ is nonnegative and some $u_c(x)f_c(x)$ is positive. Thus $g(x) > 0$ for all $x$, and thus $1/g \in R$. But then $M$ contains a unit, a contradiction. Thus $M = M_c$ for some $c \in [0,1]$.
2. Suppose $b \neq c$. Note that $x-b \in M_b$ but $(x-b)(c) = c-b \neq 0$, so that $x-b \notin M_c$. Thus $M_b \neq M_c$.
3. Suppose $M_c = (x-c)$. Then in particular, $|x-c| = f(x)(x-c)$ for some $f(x) \in R$. For $x \neq c$, we have $f(x) = \dfrac{|x-c|}{x-c}$. Note that as $x$ approaches $c$ from the right, $f(x)$ approaches $-\infty$, while $f(x)$ approaches $\infty$ as $x$ approaches $c$ from the left. In particular, no extension of $f$ is in $R$. Thus $M_c$ is not generated by $x-c$.
4. Suppose $M_c = (A)$ is finitely generated, with $A = \{ a_i(x) \ |\ 1 \leq i \leq n \}$. Let $f = \sum |a_i|$; then $\sqrt{f}$ is continuous on $[0,1]$. Moreover, we have $\sqrt{f} \in M_c$. Thus $\sqrt{f} = \sum r_ia_i$ for some continuous functions $r_i \in R$. If we let $r = \sum |r_i|$, we have $\sqrt{f}(x) = \sum r_i(x)a_i(x)$ $\leq \sum |r_i(x)||a_i(x)|$ $\leq r(x)f(x)$.

Note that for each $b \neq c$, there must exist a function $a_i$ such that $a_i(b) \neq 0$, as otherwise we have $h(b) = 0$ for all $h \in M_c$, and in particular, $x-c \in M_c$. Thus $c$ is the only zero of $f$. From $\sqrt{f}(x) \leq r(x)f(x)$, for $x \neq c$ we have $r(x) \geq 1/\sqrt{f(x)}$. As $x$ approaches $c$, $f(x)$ approaches 0, so that $1/\sqrt{f(x)}$ is unbounded. But then $r(c)$ does not exist, a contradiction since $r(x) \in R$. Thus $M_c$ is not finitely generated.