**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.4 Exercise 7.4.33**

Solution:

(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.