If you find any mistakes, please make a comment! Thank you.

Characterization of maximal ideals in the ring of all continuous real-valued functions


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.


Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

This Post Has One Comment

  1. I think that on 1), f[(c-e, c+e)] = 0 is not correct since you assume f(c)!=0. I think that you mean f[(c-e, c+e)] >= 0 . Thank you so much for your help in any of those exercises available, including this one that I couldn't figure out alone.

Leave a Reply

Close Menu