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.

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  • Correction  On November 17, 2010 at 4:42 pm

    Your proof in part 3 is wrong since the constant map 1 is not in M_c. However, you can take the function |x-c| and by similar argument to conclude that f has to be discontinuous.

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