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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 $M_c = \{f \in R |f(c) = 0\}.$

(a) Prove that if $M$ is any maximal ideal of $R$ then there is a real number $c \in [0,1]$ such that $M = M_c.$
(b) Prove that if $b$ and $c$ are distinct points in $[0,1] $ then $M_b\neq M_c.$

(c) Prove that $M_c$ is not equal to the principal ideal generated by $x — c.$

(d) Prove that $M_c$ is not a finitely generated ideal.

(a) I know that $M_c$ is maximal ideal of $R.$

Since $M$ is a maximal ideal of $R$ then $M$ is prime ideal. I am unsure how to proceed.

(b) Assume that $b\neq c$ and $M_b=M_c.$ Let $f \in M_c.$ Then $f \in M_b.$ Thus, $ M_c$ is not maximal ideal of $R.$

(c) and (d) I have no idea how to start it.

Help me, please!

Liza
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  • Too many questions in a post. Besides, the answers to the first two are available many times on the site. Look for those answers yourself and consider dropping those two questions. – rschwieb Dec 09 '15 at 10:40
  • See http://www.math.lsa.umich.edu/~kesmith/593hmwk2-2014-solutions.pdf – Knt Aug 01 '19 at 03:49

1 Answers1

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Hint for (c):

Let $f \in M_c$. One is tempted to consider $g(x)=\dfrac{f(x)}{x-c}$ so that $f(x)=(x-c)g(x)$. But we need to give a value to $g(c)$. The natural candidate if $g(c)=f'(c)$, but $f$ is not necessarily differentiable at $x=c$. The simplest counterexample that comes to mind is $f(x)=|x-c|$. Now prove directly that there is no continuous $g$ such that $|x-c|=(x-c)g(x)$.

lhf
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