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(a) If M is any maximal ideal of $R$, then $\exists c\in[0,1]$ s.t. $M=M_C=\{f|f(c)=0\}$

So somehow I need to prove that every maximum ideal is the kernal of an evaluation at some $c \in [0,1]$....

(b) $x \neq y \to M_x \neq M_y$

This seems intuitive but I need help showing it rigorously. If $x \neq y$ then of course there will be at least one function that has a zero at $x$ but not $y$ and vice versa...

(c) $M_c \neq <x-c>$

$sin(2 \pi x/c) \in M_c$ and $sin(2 \pi x/c) \notin <x-c>$... this is the only one I think that i've got

(d) $M_c$ is not finitely generated

Perhaps I would proceed by contradiction and assume it is finitely generated by $<f_1,f_2,....,f_k>$ and then try to construct a $f \in M_c$ s.t. $f \notin f_1R+f_2R+....f_kR$

Insight greatly appreciated!!

Math is hard
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  • I'm not entirely sure on how to prove (a) and (d) yet, but here's a partial answer for the other two items:

    Regarding (b), as you claim, if $x \neq y$, then $h(t) = t - x$ is in $M_x$ but not in $M_y$. As for (c), you're correct when $c \neq 0$. In the remaining case the function you propose is ill-defined, but you can fix it effortlessly by taking $\sin(t)$.

     – qualcuno Jul 08 '18 at 00:21
  • https://math.stackexchange.com/questions/375400/maximal-ideals-in-the-ring-of-real-functions-on-0-1 – Theo Bendit Jul 08 '18 at 00:32

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