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Let $R$ be the ring of continuous functions on $[0,1]$ and let $I=\{f \in R\mid f(1/3) = f(1/2) = 0\}$. Then $I$ is an ideal but is not prime.

Please help me prove this, thank you.

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closed as too localized by Nate Eldredge, Akhil Mathew Apr 11 '11 at 2:32

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Why do you want other people to do your homework? You have already asked about 4 homework questions in just a few hours and haven't even mentioned why you're having problems with the exercises. – Adrián Barquero Apr 11 '11 at 0:36
Basically you're just posting all of your homework problems and asking "Please do my homework for me". That does not seem very nice of you. The people in here are using their time to help others and to learn, but you need to do your homework by yourself. There are some other persons who actually ask homework problems after having tried them, but this does not seem to be the case with you. – Adrián Barquero Apr 11 '11 at 0:42
I have tried it out, but the steps which I have aren't so good – mary Apr 11 '11 at 0:55
This is what I have so far: If f, g are elements of Ideal I, and h an element R, then (hf - g)(1/3) = 0 and (hf-g)(1/2) = 0, and I is an ideal. – mary Apr 11 '11 at 0:57
Dear user8917, I'm closing this question for now; please see the FAQ for how to ask a homework question. – Akhil Mathew Apr 11 '11 at 2:33

Hint: Let $f$ and $g$ be elements of $I$. What is $(f+g)(1/3)$? How about $(f+g)(1/2)$? How about for $hf$ instead of $f + g$, where $h$ is an arbitrary function in $R$?. You should be able to use this information to show that $I$ is an ideal.

As for proving that the ideal is not prime, note that $f\in R$ only if $f$ vanishes at both $1/2$ and $1/3$. Try to think of two functions $f,g\in R$ such that this is true for $fg$ but not $f$ or $g$.

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