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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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    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.2011-04-11
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    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.2011-04-11
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    I have tried it out, but the steps which I have aren't so good2011-04-11
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    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.2011-04-11
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    However, I don't know how to show that I is not prime2011-04-11
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    @user8917: If you know what a *prime ideal* is, you just have to check $I$ satisfies the definition. **Hint**: Use the *zero-product rule*.2011-04-11
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    Dear user8917, I'm closing this question for now; please see the FAQ for how to ask a homework question.2011-04-11
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    Is this really tooo localized!2013-04-28

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