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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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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