## Characterization of maximal ideals in the ring of all continuous real-valued functions on [0,1]

Let be the ring of all continuous functions from the closed interval to and for each , let be the maximal ideal of all functions such that .

- Prove that if is a maximal ideal, then there exists a number such that .
- Prove that if , then .
- Prove that is not equal to .
- Prove that is not finitely generated.

[With loads of help from Ravi Vakil’s notes.]

- Let be a maximal ideal, and suppose for all . Then for all such , there exists a function such that . Without loss of generality, we may assume that . Since is continuous, there exists a positive real number such that . Clearly the set covers . Since is compact, this cover has a finite subcover; say is finite and covers . Now for each , define if , if , and 0 otherwise. Evidently and vanishes outside of . Consider . Since , we have . However, for all , is positive since each is nonnegative and some is positive. Thus for all , and thus . But then contains a unit, a contradiction. Thus for some .
- Suppose . Note that but , so that . Thus .
- Suppose . Then in particular, for some . For , we have . Note that as approaches from the right, approaches , while approaches as approaches from the left. In particular, no extension of is in . Thus is not generated by .
- Suppose is finitely generated, with . Let ; then is continuous on . Moreover, we have . Thus for some continuous functions . If we let , we have .
Note that for each , there must exist a function such that , as otherwise we have for all , and in particular, . Thus is the only zero of . From , for we have . As approaches , approaches 0, so that is unbounded. But then does not exist, a contradiction since . Thus is not finitely generated.

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## Comments

Your proof in part 3 is wrong since the constant map 1 is not in M_c. However, you can take the function |x-c| and by similar argument to conclude that f has to be discontinuous.

Thanks!