Let be the ring of all real valued continuous functions on the closed unit interval. If is a maximal ideal of , prove that there exists a real number such that

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- January 25th 2010, 11:15 PMChandru1Maximal Ideal
Let be the ring of all real valued continuous functions on the closed unit interval. If is a maximal ideal of , prove that there exists a real number such that

- January 26th 2010, 01:46 PMOpalg
First, notice that each of the sets is a maximal ideal in . In fact, if 1 denotes the constant function 1 then for any we can write . Since , it follows that (which is trivially an ideal) has codimension 1 in and is therefore maximal.

Now let be a maximal ideal of . If there exists such that each element of vanishes at , then is contained in (and hence by maximality is equal to) , and we are done.

So suppose that there is no such . Then for each there exists such that . The set is an open neighbourhood of t. These neighbourhoods cover the unit interval, so by compactness there is a finite subcover, given by elements of . Then is an element of that is strictly positive throughout the interval and hence invertible. But an ideal containing an invertible element must by the whole ring. So is not a proper ideal. - January 27th 2010, 12:32 AMChandru1Not Clear
Hi--

I am not able to understand what do u mean by Codimension. By the way the proof is excellent, shows how algebra and analysis are closely related.

" Algebra is properly speaking the analysis of equations". - January 27th 2010, 01:23 AMOpalg
To say that a subspace M of a vector space R has codimension 1 means that the subspace generated by adding just one additional element to M is the whole space. (Equivalently, the quotient space R/M has dimension 1.) So if M has codimension 1 in R then M is a maximal proper subspace of R. In the particular case where R is a ring and M is an ideal, this obviously implies that M is a maximal ideal.