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Math Help - Maximal Ideal

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    Maximal Ideal

    Let R be the ring of all real valued continuous functions on the closed unit interval. If M is a maximal ideal of R, prove that there exists a real number  \gamma, \ 0 \leq \gamma \leq 1 such that M=M_{\gamma}=\{ f(x) \in R \ | \ f(\gamma)=0\}
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    Quote Originally Posted by Chandru1 View Post
    Let R be the ring of all real valued continuous functions on the closed unit interval. If M is a maximal ideal of R, prove that there exists a real number  \gamma, \ 0 \leq \gamma \leq 1 such that M=M_{\gamma}=\{ f(x) \in R \ | \ f(\gamma)=0\}
    First, notice that each of the sets M_{\gamma} is a maximal ideal in R . In fact, if 1 denotes the constant function 1 then for any f\in R we can write f = (f-f(\gamma)1) + f(\gamma)1. Since f-f(\gamma)1\in M_{\gamma}, it follows that M_{\gamma} (which is trivially an ideal) has codimension 1 in R and is therefore maximal.

    Now let M be a maximal ideal of R. If there exists \gamma\in[0,1] such that each element of M vanishes at \gamma, then M is contained in (and hence by maximality is equal to) M_{\gamma}, and we are done.

    So suppose that there is no such \gamma. Then for each t\in[0,1] there exists f_t\in M such that f_t(t)\ne0. The set \{s\in[0,1]:f_t(s)\ne0\} is an open neighbourhood of t. These neighbourhoods cover the unit interval, so by compactness there is a finite subcover, given by elements f_{t_n} of M. Then \textstyle\sum f_{t_n}^2 is an element of M that is strictly positive throughout the interval and hence invertible. But an ideal containing an invertible element must by the whole ring. So M is not a proper ideal.
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    Not 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".
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  4. #4
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    Quote Originally Posted by Chandru1 View Post
    I am not able to understand what do u mean by Codimension.
    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.
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