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Math Help - characterizing the maximal ideals of a function ring

  1. #1
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    characterizing the maximal ideals of a function ring

    Hello!!!
    So much homework :-\
    I need to show that every maximal ideal in the ring R, which is made of all continuous functions [0,1]-->R, is of the form M(c)={f in R| f(c)=0).

    I think I have a kinda of a topological proof formed, and of course it can be specifically used here (the proof is for any hausdorff-compact space), but I'm seeking a more algebraic proof, which I'm sure exists since most of us haven't studied topology yet.
    I was to show that if M is a maximal ideal in R, then there must exist a point c in [0,1] which satisfies f(c)=0 for all f in M. Therefore, I start by contradically assuming that for all x in [0,1] there exists a function f_x in M such that f_x(x)=0. I cannot reach a contradiction though...

    Thank you in advance...
    Tomer.
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  2. #2
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    Quote Originally Posted by aurora View Post
    Hello!!!
    So much homework :-\
    I need to show that every maximal ideal in the ring R, which is made of all continuous functions [0,1]-->R, is of the form M(c)={f in R| f(c)=0).

    I think I have a kinda of a topological proof formed, and of course it can be specifically used here (the proof is for any hausdorff-compact space), but I'm seeking a more algebraic proof, which I'm sure exists since most of us haven't studied topology yet.
    I was to show that if M is a maximal ideal in R, then there must exist a point c in [0,1] which satisfies f(c)=0 for all f in M. Therefore, I start by contradically assuming that for all x in [0,1] there exists a function f_x in M such that f_x(x)=0. I cannot reach a contradiction though...

    Thank you in advance...
    Tomer.

    Use the fact that if M is a maximal ideal in the ring R, then R/M is a field. This is easy to see. Using I as the ideal in the problem and f in C, then I + f = I iff f is in I, that is f(a) = 0 for some a in  [0,1], thus if I + f is a distinct coset, then f is not in I, hence f(x) is nonzero for any x in [0,1], hence 1/f(x) is continuous and is, of course, nonzero on [0,1] and 1/f + I is a distinct coset and (f+I)(1/f+I) = 1 + I, so that R/M is a field.
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  3. #3
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    Quote Originally Posted by aurora View Post
    I need to show that every maximal ideal in the ring R, which is made of all continuous functions [0,1]-->R, is of the form M(c)={f in R| f(c)=0).

    I think I have a kinda of a topological proof formed, and of course it can be specifically used here (the proof is for any hausdorff-compact space), but I'm seeking a more algebraic proof, which I'm sure exists since most of us haven't studied topology yet.
    I was to show that if M is a maximal ideal in R, then there must exist a point c in [0,1] which satisfies f(c)=0 for all f in M. Therefore, I start by contradically assuming that for all x in [0,1] there exists a function f_x in M such that f_x(x)=0. [You mean f_x(x)≠0, of course.] I cannot reach a contradiction though...
    That's a good start. But you will have to resort to analysis at some stage because this is a result about a ring of continuous functions (and it's no longer true if you drop the continuity condition).

    So for each x in [0,1] you have a function f_x in M such that f_x(x)≠0. Replacing f_x by –f_x if necessary, you can even assume that f_x(x)>0. The set U_x = \{y:f_x(y)>0\} is then an open neighbourhood of x. By the compactness of the unit interval, finitely many of these sets U_x, say U_{x_1},\ldots,U_{x_n} will cover the unit interval. Then the function f_{x_1}+\ldots+f_{x_n} in M will be strictly positive throughout the interval and hence invertible. Hence M is the whole of R. Contradiction!
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  4. #4
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    Galois, thanks. I know that R/I is a field, and have even used it in my topological proof.
    Opalg, thank you. However, doesn't that count for a topological proof? The whole use of the terms open sets, neighborhoods and coverings is a bit problematic here, since as I've said, most of us shouldn't be familiar with these terms (I specifically have taken topology but others did not).

    After I show there exists some point c in [0,1] so that f(c)=0 for all f in M, it's pretty easy to show that M=M_c, though I still needed to use the fact that R/M is a field.

    But so far, I have written down a topological proof Isn't there a proof not dealing with topological concepts?
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  5. #5
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    Quote Originally Posted by aurora View Post
    But so far, I have written down a topological proof Isn't there a proof not dealing with topological concepts?
    I don't think so. As I said above, the result is false if you drop the continuity condition. The maximal ideal space of the ring of all functions on the unit interval is a huge space, far bigger than the unit interval.

    So you have to bring continuity into the proof somehow, and that implies that you need to use some analysis or topology.
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  6. #6
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    Well, I guess I'll sattle with this one then. :-)

    I thank you guys very much.
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