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Homework Help: More number theory

  1. Mar 5, 2010 #1
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    For i) I said for a in [0,1], then the group of units are = {f in R | f(a) =/= 0}

    i.e a continous function f on [0,1] would have a continous function g on [0,1] such that f.g=1

    but the function would have to be g = 1/f, but this wouldnt be continous if f(a) = 0

    for ii) I have to show for a,b in R then if ab=0 then either a = 0 or b = 0

    but I havn't gotton any further here since I'm having a hard time finding functions a and b where the product is 0

    for iii) I believe I have to construct a surjective ring homomorphism ( f: R -> F ) and show that the kernal is equal to the Ideal.

    But I'm not sure how what field F to choose for this

    Or I have to show that the quotient group R/I is a field.

    Thanks for any help!
     
    Last edited: Mar 5, 2010
  2. jcsd
  3. Mar 5, 2010 #2

    Dick

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    The statement of (i) doesn't have any 'a' in it. You aren't thinking very abstractly. Yes, the inverse of f would be 1/f. What conditions on f do you need to ensure that exists and is continuous?
     
  4. Mar 6, 2010 #3
    the condition that f is a polynomial?
     
  5. Mar 6, 2010 #4

    Dick

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    Is that just a random guess?
     
  6. Mar 6, 2010 #5
    No it wasn't, I looked up a more precise definition of the conditions for continuity and it stated that all polynomials are continuous.

    Otherwise I'm not sure what you mean sorry :/
     
  7. Mar 6, 2010 #6

    Dick

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    Ok. Maybe I'm just being overly fussy about your original answer because you didn't specify what 'a' is supposed to be. Is it some point in [0,1], or is it ANY point in [0,1]?
     
  8. Mar 6, 2010 #7
    Any point.
     
  9. Mar 6, 2010 #8

    Dick

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    Ok. Be careful when you get to part (iii). There 'a' means 'some point'. So f is a nonzero continuous function on [0,1]. So 1/f exists. If you can say why 1/f is continuous then you've show f is a unit.
     
  10. Mar 6, 2010 #9
    For part iii) I've said

    that I is a proper ideal since the constant function f(x)=1 is not an element of the ideal.

    So R/I is isomorphic to the image of the functions, and the Im(f) = the real numbers

    Since the real numbers are a field then I is maximal. Is this ok?
     
  11. Mar 6, 2010 #10

    Dick

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    It's not very clear. What exactly is the ring homomorphism from R (the ring) to F (the reals) that I is the kernel of?
     
  12. Mar 6, 2010 #11
    phi : R -> R

    f -> f(a)=0

    Pretty sure that's a ring homomorphism
     
  13. Mar 6, 2010 #12

    Hurkyl

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    "f(a)=0" is a truth-value (well, a truth-value-valued equation) not a real number....
     
  14. Mar 6, 2010 #13
    hmm so what could I change it to, just 0? Or do I need to change my co-domain?
     
  15. Mar 6, 2010 #14

    Dick

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    You just aren't expressing yourself very well. What is the real number phi(f)?
     
  16. Mar 6, 2010 #15
    I'm having a hard time getting my head around the ring of functions, usually it's something simpler.

    Perhaps it's just

    phi : R -> R

    f -> f(a)
     
  17. Mar 6, 2010 #16

    Dick

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    Yes, it is. Is I the kernel? You said before you were only 'pretty sure' that it was a homomorphism. Can you check that it satisfies the properties of a homomorphism, so you can say you are 'really sure'?
     
  18. Mar 6, 2010 #17
    For part ii) I can only think of the fact that

    'Theorem: Let R be an integral domain and let R[x] be the ring of polynomials in powers of x whose coefficients are elements of R. Then R[x] is an integral domain if and only if R is.'

    These polynomials R[x] are functions that are most likely continous on [0,1], so if I can choose an R that isn't an integral domain then by the theorem the continuous functions on [0,1] are not an integral domain?

    Just a thoguht as I'm having trouble with this..
     
  19. Mar 6, 2010 #18

    Dick

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    The problem is not about polynomials, it's about continuous functions, which don't have to be polynomials. Surely you can think of two nonzero continuous functions on [0,1] whose product is zero.
     
  20. Mar 6, 2010 #19
    You make it sound trivial but nothings jumping out at me
     
  21. Mar 6, 2010 #20

    Dick

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    How about defining the functions piecewise?
     
  22. Mar 6, 2010 #21
    hmm if they're piecewise then how to make sure they both stay continuous on [0,1] so that the product is zero?

    is it still continuous if the endpoints match up?

    so f(x) = x for x<0.5

    and g(x) = x for x>=0.5

    is sufficient?
     
  23. Mar 6, 2010 #22

    Dick

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    You haven't defined what f(x) is for x>=0.5. If you mean f(x)=0 for x>=0.5, then f(x) isn't continuous. Making sure the pieces match up so it's continuous is YOUR job.
     
  24. Mar 6, 2010 #23
    how about for some small epsilon e > 0

    f(x) = e for x<0.5 and f(x) = 0 for x>=0.5, would that be continuous?
     
  25. Mar 6, 2010 #24

    Dick

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    What do you think? How small would e have to be to make it continuous? This isn't supposed to be anything horribly exotic. Just pick a nonzero function on [0,.5] such that f(.5)=0.
     
  26. Mar 6, 2010 #25
    oh so something like f(x) = -x+0.5 for x<0.5, f(x)=0 for x>=0.5

    g(x) = x - 0.5 for x>0.5, g(x) = 0 for x<=0.5
     
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