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Principal Ideals and associates

  1. May 15, 2010 #1
    I'm wondering something about principal ideals which I'm using to prove something.
    K-field, let f,h be non-constant in K[X]. If f and h are associates, does it follow (f) = (h) ?
    I tried to just prove it myself but I'm not sure if it's correct.
    f, h associates means f=ch for some unit c. Then (f) = {fg : g in K[X]} = {hcg : g in K[X]}. Now I want to put " = (h) " but I'm not sure if that's correct. I think it is, because g runs over all of K[X], and cK[X] = K[X], so {hcg : g in K[X]} = {hq : q in K[X]}. So is my statement true?
     
  2. jcsd
  3. May 15, 2010 #2

    lavinia

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    (f) = {fg : g in K[X]} = {hcg : g in K[X]} shows that any element in the ideal generated by f is in the ideal generated by g. Since c is a unit the converse is also true.
     
  4. May 15, 2010 #3
    Did you mean h?

    Also, is the equality {hcg : g in K[X]} = {hq : q in K[X]} incorrect?
     
  5. May 15, 2010 #4

    lavinia

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    yes I meant h
     
  6. May 18, 2010 #5
    a = bu, where u is a unit. also from this you get b = au^-1. obviously a is in (b), which means that (a) is a subset of (b), and b is in (a), so (b) is a subset of (a). by double inclusion, they are equal.
     
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