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Ideals with subsets and divides

  1. Oct 22, 2008 #1
    1. The problem statement, all variables and given/known data
    Let I = <f(x)>, J =<g(x)> be ideals in F[x]. prove that I[tex]\subset[/tex]J [tex]\leftrightarrow[/tex] g(x)|f(x)

    2. Relevant equations

    3. The attempt at a solution
    If I is a subset of J then does that mean that f is in J also and by definition of an ideal g*some b in J must equal something in J so g|f? because g|f means that f=bg for some b in J
  2. jcsd
  3. Oct 22, 2008 #2

    Sort of. I would just use the fact that J is generated by g(x).
  4. Oct 23, 2008 #3
    what does it mean that J is generated by g(x)? in layman's terms
  5. Oct 23, 2008 #4
    It means J = {a(x)g(x) : a(x) in F[x]}, in other words J is the set of all "multiples" of g(x).
  6. Oct 23, 2008 #5
    So if f(x) is in J and J = {a(x)g(x): a(x) in F[x]} then f(x) = a(x)g(x) therefore g(x)|f(x)?
  7. Oct 23, 2008 #6
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