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Does an example exist

  1. Dec 11, 2009 #1
    Hi All,
    If I is an ideal of J and J is an ideal of R (the ring)... Is it possible that I is not an ideal of R. When are we guaranteed to have this transitivity, if at all? Thank you
     
  2. jcsd
  3. Dec 21, 2009 #2
    I'm still trying to work out the details, but would this be an example?

    Let Z be the ring of integers, then 2Z is an ideal of Z and 4Z is an ideal of 2Z. But 4Z is also an ideal of Z. I would guess that is you have the situation you described, then I would always be an ideal of R.

    Let R be a ring and let J be an ideal of R. We must first ensure that J is a ring in order for it to have any ideals. So as long as its closed under subtraction and multiplication, its a ring, which we can confirm. Next let I be an ideal of J. Then all elements in I must be in R, since I is an ideal of J and J is one of R. So, as long as we can take any element in R, multiply by an element in I and still stay in I, then I would say that I is an ideal of R. Give that a try and see what happens.
     
  4. Dec 21, 2009 #3
    In general this is not true: Consider [tex]I= \langle x^2 \rangle = \{ fx^2 : f\in J= \langle x \rangle \}[/tex] with both [tex]I[/tex] and [tex]J[/tex] considered as subsets of [tex]R= \mathbb{R} [x][/tex]. By definition [tex]I[/tex] ideal of [tex]J[/tex] and [tex]J[/tex] ideal of [tex]R[/tex], but [tex]I[/tex] is not an ideal of [tex]R[/tex] since [tex]ax^2 \notin I[/tex] for all [tex]a\neq 0,1[/tex] such that [tex]a\in \mathbb{R}[/tex] . As for conditions that guarantee transitivity I can't think of any but that [tex]J=R[/tex].
     
  5. Dec 25, 2009 #4
    Well, in all honesty, it's a bit silly to talk about "an ideal of an ideal" since we define (particularly in commutative algebra) ideals to be subsets of rings and proper ideals are not rings (ie. they are not subrings).
     
  6. Dec 25, 2009 #5

    Hurkyl

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    Some authors use the word "ring" to refer to rngs, so ideals of ideals makes sense (because any ideal of a rng is a rng).

    In what I've read I've never seen that usage -- although most authors will make an explicit statement they are not adopting that convention.

    I'm going to assume the opening poster is talking about rngs.
     
  7. Dec 25, 2009 #6

    Hurkyl

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    Jose, there is an error in your example: because J doesn't have a unit, the ideal of J generated by x2 is not equal to x2J.
     
  8. Dec 27, 2009 #7
    You're right, but couldn't we "close" it under addition and then it still won't have any, for example, irrational miltiples of [tex]x^2[/tex]?
     
  9. Dec 28, 2009 #8

    Hurkyl

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    I believe that is true.
     
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