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Product ideals that yield reduced quotients

  1. Jun 17, 2011 #1
    1. The problem statement, all variables and given/known data

    Let I,J be ideals in a commutative ring R. Assume that R/(IJ) is reduced (so that it contains no nilpotent elements). Prove that [itex] IJ = I \cap J [/itex]

    2. Relevant equations

    For two ideals I and J, we have
    [tex] IJ = \left\{ a_1 b_1 + \cdots + a_n b_n : a_i \in I, b_i \in J, i=1,\ldots,n \ n \in \mathbb N \right\} [/tex]

    3. The attempt at a solution

    I've been thinking about this problem for some time. I think I've narrowed it down to show a small result, but perhaps someone could comment on whether I'm thinking in the right direction.

    It seems unlikely that we will be able to use the fact that R/(IJ) is reduced to create a constructive proof, so we move to prove via contradiction. Since it is always true that [itex] IJ \subseteq I \cap J [/itex] let us assume for the sake of contradiction that the inclusion is strict, and show that there is a nilpotent element in R/(IJ).

    Now by looking at the definition of the product ideal, it seems that the best choice for a candidate element would be to find two elements [itex] s,t \in I \cap J [/itex] such that [itex] (s+t) \in (I \cap J)\setminus IJ [/itex] if such an element exists (this is what I have yet to prove). Then the projection map [itex] \pi: R \to R/(IJ) [/itex] would yield
    [tex] \pi(s+t) = (s+t) + IJ \neq IJ [/tex]
    because we chose s+t to not be in IJ. On the other hand
    [tex] [(s+t) + IJ]^2 = (s+t)^2 +IJ = s^2 + st + ts + t^2 + IJ = IJ [/tex]
    where the equality to IJ follows because s and t are both in I and J, and the ring is commutative. Hence [itex] (s+t)^2 + IJ = IJ [/itex] and so this is a nonzero nilpotent element which is a contradiction.

    I think this is all correct. However, I'm a bit stuck on showing that [itex] \exists s, t \in I \cap J, (s+t) \in (I\cap J)\setminus IJ [/itex]. Certainly, we know there exists at least one element, so it remains to show that there is a second, and that we can choose them such that the sum is not in the product ideal. Any thoughts? I'm also open to other suggestions.
     
  2. jcsd
  3. Jun 17, 2011 #2

    micromass

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    Hi Kreizhn! :smile:

    If [itex]a\in I\cap J\setminus IJ[/itex], isn't a+IJ nilpotent then?
     
  4. Jun 17, 2011 #3
    Oh yes, quite obviously. Since then [itex] (a +IJ)^2 = a^2 + IJ = a\cdot a + IJ [/itex] and [itex] a \cdot a \in IJ [/itex] since [itex] a \in I [/itex] and [itex] a \in J [/itex]. How did I miss that? Thanks.
     
  5. Jun 17, 2011 #4
    While I'm here, I have another quick question.

    I want to show that if R is a commutative ring, a in R, and [itex] f_1(x),\ldots, f_r(x) \in R[x] [/itex] then the following ideals are equal
    [tex] (f_1(x),\ldots, f_r(x),x-a) = (f_1(a),\ldots, f_r(a),x-a) [/tex]
    Now this is fairly easy if R[x] is a Euclidean domain, since I can then write
    [tex] f_i(x) = q_i(x)(x-a) + f_i(a) [/tex]
    and make appropriate substitutions. However, I am only told that R is commutative, which if memory serves, is not sufficient for R[x] to be Euclidean. Is there another way of doing this, or is the statement incorrect in that R needs to be a field?
     
  6. Jun 17, 2011 #5

    micromass

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    Hmm, you can't use the division algorithm here. But to show that

    [tex](f_1(x),\ldots, f_r(x),x-a) = (f_1(a),\ldots, f_r(a),x-a)[/tex]

    It suffices to show that

    [tex]R[x]/(f_1(x),\ldots, f_r(x),x-a) = R[x]/(f_1(a),\ldots, f_r(a),x-a)[/tex]

    which shouldn't be hard to show... (intuitively, R[x]/(x-a) is just exchanging all the x's with a's).
     
  7. Jun 17, 2011 #6
    That's actually what I'm asked to show in the next part of the question, so it seems like I'm supposed to do it a different way. :confused:

    Edit: Well, the next question actually asks me to go one step further and show that
    [tex] R[x]/(f_1(x),\ldots, f_r(x),x-a) \cong R/(f_1(a),\ldots, f_r(a)) [/tex]
    which is just using a corollary to the third iso theorem.
     
  8. Jun 17, 2011 #7

    micromass

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    Ah, I see. Well, maybe you can show that

    [tex]f_i(x)=q_i(x)(x-a)+f_i(a)[/tex]

    holds in general (so without referencing the division algorithm). This comes down to showing that if g(a)=0, then (x-a) divides g(x).
     
  9. Jun 17, 2011 #8
    I'll give both ways a shot, just for the practice. Thanks.
     
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