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Abstract Algebra: Unnecessary Information in D&F Problem Statement?

  1. Jul 2, 2013 #1

    n+1

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    1. The problem statement, all variables and given/known data

    Problem 35, Section 7.3 of Dummit and Foote:
    Let I, J, and K be ideals of R.
    (a) Prove that I(J+K) = IJ+IK and IJ+IK = I(J+K).
    (b) Prove that if J [itex]\subseteq[/itex] I then I [itex]\cap[/itex] (J + K) = J + (I [itex] \cap[/itex] K).

    2. Concern/Question

    Despite the problem statement specifically stating that I, J, and K are ideals, this information is irrelevant to the proofs (only that I, J, and K are subrings of R is needed). Am I missing something or is this really just unnecessary information?

    3. The Proofs

    (a) To prove that [itex] I(J+K) = IJ + IK [/itex], we will show that each set is a subset of the other. To see that [itex]IJ + IK \subseteq I(J+K)[/itex], note that [itex]IJ, IK \subseteq I(J+K)[/itex]; since [itex]I(J+K)[/itex] is a ring, [tex] IJ + IK \subseteq I(J+K) +I(J+K) = I(J+K).[/tex]
    To see the opposite inclusion, first consider [itex]x \in I(J+K)[/itex]; [tex] x = i_1(j_1+k_1) + ... + i_n(j_n+k_n)[/tex] for [itex] i_* \in I, j_* \in J, k_* \in K, n \in \mathbb{Z}^+[/itex]. Reorganizing this equation, we get that [tex]x=i_1j_1+...+i_nj_n + i_1k_1 + ... + i_nk_n \in IJ+IK.[/tex] Since [itex]x[/itex] was arbitrary, this gives us the opposite inclusion and, thus, the desired equality. The second proof is analogous to the above one.

    (b) Once again, the equality will be proven by showing that each set is a subset of the other. Consider [itex]J+(I\cap K)[/itex]; since [itex]J \subseteq I[/itex] and [itex]I[/itex] is a ring, [itex]J+(I \cap K) \in I[/itex]. Furthermore, every element of [itex]J + (I \cap K)[/itex] can be written as the sum of an element of J and an element of K. Since [itex]I \cap (J+K)[/itex] is the largest such subring of [itex]R[/itex] that satisfies both these properties, [itex]J+(I \cap K) \subseteq I \cap (J+K)[/itex].
    Now consider [itex]x \in I \cap (J+K)[/itex]; we can rewrite [itex]x[/itex] as [itex]j+k[/itex] for [itex]j \in J[/itex] and [itex]k \in K[/itex]. Since [itex]J \subseteq I[/itex], [itex]j \in I[/itex] as well, and so since rings are closed under subtraction, [tex] k = x-j \in I [/tex]. Therefore, every [itex]x \in I \cap (J+K)[/itex] can be written as [itex]j+k'[/itex] for [itex]k' \in I \cap K[/itex]. From this, [itex]I \cap (J+K) \subseteq J+(I \cap K)[/itex] follows and, thus, so does the desired equality.
     
  2. jcsd
  3. Jul 2, 2013 #2

    verty

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    Homework Helper

    For part (a), no second proof is needed. The claim is identical to the first. As a matter of logic, there is no second claim to be proved.

    I think the second part (of (a)) should be (I+J)K = IK + JK.
     
    Last edited: Jul 2, 2013
  4. Jul 4, 2013 #3

    n+1

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    This is correct. I can't seem to find the "edit" button for my og post, but when I do, I'll fix that typo.

    Also, bump.
     
  5. Jul 10, 2013 #4

    n+1

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    Nevermind

    NEVERMIND

    Whoops! The product of two non-ideal subrings isn't even well defined! While I don't need to mention property of ideals explicitly in my proof, I'm assuming I, J, K are ideal when using terms like I(J+K).

    Sorry about bumping the thread just to say that it doesn't need to be addressed anymore.
     
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