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Ideal Rings - Abstract Algebra

  1. Apr 3, 2009 #1
    1. The problem statement, all variables and given/known data

    Suppose R is a ring and I,J is an ideal to R.

    Show (i) I+J is ideal to R. (ii) I union J is ideal to R.


    2. Relevant equations

    none
     
  2. jcsd
  3. Apr 3, 2009 #2
    Can you show that you've attempted it?
     
  4. Apr 3, 2009 #3
    Here is what I have (I think I have (ii)):

    (i)
    Part 1: Clearly if a,b are elements of I, then a+b are elements of I (since I is ideal to R) and if a,b are elements of J then a + b are elements of J (since J is ideal to R)

    If we have one element from each ideal, say a element of I/b element of J, then we must show a+b is an element of (I+J).

    a+b is an element of R since a,b is an element of R and R is closed.

    Part 2: Let a be an element of (I + J) and r is an element of R.

    Then if a is an element of I, ar is an element of I and ra is an element of I since I is ideal.

    Similarly, if a is an element of J, ar is an element of J and ra is an element of J since I is ideal.

    In either case, ar is an element of (I + J) and ra is an element of (I + J)

    (ii)
    Part 1: Assume a, b are elements in I union J.

    Then a + b is an element of I since I is ideal and a + b is an element of J since J is ideal.

    Then a + b is in both I and J and therefore a + b is an element of I union J.

    Part 2: Assume a is and element of (I union J) and r is an element of R.

    Then, ar is an element of I and ra is and element of I since I is ideal.

    Similarly, ar is an element of J and ra is an element of J since J is ideal.

    So, ar is an element of (I union J) and ra is an element of (I union J)
     
  5. Apr 3, 2009 #4
    you'll also need to show that I+J and [tex]I\cup J[/tex] are closed under multiplication and subtraction (in fact addition follows from subtraction).

    For (i) part 1, it might help to recall that since [tex](I,+,\cdot)[/tex] is an ideal of [tex](R,+,\cdot)[/tex], [tex](I,+)[/tex] is a normal subgroup of [tex](R,+)[/tex] (since [tex](R,+)[/tex] is abelian). Simlarly for J.
     
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