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Homework Help: Sum of Ideals with Empty Intersection

  1. Jul 3, 2012 #1
    1. The problem statement, all variables and given/known data
    If a ring R contains two ideals B and C with B+C=R and B[itex]\cap[/itex]C=0, prove that B and C are rings and R[itex]\cong[/itex]B x C.


    2. Relevant equations
    B+C={all b+c|b[itex]\in[/itex]B and c[itex]\in[/itex]C}


    3. The attempt at a solution
    So far I've discovered that if the unit of R is in one of the ideals then that ideal is all of R, while the other is just {0}. Unfortunately this doesn't help much because I can't guarantee the unit of R is in either ideal. In fact I'm having trouble showing that anything is in B or C. I've been trying to find units for B and C (distinct from the unit of R since while B and C are ring,s they are not necessarily subrings). Is this a faulty approach? I can't think of any other approach...
     
  2. jcsd
  3. Jul 3, 2012 #2
    First of all, what do you mean with ring?? Is a ring with you always unital? Commutative?
    What is an ideal? Is it always a two-sided ideal?
     
  4. Jul 3, 2012 #3
    My text defines a ring to be unital but not necessarily commutative, and an ideal to be a two-sided ideal unless otherwise stated.
     
  5. Jul 3, 2012 #4
    OK, so we wish to find units in B and C. Well, we can certainly write

    [tex]1=b+c[/tex]

    for unique b and c. I claim that the b and c are units of B and C respectively. Can you show this?
     
  6. Jul 3, 2012 #5
    Well for b'[itex]\in[/itex]B and c'[itex]\in[/itex]C,

    b'1=b'(b+c)=b'b+b'c=b'b (since b'c is an element of B[itex]\cap[/itex]C, and is thus 0)

    so b'=b'b, and similar arguments show
    b'=bb'
    c'=c'c
    c'=cc'

    so that b is the unit in B and c is the unit in C.

    Given that we can write r=b+c for unique b and c, this and the rest of the problem is clear, and fairly straightforward ([itex]\phi[/itex](r)=(b,c) where b and c are the unique elements of B and C respectively s.t. r=b+c, and then injectivity and surjectivity are both pretty quick).

    I hadn't thought originally that r=b+c gave unique b and c. However, I can show this:

    Suppose r=b+c=b'+c'. Then (b-b')=(c'-c)[itex]\in[/itex](B[itex]\cap[/itex]C), so that (b-b')=(c'-c)=0 and b=b' and c=c'.

    Is there something that I should have picked up on that would have allowed me to see this immediately? Should I generally check uniqueness whenever I hear "___ can be expressed as ___"? Thanks!
     
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