1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Sum of Ideals with Empty Intersection
Loading...