# Sum of Ideals with Empty Intersection

1. Jul 3, 2012

### wnorman27

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$\cap$C=0, prove that B and C are rings and R$\cong$B x C.

2. Relevant equations
B+C={all b+c|b$\in$B and c$\in$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. Jul 3, 2012

### micromass

Staff Emeritus
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?

3. Jul 3, 2012

### wnorman27

My text defines a ring to be unital but not necessarily commutative, and an ideal to be a two-sided ideal unless otherwise stated.

4. Jul 3, 2012

### micromass

Staff Emeritus
OK, so we wish to find units in B and C. Well, we can certainly write

$$1=b+c$$

for unique b and c. I claim that the b and c are units of B and C respectively. Can you show this?

5. Jul 3, 2012

### wnorman27

Well for b'$\in$B and c'$\in$C,

b'1=b'(b+c)=b'b+b'c=b'b (since b'c is an element of B$\cap$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 ($\phi$(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)$\in$(B$\cap$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!