Sum of Ideals with Empty Intersection

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wnorman27
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Homework Statement


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.


Homework Equations


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


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