Modules over an Integral Domain

[Newtime]

Homework Statement

Let R be a ring with no zero divisors such that for any $$r,s\in R$$, there exist $$a,b \in R$$ such that $$ar+bs=0$$. Prove: $$R=K \oplus L$$ implies $$K=0$$ or $$L=0$$.

Homework Equations

Definition of direct sum of modules, integral domain...

The Attempt at a Solution

I didn't know where to start on this one. In particular, I don't see what the hypothesis has to do with anything and certainly I do not see how it's related to the conclusion. I think all I need is a nudge in the right direction.

 

[micromass]

Assume that $$R=K\oplus L$$. Take $$k\in K$$ and $$l\in L$$. Is it ever possible that $$ak=bl$$ for some $$k,l\in R$$??

If this doesn't help you, let's simplify the question: is it possible that k=l?

 

[Newtime]

Assume that $$R=K\oplus L$$. Take $$k\in K$$ and $$l\in L$$. Is it ever possible that $$ak=bl$$ for some $$k,l\in R$$??

If this doesn't help you, let's simplify the question: is it possible that k=l?

k=l only if k=l=0 since the sum is direct. Also, it is possible that ak=bl by hypothesis, but I don't see how this points toward a solution.

However, it seems clear that just before we say "thus, l=0" we would use the hypothesis that R is an integral domain and that bl=0.

 

[micromass]

Ok, you are correct. k=l only if k=0 and l=0.

Can you describe a similar restriction for ak=bl? I.e. what can you say about ak and bl if ak=bl?

 

[Newtime]

Ok, you are correct. k=l only if k=0 and l=0.

Can you describe a similar restriction for ak=bl? I.e. what can you say about ak and bl if ak=bl?

Well the obvious implication is that one is the additive inverse of the other: $$ak-bl=0$$. But this is just a trivial restatement of what you said so I must be missing something here.

 

[micromass]

Ok, let's simplify it a bit: can it be true that 2k=l?? (assume for a second that $$2\in R$$). Under what conditions can this be true??

 

[Newtime]

Ok, let's simplify it a bit: can it be true that 2k=l?? (assume for a second that $$2\in R$$). Under what conditions can this be true??

Yes but this implies $$k,l \in K \cap L=\lbrace 0 \rbrace$$ does it not?

 

[micromass]

Yes, if 2k=l, then k=0 and l=0.

Now, the general case: what if ak=bl??

 

[Newtime]

Yes, if 2k=l, then k=0 and l=0.

Now, the general case: what if ak=bl??

I still don't see it. The previous case (2k=l) was obvious because 2k stays in K since of course K is a ring and thus closed under addition (2k=k+k). But a and b are arbitrary elements of R so we don't know where the elements ak and bl end up and I don't think the problem is designed to be broken up into cases. Even so, considering the different subcases doesn't seem to help.

 

[micromass]

Maybe you can prove that K and L are in fact ideals if $$R=K\oplus L$$...
So if a is in R, then ak is in K...

 

[Newtime]

Maybe you can prove that K and L are in fact ideals if $$R=K\oplus L$$...
So if a is in R, then ak is in K...

I dismissed this idea earlier because I thought it would be more difficult to prove that K and L are ideals than the original claim. But here's what I have :

Let $$K \ni k=(k,0)\in K \oplus L, R \ni r=(k_1,l_1) \in K\oplus L$$. Then, $$rk = (k_1,l_1)(k,0)=(kk_1,0) = kk_1\in K$$. Thus, K is an ideal.

Assuming this is valid, the problem is complete! Thanks for your help.