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Modules over an Integral Domain

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Homework Statement


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


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.
 

Answers and Replies

  • #2
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Assume that [tex]R=K\oplus L[/tex]. Take [tex]k\in K[/tex] and [tex]l\in L[/tex]. Is it ever possible that [tex]ak=bl[/tex] for some [tex]k,l\in R[/tex]??

If this doesn't help you, let's simplify the question: is it possible that k=l?
 
  • #3
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Assume that [tex]R=K\oplus L[/tex]. Take [tex]k\in K[/tex] and [tex]l\in L[/tex]. Is it ever possible that [tex]ak=bl[/tex] for some [tex]k,l\in R[/tex]??

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.
 
  • #4
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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?
 
  • #5
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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: [tex] ak-bl=0 [/tex]. But this is just a trivial restatement of what you said so I must be missing something here.
 
  • #6
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Ok, let's simplify it a bit: can it be true that 2k=l?? (assume for a second that [tex]2\in R[/tex]). Under what conditions can this be true??
 
  • #7
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Ok, let's simplify it a bit: can it be true that 2k=l?? (assume for a second that [tex]2\in R[/tex]). Under what conditions can this be true??
Yes but this implies [tex]k,l \in K \cap L=\lbrace 0 \rbrace [/tex] does it not?
 
  • #8
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Yes, if 2k=l, then k=0 and l=0.

Now, the general case: what if ak=bl??
 
  • #9
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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.
 
  • #10
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Maybe you can prove that K and L are in fact ideals if [tex]R=K\oplus L[/tex]...
So if a is in R, then ak is in K...
 
  • #11
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Maybe you can prove that K and L are in fact ideals if [tex]R=K\oplus L[/tex]...
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 [tex]K \ni k=(k,0)\in K \oplus L, R \ni r=(k_1,l_1) \in K\oplus L [/tex]. Then, [tex] rk = (k_1,l_1)(k,0)=(kk_1,0) = kk_1\in K [/tex]. Thus, K is an ideal.

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

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