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Abstract Proof

  • Thread starter kuahji
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  • #1
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Let R be a ring with multiplicative identity 1R. Suppose that R is finite. The elemets xy1, xy2,...xyn are all different. So x y_i=1R for some i.

A lemma that is not proven is given. If xyi=1R & yjx=1R, then yi=yj

I need to show that yjx=1R.

Right now I haven't got much. I took the contrapositive of the lemma, but I still get stuck as I'm not sure where I could go from there with the information that I'm given.

The book gives a theorem which states Let R be a ring with identity and a, b of R. If a is a unit each of the equations ax=b & ya=b has a unique solution in R.

Then it goes on to state that if the ring is not commutative, x may not be equal to y. But yeah, I'm still stuck.
 
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Answers and Replies

  • #2
tiny-tim
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Hi kuahji! :smile:

(try using the X2 tag just above the Reply box :wink:)

Hint: what's yjxyi ? :smile:
 
  • #3
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Hello,

yjxyi=yj1R

The teacher also gave a proof for the lemma.

yi=(yjx)i=yj(xyi)=yj1R=yj

Except of course we haven't shown yjx=1R yet.

Still stuck, can't make the intuitive leap.
 
  • #4
tiny-tim
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yi=(yjx)i=yj(xyi)=yj1R=yj
Hello,

I assume you mean yi=(yjx)yi=yj(xyi)=yj1R=yj ?
Except of course we haven't shown yjx=1R yet.
But that's the definition of yj

yj is defined as the left inverse of x, and yi is defined as the right inverse …

so you're home. :smile:
 
  • #5
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Ok, sorry I don't see how it's the definition of yj.

I mean with your hint we have
yjayi=yj1R

But that is still different from yjx=1R
 
  • #6
tiny-tim
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Ok, sorry I don't see how it's the definition of yj.
Because of …
A lemma that is not proven is given. If xyi=1R & yjx=1R, then yi=yj
… yjx=1R is given. :wink:
 
  • #7
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hehe but that lemma is what we're trying to prove. We can't use it to prove itself. But I got it now, had to work some algebraic magic. The "proof" of the lemma only showed that yi=yj. We were already given xyi=1R. So we still had to show the one part. Thanks for the help.
 

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