Proving the Unproven: A Finite Ring with Identity

[kuahji]

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 xy_i=1_R & y_jx=1_R, then y_i=y_j

I need to show that y_jx=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.

 

[tiny-tim]

Hi kuahji!

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

Hint: what's y_jxy_i ?

 

[kuahji]

Hello,

y_jxy_i=y_j1_R

The teacher also gave a proof for the lemma.

y_i=(y_jx)_i=y_j(xy_i)=y_j1_R=y_j

Except of course we haven't shown y_jx=1_R yet.

Still stuck, can't make the intuitive leap.

 

[tiny-tim]

y_i=(y_jx)_i=y_j(xy_i)=y_j1_R=y_j

Hello,

I assume you mean y_i=(y_jx)y_i=y_j(xy_i)=y_j1_R=y_j ?

Except of course we haven't shown y_jx=1_R yet.

But that's the definition of y_j …

y_j is defined as the left inverse of x, and y_i is defined as the right inverse …

so you're home.

 

[kuahji]

Ok, sorry I don't see how it's the definition of y_j.

I mean with your hint we have
y_jay_i=y_j1_R

But that is still different from y_jx=1_R

 

[tiny-tim]

Ok, sorry I don't see how it's the definition of y_j.

Because of …

A lemma that is not proven is given. If xy_i=1_R & y_jx=1_R, then y_i=y_j

… y_jx=1_R is given.

 

[kuahji]

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 y_i=y_j. We were already given xy_i=1_R. So we still had to show the one part. Thanks for the help.