Need help with invertibility proof

• Mar 24th 2013, 05:11 AM
rain1
Need help with invertibility proof
Hello.

I would use some help with following task:
Prove that when in ring R products xy and yx are invertible then elements x and y are also invertible.

• Mar 24th 2013, 11:13 AM
Siron
Re: Need help with invertibility proof
What does it mean by definition that $\displaystyle xy$ is invertible?

If you have the definition then you can use the fact that the multiplication is an associative operation.
• Mar 24th 2013, 11:15 AM
rain1
Re: Need help with invertibility proof
It means that there exists (xy)^-1 so that (xy)^-1 * xy = 1
• Mar 24th 2013, 11:29 AM
Siron
Re: Need help with invertibility proof
Quote:

Originally Posted by rain1
It means that there exists (xy)^-1 so that (xy)^-1 * xy = 1

Yes, $\displaystyle xy$ is invertible is there exists an unique $\displaystyle z \in R$ such that $\displaystyle (xy)z=1$.
Now, use the associtativity.
• Mar 24th 2013, 11:35 AM
rain1
Re: Need help with invertibility proof
You mean like this?$\displaystyle (xy)^{-1}\cdot (xy) = x\cdot (y^{-1}\cdot y)\cdot x^{-1} = x\cdot 1x^{-1} = 1 \cdot 1 =1$? Thats all?
• Mar 24th 2013, 02:13 PM
Siron
Re: Need help with invertibility proof
Quote:

Originally Posted by rain1
You mean like this?$\displaystyle (xy)^{-1}\cdot (xy) = x\cdot (y^{-1}\cdot y)\cdot x^{-1} = x\cdot 1x^{-1} = 1 \cdot 1 =1$? Thats all?

In fact, what I mean is the following.
Suppose $\displaystyle xy$ is invertible then there exists an unique $\displaystyle z \in R$ such that $\displaystyle (xy)z=1$. Since $\displaystyle 1=(xy)z=x(yz)$ (here I use the associativity), we find that $\displaystyle x$ is invertible with $\displaystyle yz$ as the inverse element.