Suppose we have a ring (could be infinite) without zero-divisors. I have to prove that if $xy=1$ then also $yx=1$ for some $x$ and $y$ in the ring. I really need hints for this, because it seems I just cant figure it out. Thank you.
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$\begingroup$$$x=(xy)x=x(yx)$$
Then $x(1-yx)=0$.
$\endgroup$ 1 $\begingroup$I always liked this proof:
Let $xy = 1$. Then $(yx)(yx) = y(xy)x = yx$, thus $yx$ is an idempotent. Therefore, since we are in an integral domain, $yx$ must be either $0$ or $1$. We have that $yx \neq 0$, because else either $x$ or $y$ is zero, implying $xy = 0$, which is absurd.
Addendum: The only idempotents in an integral domain are $0$ and $1$.
Let $e^2 = e$. Then $e^2 - e = e(e - 1) = 0$, implying that $e = 0$ or $e - 1 = 0$.
$\endgroup$ 3 $\begingroup$Hints:
- If $xy=1$, then $xyx=x$.
- If $xyx=x$, then $x(yx)=x1$.
- Cancellation law? Why ok?
I call a ring without zero divisors a domain. It need not have a multiplicative identity; if it does, I call it a domain with an identity.
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