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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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4 Answers 4

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$$x=(xy)x=x(yx)$$

Then $x(1-yx)=0$.

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  • $\begingroup$ Here we've used the distributive property of our ring, of course. $\endgroup$
    – user459879
    Apr 10, 2019 at 15:11
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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 $e - 1 = 0$.

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    $\begingroup$ It's not necessarily an integral domain because it's not said that the ring is commutative. The proof is fine nevertheless. $\endgroup$
    – lhf
    Jul 19, 2013 at 1:46
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    $\begingroup$ What do you actually call a ring without zero divsors, in general? I just call them integral domains, but that might be confusing without giving an explicit definition. $\endgroup$ Jul 19, 2013 at 1:49
  • $\begingroup$ Usually one requires $1\neq 0$ in an integral domain, i.e., the ring should be non-zero. So your proof is fine except in the case $1=0$, in which case, there's nothing to prove. $\endgroup$ Jul 19, 2013 at 21:47
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Hints:

  • If $xy=1$, then $xyx=x$.
  • If $xyx=x$, then $x(yx)=x1$.
  • Cancellation law? Why ok?
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  • $\begingroup$ could you explain why you can use cancellation law? because then you would have to multiply from the left with the inverse of $x$, but we dont know if $x$ has an inverse. $\endgroup$
    – Badshah
    Jul 16, 2013 at 21:42
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    $\begingroup$ @Badshah: If $ab=ac$ where $a\neq0$, then $a(b-c)=0$ and we can deduces $b-c=0$ as we are in an integral domain. No need to have an inverse!! $\endgroup$ Jul 16, 2013 at 22:03
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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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  • $\begingroup$ I've seen domains defined as commutative with unity as well, for example see "An introduction to Abstract Algebra" by Derek J.S Robinson. In the other direction, integral domains have been defined simply as rings without zero divisors, as seen in "Basic Abstract Algebra", Bhattacharya. This situation kind of reminds me of $\subseteq$ vs. $\subset$, which might be the same thing or not, depending on the text. In this situation, however, we can use $\subseteq$ and $\subsetneq$ - these are not ambigious. I wish there were similar conventions for rings without zero divsors. $\endgroup$ Jul 23, 2013 at 0:05

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