# ring without zero-divisors

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.

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

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

• Here we've used the distributive property of our ring, of course. – Wesley Strik Apr 10 at 15:11

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 $e - 1 = 0$.

• It's not necessarily an integral domain because it's not said that the ring is commutative. The proof is fine nevertheless. – lhf Jul 19 '13 at 1:46
• 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. – Nonoffensive name Jul 19 '13 at 1:49
• 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. – Keenan Kidwell Jul 19 '13 at 21:47

Hints:

• If $xy=1$, then $xyx=x$.
• If $xyx=x$, then $x(yx)=x1$.
• Cancellation law? Why ok?
• 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. – Badshah Jul 16 '13 at 21:42
• @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!! – Jyrki Lahtonen Jul 16 '13 at 22:03

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.

• 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. – Nonoffensive name Jul 23 '13 at 0:05