# Proving units in a ring

Suppose $R$ is a ring with no zero divisors and with identity $1_R$ not equal to $0_R$. Suppose that $a,b$ are in $R$ and that $ab$ is a unit. Prove that $b$ is a unit.

My thoughts: I know a unit is basically a unit that (for this example) would mean $abu = 1_R$ for some nonzero $u$ in $R$. I am really stuck after that. Not seeing a clear path to manipulate the variables to prove b is a unit by itself.

-
Well, write it this way: $\;1=u(ab)=(ua)b\implies b\;$ is a unit...:) – DonAntonio Apr 27 '14 at 19:00
@Don That yields only a left inverse $\,c = ua\,$ for $\,b.\,$ But conjugation shows it is a right inverse too - see my answer. – Bill Dubuque Apr 27 '14 at 23:25

As the others have pointed out the calculation $$1=u(ab)=(ua)b$$ shows that $ua$ is a left inverse to $b$. Consider the product $b(ua)$. We have $$ua=1(ua)=((ua)b)(ua)=(ua)(b(ua)),$$ so $$(ua)(1-b(ua))=0.$$ As the ring has no zero divisors this implies that either $ua=0$ or $b(ua)=1$. But if $(ua)=0$, then $1=(ua)b=0$ which is a contradiction. The claim follows.

-
+1 for actually using the fact the ring has no zero divisors... – 6005 Apr 27 '14 at 19:30
@Goos: Well, the claim is false without something extra, and that was the only piece available :-) All: Sorry about spoiling the problem. I really should have come up with an appropriate hint. – Jyrki Lahtonen Apr 27 '14 at 19:45
yes, but there were two maybe three people claiming "solutions" without using that fact. So +1 for a legitimate proof that it is a unit and not just having a left inverse. – 6005 Apr 27 '14 at 20:10
@Jyrki Re: hints. See my answer for one way I often hint at it. It's not easy to give a hint for this without spilling the beans. – Bill Dubuque Apr 27 '14 at 23:05

As discussed in the comments, since $ab$ is a unit then $uab = 1$ for some $u \in R$, so $ua$ is a left inverse for $b$. It remains to show that $ua$ is also a right inverse for $b$, i.e., $bua = 1$. Taking the equation $1 = uab$ and multiplying both sides by $ua$ on the right, we have $$ua = uabua \implies 0 = ua - uabua = ua(1 - bua) \, .$$ Since $R$ has no zero divisors, then either $ua = 0$ or $1 - bua = 0$. But again, $R$ has no zero divisors, so we must have $1 - bua = 0$, hence $1 = bua$. Thus $ua$ is a two-sided inverse for $b$.

-

Hint $\$ Conjugate a one-sided inverse $\,bc=1\,$ to the other side via $\ (bc\!-\!1)b\, =\, b(cb\!-\!1)$

-