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Let $R$ be a ring with unity and $a, b \in R$. Assume $a$ and $b$ are not zero divisors. Show that $a$ and $b$ are units, if $ab$ is a unit.

Clearly, $a$ has a right inverse etc., but $R$ is not required to be commutative. Evidently, it's necessary to use the condition that the elements are not zero divisors, but I'm not getting anywhere.

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Since $ab$ is a unit, there is an element $c\in R$ such that $abc=1=cab$. To show that $a$ is a unit, it is enough to show that $bca=1$, for then $bc$ is both a left and right inverse of $a$.

Starting with $abc=1$ and multiplying both sides on the right by $a$, we obtain $abca=a$, or $a(bca-1)=0$. Since $a$ is not a zero divisor, this implies that $bca=1$.

A similar argument, starting with $cab=1$, shows that $b$ is also a unit.

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  • $\begingroup$ One of those "speed answering" things just happened here XD $\endgroup$ – Patrick Da Silva Oct 6 '16 at 3:49
  • $\begingroup$ We even picked the same name for the inverse! $\endgroup$ – carmichael561 Oct 6 '16 at 3:49
  • $\begingroup$ Well that's not big news $\endgroup$ – Patrick Da Silva Oct 6 '16 at 3:50
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So there exists $c \in R$ such that $(ab)c = c(ab) = 1$. Therefore $a(bc) = 1$ and $(ca)b = 1$. If $(bc)a \neq 1$, then $d \overset{def}= bca-1$ is such that $db = (bca-1)b = b( c(ab)) - b = b-b = 0$, thus $b$ is a zero divisor, a contradiction. A similar argument shows $a$ is a unit.

Hope that helps,

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