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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.

Caleb Stanford
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D-Man
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3 Answers3

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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.

Jyrki Lahtonen
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    +1 for actually using the fact the ring has no zero divisors... – Caleb Stanford Apr 27 '14 at 19:30
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    @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
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    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. – Caleb Stanford 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
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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 $b$.

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

Bill Dubuque
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