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Suppose that R is a ring with unity, $a, b \in R$ and that neither a nor b is a zero divisor. If ab unit, then a and b are units.

I know that a ring with unity means that

$$\exists 1_{R} \in R:\forall a \in R: 1_{R}a = a = a1_{R}.$$

For an element $a \in R$ to be a zero divisor we must have $a \neq 0_{R}$ and there must exist an element $\hat{a} \neq 0_{R}$ such that $a \hat{a} = 0_{R}$ or $\hat{a}a = 0_{R}$. Therefore, the meaning of not a zero divisor would be that given $a \neq 0_{R}$ we must have for every $\hat{a} \neq 0_{R}$ that $a\hat{a} \neq 0_{R}$ and $\hat{a}a \neq 0_{R}$. Similarly, for b we then have $b\hat{b} \neq 0_{R}$ and $\hat{b}b \neq 0_{R}$.

For $ab$ to be a unit there exists $(ab)^{-1} \in R$ such that $(ab)^{-1}ab = 1_{R} = ab(ab)^{-1}$.

Now, I have to show that $\exists a^{-1}, b^{-1} \in R$ such that $a^{-1}a = 1_{R} = aa^{-1}$ and $b^{-1}b = 1_{R} = bb^{-1}$.

I have no starting point thus far so any hint(s) are greatly appreciated.


marked as duplicate by André 3000, Community Oct 12 '18 at 18:40

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


You already have $a\underbrace{b(ab)^{-1}}_{a^{-1}?} = 1_R$. The question is, can we also show that $b(ab)^{-1}a = 1_R$ to satisfy the other condition for an inverse?

This is where the zero divisor condition comes in. Consider the following product: $$ab(ab)^{-1}a = 1_R a = a \implies ab(ab)^{-1}a - a = 0_R.$$ Then, we have $a(b(ab)^{-1}a - 1_R) = 0_R$. Since $a$ is not a zero divisor, we may conclude that $b(ab)^{-1}a - 1_R = 0$, which is what we need. Hence $a^{-1} = b(ab)^{-1}$.

A very similar method works to show that $b^{-1} = (ab)^{-1}a$.

  • $\begingroup$ How do you conclude for the right inverse $aa^{-1} = 1_{R}$ that $a^{-1} = b(ab)^{-1}$ without any further justification? Surely we cannot simply just multiply $ab(ab)^{-1} = 1_{R}$ on the left with $a^{-1}$? $\endgroup$ – salad salad Oct 12 '18 at 16:45
  • 1
    $\begingroup$ A two-sided inverse for $a$ is an element $x$ of the ring such that $ax = xa = 1_R$. If you can find such an $x$, it is a two-sided inverse for $a$, by definition. As it turns out, these inverses are unique, and we can refer to them using the $a^{-1}$ notation without fear of confusion. My answer shows how replacing $x$ with $b(ab)^{-1}$ will satisfy this definition. In other words, $b(ab)^{-1}$ is the unique inverse of $a$. So, no, it's not about multiplying both sides by $a^{-1}$, it's about showing $b(ab)^{-1}$ meets the definition of $a^{-1}$. $\endgroup$ – Theo Bendit Oct 12 '18 at 16:58

$(ab)u=1$ implies that $bu$ is a right unit of $a$, you have $a(bu)a=a$ implies that $a(bua-1)=0$, since $a$ is not a divisor of zero, $bua=1$


Since $ab$ is a unit, there is some $r$ such that $rab=abr=1_R$. Thus $br=a^{-1}$ (since $abr=1_R$) and $ra=b^{-1}$ (since $rab=1_R$).

  • 4
    $\begingroup$ This only gives you a one-sided inverse, doesn't it? Surely a little more justification is necessary to show $bra = 1_R$? $\endgroup$ – Theo Bendit Oct 12 '18 at 15:57

Hint $\ $ It is the special case $\,d=1\,$ of the following (and its right-side analog)

Lemma $\ $ If $\,\rm\color{#0a0}{cancellable}$ $\,c\,$ left-divides $\,d\,$ then $\,c\,$ right-divides $\,d\ $ if $\ \color{#c00}{c\ \&\ d\ \rm commute}$

Proof $\quad cb = d\,\overset{\large \times\ c}\Rightarrow\, cbc = \color{#c00}{dc=cd}\,\Rightarrow\, bc = d\,$ by $\,c \ \ \rm\color{#0a0}{cancellable}$

  • $\begingroup$ Note that it can also be viewed as descent of commutativity from a multiple. $\endgroup$ – Bill Dubuque Oct 12 '18 at 16:59

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