# Ring with unity. Show that if ab units, then a and b are units. [duplicate]

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

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 $$b^{-1} = (ab)^{-1}a$$.

• 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 $a^{-1}$? – salad salad Oct 12 '18 at 16:45
• 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 $a^{-1}$. – 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 $$rab=1_R$$).

• This only gives you a one-sided inverse, doesn't it? Surely a little more justification is necessary to show $bra = 1_R$? – 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 $$\,c \ \ \rm\color{#0a0}{cancellable}$$

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