Ring with unity. Show that if ab units, then a and b are units.

Question

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.

Answer 1

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

Answer 2

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 $rab=1_R$).

EDIT. My answer was wrong: I assumed that $r$ must be a two-sided inverse, which is not necessarily true. I didn't even use the fact $a,b$ are not zero divisors.

Instead of deleting my answer, let me give an example that shows it is necessary to assume $a$ and $b$ are not zero divisors, which I think it also interesting.

Let $L$ be the ring $\Bbb{R}^\Bbb{N}$, with pointwise addition and multiplication. Let $R$ be a ring of linear transormations of $L$, where multiplication is composition (so $r_1r_2$ is $r_1 \circ r_2$). Now, take $b$ to be the transformation $(r_1,r_2,...)\mapsto (0,r_1,r_2,...)$, and take $a$ to be the transformation $(r_1,r_2,...)\mapsto(r_2,r_3,...)$. Then $ab=\mathrm{id}$ is a unit, but neither $a$ nor $b$ are units (since an element in $R$ is a unit if and only if it is bijective, and it is easy to see $a$ and $b$ are not).

Answer 3

$(ab)u=1$ 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, $bua=1$

Answer 4

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 $\,c \ \ \rm\color{#0a0}{cancellable}$