Proving units in a ring 
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.
 A: 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 $b$.
A: 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.
A: Hint $\ $ Conjugate a one-sided inverse $\,bc=1\,$  to the other side via $\ (bc\!-\!1)b\, =\, b(cb\!-\!1)$
