Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.1 Exercise 7.1.28
Let $R$ be a ring with $1 \neq 0$. A nonzero element $a \in R$ is called a left zero divisor in $R$ if there is a nonzero element $x \in R$ such that $ax = 0$. Symmetrically, $b \neq 0$ is called a right zero divisor in $R$ if there is a nonzero element $y \in R$ such that $yb = 0$. (So a zero divisor is an element which is either a left or a right zero divisor, or both.) An element $u \in R$ is said to have a left inverse in $R$ if there is some $s \in R$ such that $su = 1$. Symmetrically, $v$ has a right inverse if there exists $t \in R$ such that $vt = 1$.
(1) Prove that $u \in R$ is a unit if and only if it has both a right and a left inverse.
(2) Prove that if $u$ has a right inverse then $u$ is not a right zero divisor.
(3) Prove that if $u$ has more than one right inverse then $u$ is a left zero divisor.
(4) Prove that if $R$ is a finite ring then every element that has a right inverse is a unit (i.e. has a two sided inverse).
Solution:
(1) First, suppose $u \in R$ is a unit. Then there exists $v \in R$ such that $uv = vu = 1$; thus $v$ is both a left and a right inverse for $u$.
Suppose now that $uv = 1$ and $wu = 1$. (That is, that $v$ is a right inverse of $u$ and $w$ is a left inverse.) Now $$v = 1 \cdot v = w \cdot u \cdot v = w \cdot 1 = w,$$ so that $v = w$. Thus $u$ has a two sided inverse, and thus is a unit.
(2) Suppose $u \neq 0$ has a right inverse and is a right zero divisor; that is, $uv = 1$ for some $v \in R$ and $zu = 0$ for some $z \in R$, where $z \neq 0$. Then $0 = zu$ implies $0 = zuv$, so that $z = 0$, a contradiction. So if an element has a right inverse, it is not a right zero divisor.
(3) Suppose $u \in R$ has distinct right inverses $v$ and $w$, that is, $uv = uw = 1$, and $v \neq w$. Now $u \neq 0$ and $v-w \neq 0$, but $$u(v-w) = uv-uw = 1-1 = 0,$$ so that $u$ is a left zero divisor.
(4) Let $R$ be a finite ring. Let $u \in R$ have a right inverse; say $uv = 1$. Now define $\varphi : R \rightarrow R$ by $\varphi(x) = xu$. Suppose $\varphi(x) = \varphi(y)$; then $xu = yu$, so that $xuv = yuv$, and we have $x = y$. Thus $\varphi$ is injective. Since $R$ is finite, in fact $\varphi$ is surjective. Thus $\varphi(x) = 1$ for some $x \in R$, and we have $xu = 1$. Since $u$ has a left and right inverse, by part 1 above, $u$ is a unit in $R$.
