On the Problem of Elements in a Ring Having More than One One-Sided Inverse

Lemma 1. Let $R$ be a ring with identity, and let $a \in R$. Suppose that $a$ has a left inverse in $R$. Then the following are equivalent:

1. $a$ has more than one left inverse.

2. $a$ is not a unit.

3. $a$ is a right zero divisor of some element.

Proof: The implication $(1) \Rightarrow (2)$ is obvious.

We now prove $(2) \Rightarrow (3)$. Let $b$ be a left inverse of $a$. Since $a$ is not a unit, we must have

$$
ba = 1 \quad \text{but} \quad ab \neq 1.
$$

Hence

$$
(ab - 1)a = aba - a = a - a = 0.
$$

Since

$$
ab - 1 \neq 0 \quad \text{and} \quad a \neq 0,
$$

it follows that $a$ is a right zero divisor of the element $ab - 1$.

Finally, we prove $(3) \Rightarrow (1)$. Suppose that there exists $u \neq 0$ such that

$$
ua = 0.
$$

Let $b$ be a left inverse of $a$. Then

$$
(b+u)a = ba + ua = 1 + 0 = 1.
$$

Thus $b+u \neq b$ is also a left inverse of $a$. Therefore $a$ has more than one left inverse.

Hence $(1)$, $(2)$, and $(3)$ are equivalent.

Theorem (Kaplansky’s Theorem). Let $R$ be a ring with identity, and let $a \in R$. If $a$ has more than one left inverse in $R$, then $a$ has infinitely many left inverses.

Proof: Let $X$ be the set of all left inverses of $a$. Since $a$ has more than one left inverse, no element of $X$ can be a right inverse of $a$.

Fix $a_0 \in X$. Then

$$
a_0a = 1.
$$

For any $x \in X$, we have

$$
(ax - 1 + a_0)a = axa - a + a_0a = a - a + 1 = 1.
$$

Hence

$$
ax - 1 + a_0 \in X.
$$

Therefore we may define a map

$$
f : X \to X
$$

by

$$
f(x) = ax - 1 + a_0.
$$

We first show that $f$ is injective. Suppose that

$$
f(x) = f(y).
$$

Then

$$
ax - 1 + a_0 = ay - 1 + a_0,
$$

so

$$
ax = ay.
$$

Multiplying both sides on the left by $a_0$, we obtain

$$
x = y.
$$

Thus $f$ is injective.

Next we show that $f$ is not surjective. We claim that

$$
f(x) \neq a_0
$$

for every $x \in X$.

Assume to the contrary that there exists $x \in X$ such that

$$
f(x) = a_0.
$$

Then

$$
ax - 1 + a_0 = a_0,
$$

and hence

$$
ax = 1.
$$

So $x$ is also a right inverse of $a$, a contradiction.

Therefore $f$ is not surjective.

Finally, if $X$ were finite, then every injective self-map of $X$ would be surjective. This contradicts the fact that $f$ is injective but not surjective. Hence $X$ must be infinite.

Therefore $a$ has infinitely many left inverses.

Proposition 1. Let $R$ be a finite ring with identity. Then for every element of $R$, if it has a left inverse, then that left inverse must be unique.

Proof: We argue by contradiction. Suppose that $a \in R$ has more than one left inverse. Then by Kaplansky’s theorem, $a$ has infinitely many left inverses. This would force $R$ to be infinite, contradicting the assumption that $R$ is finite.

Therefore every element of $R$ has at most one left inverse.

Proposition 2. Let $R$ be a ring with identity, and let $a,b \in R$ satisfy

$$
ba = 1 \quad \text{but} \quad ab \neq 1.
$$

Then $a$ has infinitely many left inverses.

Proof: From the assumption, $a$ is not a unit. By the previous proposition, $a$ has more than one left inverse. Therefore, by Kaplansky’s theorem, $a$ has infinitely many left inverses.

The cover image of this article was taken in Irkutsk Oblast, Russia.