On a proof of Kaplansky’s Theorem

On a proof of Kaplansky's Theorem

In a ring with identity, an element without a left inverse but with at least one right inverse has infinitely many right inverses.

Let $b$ be an element of the ring. Denoting "multiplication" by juxtaposition, the left inverses of $b$ are the solutions of the equation

(0) $x : \left( {{xb} = 1}\right)$

we assume that (0) has no solutions. The right inverses of $b$ are the solutions of the equation

$x : \left( {b - x = 1}\right)$

to ${a}_{k}$ for $0 \leq k < N$ be $N$ distinct solutions

(2) ${b}{a}{k} = 1$ ${a}{h} = {a}{k} \Rightarrow h = k$

We shall show that if $N > 0$ , (1) has at least $N + 1$ solutions. From now on, we assume $N > 0$ , i.e. the existence of $a_0$ .

Consider the quantities ${c}_{k}$ given by

[{c}{k} = {a}{0} + 1 - {a}_{k}b]

Lemma 0 Each $c_{k}$ is a right inverse of $b$

Proof We observe for any $k$ , $0 \leq k < N$

$$ \begin{array}{l} b c _ {k} \ = {(4) } \ b (a _ {0} + 1 - a _ {k} b) \ = {\mathrm {r i n g p r o p e r t i e s} } \ b a _ {0} + b ^ {\prime} - b a _ {k} b \ = \left{ \begin{array}{l l} (2) & \text {f w i c e} \end{array} \right} \ 1 + b - 1 b \ = {r i n g p r o p e r t h i e s } \ 1. \ \end{array} $$

(End of Proof.)

Lemma 1 All the $c_{k}$ are distinct.

Proof We observe for any $h,k,0 \leq k < N$

$$ \begin{array}{l} c _ {n} = c _ {k} \ = {(4 _ {1}) } \ a _ {0} + 1 - a _ {h} b = a _ {0} + 1 - a _ {k} b \ = {\text {r i n g p r o p e r t i e s} } \ a _ {h} b = a _ {k} b \ \Rightarrow \quad {\text {L e i b n i z} } \ a _ {h} b a _ {0} = a _ {k} b a _ {0} \ = {2) w i t h k := 0, f w i c e } \ a _ {k} ^ {1} = a _ {k} ^ {1} \ = {\mathrm {r i n g p r o p e r t i e s} } \ a _ {h} = a _ {k} \ \Rightarrow \quad {(3) } \ h = k \ \end{array} $$

(End of Proof.)

Lemma 2 Each $c_{k}$ differs from $a_{0}$

Proof We observe for any $k$ , $0 \leq k < N$

$$ \begin{array}{l} a _ {0} \neq c _ {k} \ = {4 } \ a _ {0} \neq a _ {0} + 1 - a _ {k} b \ = {r i n g p r o p e r t i e s } \ a _ {k} b \neq 1 \ = {0) h a s n o s o l u t i o n s } \ t r u e \ \end{array} $$

(End of Proof.)

Consider now $a_0$ and the N values $c_{k}$ From (2) and Lemma 0 we conclude that these $N + 1$ values are all right inverses of b. From Lemmata 1 and 2 we conclude that these $N + 1$ values are all distinct. QED

The above has been triggered by "A Pigeonhole Proof of Kaplansky's Theorem" by Ira Rosenholtz, published in The American Mathematical Monthly, Volume 99, Number 2, February 1992, p. 132-133. A copy of Rosenholtz's contorted argument is included. It was a pure coincidence that I encountered this article in the AMM a few hours after I had announced in my course "Mathematical Methodology" that next week I would lecture on the Pigeonhole Principle "because it is so frequently misused". Besides the spurious use

A Pigeonhole Proof of Kaplansky's Theorem

Ira Rosenholtz

The purpose of this little note is to sketch a simple proof of the following result, which Kaplansky has referred to as his "infamous little exercise". (See [1], [2], [3], [4].)

Theorem (Kaplansky). Suppose that an element in a ring with identity has two right inverses. Then it has infinitely many right inverses.

The proof consists of the following two lemmas. It is analogous to solving linear differential equations and is a nice application of the pigeonhole principle.

Lemma 1 (The Homogeneous Solution). If $b$ has $N$ right inverses with $N$ at least 2, then the equation $bx = 0$ has at least $(N + 1)$ solutions.

Proof of Lemma 1: Suppose $b$ has distinct right inverses $a_1, a_2, \ldots, a_N$ . Then $a_1 - a_1, a_2 - a_1, \ldots, a_N - a_1$ are $N$ distinct solutions of $bx = 0$ . We will show that the set ${1 - a_1b, 1 - a_2b, \ldots, 1 - a_nb}$ contains at least one additional solution of $bx = 0$ .

Clearly all of the elements of this set are solutions. If there were not a new solution in this set, then for each $j$ there is a $k$ so that $1 - a_{j}b = a_{k} - a_{1}$ . However, $1 - a_{j}b$ cannot equal $a_{1} - a_{1} = 0$ , because then $a_{j}$ would be a left inverse for $b$ , and in this case it is easy to see that $b$ has only one right inverse, a contradiction. Thus, since there are $N(1 - a_{j}b)$ 's (the pigeons) and only $(N - 1)$ acceptable $(a_{k} - a_{1})$ 's (the pigeon-holes), by the pigeon-hole principle we must have that for some $m \neq n$ , $1 - a_{m}b = 1 - a_{n}b$ . But then $a_{m}b = a_{n}b$ , and multiplying this on the right by $a_{1}$ , we get $a_{m} = a_{n}$ , a contradiction.

Lemma 2 (The Non-Homogeneous Solution). If $b$ has $N$ right inverses with $N$ at least 2, then $b$ has $(N + 1)$ right inverses.

Proof of Lemma 2: By Lemma 1, $bx = 0$ has $(N + 1)$ distinct solutions $x_{1}, x_{2}, \ldots, x_{N + 1}$ . But then if $a_{1}$ is a right inverse of $b$ , then ${a_{1} + x_{1}, a_{1} + x_{2}, \ldots, a_{1} + x_{N + 1}}$ is a set of $(N + 1)$ distinct right inverses of $b$ .

of the Pigeonhole Principle, it contains a nested reductio ad absurdum!

Austin, 2 February 1992

prop. dr. Edsger W. Dgksha Departmen t of Compu er Scicnces The University of Texas at Austin Austin, TX 78712-1188, USA