infinitude of inverses

Proposition 1.

Let $R$ be a ring with 1.

1. If $a\in R$ has a right inverse but no left inverses, then $a$ has infinitely many right inverses.

2. If $a\in R$ has more than one right inverse, then $a$ has infinitely many right inverses.

Proof.

1. Let $ab=1$. Define $b_0=b,b_1=1-b_{0}a+b_0,\mathrm{\dots },b_{i+1}=1-b_{i}a+b_i,\mathrm{\dots }$ Then, by induction, we see that$a{b}_i=a-a{b}_{i-1}a+a{b}_{i-1}=a-a+1=1$. Next we want to show that $b_i\ne b_j$ if $i\ne j$. Suppose $i>j$ and$b_i=b_j$. Again by induction, we have

   $$b_j=b_i=1+(1-a)+\mathrm{\cdots }+{(1-a)}^{i-j-1}+b_j{(1-a)}^{i-j}$$(1)

   If we let $c=1+(1-a)+\mathrm{\cdots }+{(1-a)}^{i-j-1}$ then $(1-a)c=c(1-a)=(1-a)+{(1-a)}^2+\mathrm{\cdots }+{(1-a)}^{i-j}=c-1+{(1-a)}^{i-j}$.So Equation 3 can be rewritten as $c=b_j-b_j{(1-a)}^{i-j}=b_j(1-{(1-a)}^{i-j})=b_{j}ca$. Then $c{b}_j=b_{j}ca{b}_j=b_{j}c$.Now, note that for $m\le n$, ${(1-a)}^n{b}_j^m={(1-a)}^{n-m}{(b_j-1)}^m$. This implies that

   $c{b}_j^{i-j-1}$$=$$b_j^{i-j-1}+(b_j-1)b_j^{i-j-2}+\mathrm{\cdots }+{(b_j-1)}^{i-j-1}$

   $=$$g(b_j)+{(b_j-1)}^{i-j-1}.$

   On the other hand, we also have

   $c{b}_j^{i-j-1}$$=$$b_{j}c{b}_j^{i-j-2}$

   $=$$b_j(b_j^{i-j-2}+{(b_j-1)}^{i-j-3}+\mathrm{\cdots }+(1-a){(b_j-1)}^{i-j-2})$

   $=$$g(b_j)+b_j(1-a){(b_j-1)}^{i-j-2}.$

   So combining the above two equations, we get ${(b_j-1)}^{i-j-1}=b_j(1-a){(b_j-1)}^{i-j-2}$. Let $d={(b_j-1)}^{i-j-2}$, then $(b_j-1)d=b_j(1-a)d=b_{j}d-b_{j}ad$. Simplify, we have $d=b_{j}ad$. Expanding $d$, then

   $b_j^{i-j-2}+\mathrm{\cdots }+{(-1)}^{i-j-2}$$=$$(b_{j}a)(b_j^{i-j-2}+\mathrm{\cdots }+{(-1)}^{i-j-2})$

   $=$$b_{j}a{b}_j^{i-j-2}+\mathrm{\cdots }+b_{j}a{(-1)}^{i-j-2}$

   $=$$b_j^{i-j-2}+\mathrm{\cdots }+{(-1)}^{i-j-2}{b}_{j}a.$

   Then $1=b_{j}a$ and we have reached a contradiction.

2. For the next part, notice that if $b$ and $c$ are two distinct right inverses of $a$, then neither one of themcan be a left inverse of $a$, for if, say, $ba=1$, then $c=(ba)c=b(ca)=b$. So we can apply the same techniqueused in the previous portion of the problem. Note that if $b_{j}a=1$, then

   $$1=b_{j}a=(1-b_{j-1}a+b_{j-1})a=a-b_{j-1}{a}^2+b_{j-1}a.$$

   Multiply $b_{j-1}$ from the right, we have

   $$b_{j-1}=a{b}_{j-1}-b_{j-1}{a}^2{b}_{j-1}+b_{j-1}a{b}_{j-1}=1-b_{j-1}a+b_{j-1}$$

   Thus $b_{j-1}a=1$. Keep going until we reach $ba=1$, again a contradiction.

∎

Remark. The first part of the above proposition implies that a finite ring is Dedekind-finite.