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 $b_{0}=b,b_{1}=1-b_{0}a+b_{0},\dots,b_{i+1}=1-b_{i}a+b_{i},\ldots$ Then, by induction, we see that $ab_{i}=a-ab_{i-1}a+ab_{i-1}=a-a+1=1$ . Next we want to show that $b_{i}\neq b_{j}$ if $i\neq j$ . Suppose $i>j$ and $b_{i}=b_{j}$ . Again by induction, we have

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

If we let $c=1+(1-a)+\cdots+(1-a)^{i-j-1}$ then $(1-a)c=c(1-a)=(1-a)+(1-a)^{2}+\cdots+(1-a)^{i-j}=c-1+(1-a)^{i-j}$ . 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 $cb_{j}=b_{j}cab_{j}=b_{j}c$ . Now, note that for $m\leq n$ , $(1-a)^{n}b_{j}^{m}=(1-a)^{n-m}(b_{j}-1)^{m}$ . This implies that

$\displaystyle c{b}_{j}^{\phantom{j}i-j-1}$ $\displaystyle=$ $\displaystyle{b}_{j}^{\phantom{j}i-j-1}+(b_{j}-1){b}_{j}^{\phantom{j}i-j-2}+% \cdots+(b_{j}-1)^{i-j-1}$

$\displaystyle=$ $\displaystyle g(b_{j})+(b_{j}-1)^{i-j-1}.$

On the other hand, we also have

$\displaystyle c{b}_{j}^{\phantom{j}i-j-1}$ $\displaystyle=$ $\displaystyle b_{j}c{b}_{j}^{\phantom{j}i-j-2}$

$\displaystyle=$ $\displaystyle b_{j}({b}_{j}^{\phantom{j}i-j-2}+(b_{j}-1)^{i-j-3}+\cdots+(1-a)(% b_{j}-1)^{i-j-2})$

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

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

$\displaystyle{b}_{j}^{\phantom{j}i-j-2}+\cdots+(-1)^{i-j-2}$ $\displaystyle=$ $\displaystyle(b_{j}a)({b}_{j}^{\phantom{j}i-j-2}+\cdots+(-1)^{i-j-2})$

$\displaystyle=$ $\displaystyle b_{j}a{b}_{j}^{\phantom{j}i-j-2}+\cdots+b_{j}a(-1)^{i-j-2}$

$\displaystyle=$ $\displaystyle{b}_{j}^{\phantom{j}i-j-2}+\cdots+(-1)^{i-j-2}b_{j}a.$

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 them can be a left inverse of $a$ , for if, say, $ba=1$ , then $c=(ba)c=b(ca)=b$ . So we can apply the same technique used 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

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

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.

Title	infinitude of inverses
Canonical name	InfinitudeOfInverses
Date of creation	2013-03-22 18:17:30
Last modified on	2013-03-22 18:17:30
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	4
Author	CWoo (3771)
Entry type	Theorem
Classification	msc 16U99

	$\displaystyle c{b}_{j}^{\phantom{j}i-j-1}$	$\displaystyle=$	$\displaystyle{b}_{j}^{\phantom{j}i-j-1}+(b_{j}-1){b}_{j}^{\phantom{j}i-j-2}+% \cdots+(b_{j}-1)^{i-j-1}$
		$\displaystyle=$	$\displaystyle g(b_{j})+(b_{j}-1)^{i-j-1}.$

$\displaystyle c{b}_{j}^{\phantom{j}i-j-1}$	$\displaystyle=$	$\displaystyle b_{j}c{b}_{j}^{\phantom{j}i-j-2}$
	$\displaystyle=$	$\displaystyle b_{j}({b}_{j}^{\phantom{j}i-j-2}+(b_{j}-1)^{i-j-3}+\cdots+(1-a)(% b_{j}-1)^{i-j-2})$
	$\displaystyle=$	$\displaystyle g(b_{j})+b_{j}(1-a)(b_{j}-1)^{i-j-2}.$

$\displaystyle{b}_{j}^{\phantom{j}i-j-2}+\cdots+(-1)^{i-j-2}$	$\displaystyle=$	$\displaystyle(b_{j}a)({b}_{j}^{\phantom{j}i-j-2}+\cdots+(-1)^{i-j-2})$
	$\displaystyle=$	$\displaystyle b_{j}a{b}_{j}^{\phantom{j}i-j-2}+\cdots+b_{j}a(-1)^{i-j-2}$
	$\displaystyle=$	$\displaystyle{b}_{j}^{\phantom{j}i-j-2}+\cdots+(-1)^{i-j-2}b_{j}a.$