Suppose $R$ is ring with unit and $a$ is a nonzero element of the ring. If there are two distinct elements $b,c$ such that $ab=ac=1$, show that there are infinite elements $x$ in $R$ such that $ax=1$.
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Is the ring assumed to be commutative? – ajotatxe Oct 17 '14 at 14:10
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Obviously not, @ajotatxe, as then a left inverse = right inverse and thus $;a;$ is a unit and there is no question. – Timbuc Oct 17 '14 at 14:24
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Look at the elements
$$x_n:=b+\left(1-ca\right)a^n\;,\;\;n=1,2,3,...$$
First, we have that
$$ax_n=ab+(a-aca)a^n=1+(a-a)a^n=1$$
Second, assume $\;n>m\;$ and
$$x_n=x_m\implies(1-ca)a^n=(1-ca)a^m\implies (1-ca)a^{n-m}=1-ca\;\text{(why?)}\implies$$
$$\left[(1-ca)a^{n-m-1}+c\right]a=1$$
and this would mean $\;a\;$ has a left inverse and it is thus a unit so $\;b=c\;$, contradiction.
We thus get $\;\{x_n\}_{n\in\Bbb N}\;$ is an infinite set of right inverses of $\;a\;$ .
Timbuc
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Thanks @HagenvonEitzen. This is a more or less known theorem by Kaplansky. – Timbuc Oct 17 '14 at 14:25