Basic decomposition in a ring with unity

Question

Consider a ring $(R,+,\cdot)$ with unity and $r_1,r_2$ two elements of $R$ with the property that $r_1-r_2$ is an idempotent element. Show that $$ r_1^2-r_2^2=1 \qquad \text{iff}\qquad (r_1-r_2)(r_1+r_2)=1. $$ I started with the observation that, using the hypothesis, from $(r_1-r_2)(r_1+r_2)=1$ one can deduce that $$ (r_1-r_2)^2(r_1+r_2)=1 \Rightarrow (r_1-r_2)(r_1-r_2)(r_1+r_2)=1 \Rightarrow r_1-r_2=1. $$ Turning back to $(r_1-r_2)(r_1+r_2)=1$, we obtain that $r_1+r_2=1$. Finally, we have that $$ r_1^2-r_2^2=r_1^2-r_1r_2+r_1r_2-r_2^2=r_1(r_1-r_2)+(r_1-r_2)r_2=r_1+r_2=1. $$ Is this little argument correct? How can I show the reverse?

Answer 1

Your computation is correct. To check the other direction, you can proceed as follows. Let $e=r_1-r_2$. Then $e$ is idempotent, i.e. $e^2=e$.

Suppose that $r_1^2-r_2^2=1$. This means that $(r_2+e)^2-r_2^2=1$, which expands to

$$ er_2+r_2e+e=1 \tag{1} $$

Multiplying by $e$ on the left, we deduce $e^2r_2+er_2e+e^2=e$, or $er_2+er_2e=0$. Similarly, multiplying by $e$ on the right, we deduce $er_2e+r_2e=0$. So $r_2e=-er_2e=er_2$, and we see that $e$ and $r_2$ commute. So $r_1$ and $r_2$ commute also and $(r_1-r_2)^2=r_1^2-r_2^2$.