TITLE: In a von Neumann regular commutative ring with unity every finitely generated ideal is principal
QUESTION [7 upvotes]: Let $R$ be a commutative, von Neumann regular ring with unity. How to show that every finitely generated ideal in $R$ is principal?

I can see, in view of mathematical induction, it suffices to show that any ideal generated by two elements of $R$ must be principal.
Let $I=(a,b)$ be an ideal of $R.$ Since $I$ is commutative with unity, $I=\{xa+yb:x,y\in R\}.$ Also since $R$ is regular there exist $r,s\in R$ such that $a=ara=ra^2$ and $b=bsb=sb^2.$
However I cannot figureout which element would generate $I.$ Please help.

REPLY [7 votes]: Since $ara=a$, the element $e=ar$ is idempotent (i.e., $e^2=e$) and generates the same ideal as $a$.
Similarly, $f=bs$ is idempotent and generates the same ideal as $b$.
These idempotent elements are easier to deal with than $a$ and $b$, and in particular
$$e+f-ef = ar + bs -arbs$$
is another idempotent element that generates the ideal $\langle e,f\rangle=\langle a,b \rangle$,
since $e(e+f-ef)=e$ and $f(e+f-ef)=f$.