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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.

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    $\begingroup$ It seems that when you write "regular" you mean "von Neumann regular". If this is the case, it would be good to update your post accordingly, since ''regular'' by itself has a very different meaning in commutative algebra. $\endgroup$ – Alex Wertheim Feb 15 at 5:20
  • $\begingroup$ This can be strengthened to: in any von Neumann regular ring (commutative or not) every f.g. right ideal is a summand of $R_R$ (and every f.g. left ideal is a summand in $_RR$.) $\endgroup$ – rschwieb Feb 15 at 13:48
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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 $f(e+f-ef)=f$.

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