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.

  • 2
    $\begingroup$ 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

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 $f(e+f-ef)=f$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.