If $I$ is an ideal of the ring $R$ then show that if $R$ has no non-zero divisors then $R/(l(I)\cap r(I))$ has the same property.

Here $l(X):=\{r\in R:rx=0,\forall x\in X\}$ and $r(X):=\{r\in R:xr=0,\forall x\in X\}.$

My Attempt: Since $R$ has non-zero divisors, this means that there exists $a\in R$ with $a\not =0$ such that $ab=0\implies b=0.$ We want to show that the quotient ring $R/(l(I)\cap r(I))$ also contains non-zero divisors. By definition we first observe that $$J=l(I)\cap r(I)=\{r\in R: rx=0\text{ and }xr=0,\forall x\in I\}.$$ Since we take $R/J=\{r+J:r\in R\}.$ Perhaps $a+J$ is a non-zero divisor for $R/J.$ First note that $a+J\not = J.$ Next consider any $b\in R$ and $(a+J)(b+J)=0\implies ab+J=0\implies ab\in J.$ This means that $$abx=0\text{ and }xab=0,\forall x\in I.$$ Thus $$bx=0,\forall x\in I.$$ Similarily by considering $(b+J)(a+J)=0$ we can deduce that $xb=0,\forall x\in I.$ We want to show that $b+J=J.$ For that let $r\in J$ then for any $x\in I$ we have $(b+r)x=bx+rx=bx=0$ and also $x(b+r)=xb+xr=0$ so $b+J=J$ hence $a+J$ is indeed a non-zero divisor for $R/J.$

I am learning ring theory and so I am not sure whether my solution is correct. Thus any feedback will be much appreciated.


If $R$ has no non-zero divisors then $l(I)=(0)=r(I)$, so $$R/r(I)\cap l(I)=R/(0)\cong R$$ which by assumption has no non-zero divisors.

  • $\begingroup$ But $R$ is not commutative and also does not necessarily have an identity element so how can it be an integral domain? $\endgroup$ – model_checker Feb 24 '18 at 15:04
  • 1
    $\begingroup$ You're right, I forgot that $R$ may not be commutative. But isn't part of the definition of a ring that it has an multiplicative identity element? Also, if either $l(I)$, or $r(I)$ were not $(0)$ you would have a non-zero zero-divisor. $\endgroup$ – cansomeonehelpmeout Feb 24 '18 at 15:09
  • $\begingroup$ How can this answer have been upvoted? $\endgroup$ – egreg Feb 24 '18 at 15:32
  • $\begingroup$ @egreg I also want to know why $\endgroup$ – cansomeonehelpmeout Feb 24 '18 at 15:33
  • $\begingroup$ @cansomeonehelpmeout You should delete it, in my opinion. $\endgroup$ – egreg Feb 24 '18 at 15:41

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.