# 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.

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.
• But $R$ is not commutative and also does not necessarily have an identity element so how can it be an integral domain? – model_checker Feb 24 '18 at 15:04
• 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. – cansomeonehelpmeout Feb 24 '18 at 15:09