Let $R$ be a ring with unity and let $I$ be an ideal of $R$. Prove that $l_{R/I}(a+I)=(l_R(a)+I)/I.$ Let $R$ be a ring with unity, $a\in R$ and let $I$ be an ideal of $R$.  Define the following annihilators as $$l_{R/I}(a+I)=\{x+I\in R/I:(x+I)(a+I)=I\}$$ and $$l_R(a)=\{r\in R:xa=0_R\}.$$  Prove that $$l_{R/I}(a+I)=(l_R(a)+I)/I.$$
Proof:  From the definition of $l_{R/I}(a+I)$, it follows that  $(x+I)(a+I)=xa+I=I=0_R+I$ and $x\in l_R(a)+I$ so that $x+I\in (l_R(a)+I)/I.$
I am failing to prove the reverse.   I don't know the flaws is in the first proof too.
 A: The reason you are having problems with the converse is because the converse is false.
Let $R=\mathbb Z/8\mathbb Z$.  Then for $a=2+8\mathbb Z$, $\ell_R(a)=\frac{4\mathbb Z}{8\mathbb Z}$. 
Set $\ell_R(a)=I$. Then $(\ell_R(a)+I)/I=I/I$, the zero ideal of $R/I$.
But $\ell_{R/I}(a+I)$ clearly contains $a+I$, which is not zero in $R/I$.

There are definitely flaws of unclarity in what you've written.


*

*You write $(x+I)(a+I)=xa+I=I$... but why?  Is it because you have assumed $x\in \ell_R(a)$? Or is it because you have assumed $x+I\in \ell_{R/I}(a+I)$?

*Then you wrote $x\in \ell_R(a)+I$, which doesn't fit with what you wrote above. What you have written says $x-y\in I$ where $y$ annihilates $a$.  I don't even see why you concluded $x\in \ell_R(a)+I$ at all from the preceding line.
Anyhow, the fact that $(\ell_R(a)+I)/I\subseteq \ell_{R/I}(a+I)$ is completely trivial, if that's what you intended in point $1$ above. Your first line before "and" has the computations that would show that if $x\in \ell_R(a)$ then $x+I$ annihilates $a+I$, so that $x+I\in \ell_{R/I}(a+I)$.
