I posted a question yesterday on how to prove the following :" $Ra/Rab$ $\cong_R$ $R/Rb$ " and the answer I got said the homomorphism $\phi :$$R$ $\rightarrow$ $Ra/Rab$ : $x$ $\mapsto$ $ax + Rab$ would be the solution, as by the isomorphism theorem the Kernel of this map is $Rb$ and thus the isomorphism in the title holds.
Now I have been thinking about the question and I am still stuck. If the mapping is a homomorphism and we take an element $x'$ of $Rb$, $\phi(x')$ would be $ax'= ax''b$, but if $Rb$ is the kernel of $\phi$, $ax''b$ has to be divisible by $ab$ and since we don't know if $R$ is commutative we can't say for a fact that $ab$ $|$ $ax''b$ or any element of $Rb$ ?
Here is the link to the original question
Thanks for reading my question.