Problem Statement
Suppose $R$ is a ring and $\mathcal{J} \subset R$ is a (two-sided) ideal. The quotient additive group $R/\mathcal{J}$ is the set of cosets, which in the additive notation, have the form $r + \mathcal{J}$ for $r \in R$
Prove that: $$(r + \mathcal{J})(s + \mathcal{J}) = rs + \mathcal{J}$$ Well defines an operation on $R/\mathcal{J}$
My attempt
So for this problem we want to show that if $r + \mathcal{J} = r' + \mathcal{J}$ and $s + \mathcal{J} = s' + \mathcal{J}$ then that means that $rs + \mathcal{J} = r's' + \mathcal{J}$
To that end I noted that for any $a \in R$ the equation $a + \mathcal{J} = a' + \mathcal{J}$ implies that:
- $a,a'$ belong to the same left coset of $\mathcal{J}$ and
- $(-a) + a' = a' - a \in \mathcal{J}$
Since this applies to both $r$ and $s$ we can utilize these facts and note that both $$(r' - r),\ (s' -s) \in \mathcal{J}$$ Since $\mathcal{J}$ is a (two-sided) ideal we can say that $$(r'-r)(s'-s) = r's' - r's - s'r + rs \in \mathcal{J}$$ Now I was hoping that by utilizing these facts and through expanding we would end up with something like $r's' - rs$ which we can clearly see is an element of $\mathcal{J}$ which would be equivalent to the statement that $rs + \mathcal{J} = r's' + \mathcal{J}$. However, of course that's not what we end up with.
So, I'm wondering am I on the right track but just potentially missing some small property or having some oversight? Or is my approach totally wrong and I need to rethink this problem from a different angle? I appreciate any guidance here.