$R/J \cong (I+J)/J$?

Question

If $R$ is an ring and $I$,$J$ are ideals then $\frac{R}{J}$ is a $R$-module. Is it true that $I\frac{R}{J} \cong (I+J)/J$? Also, is there a difference between $(I+J)/J$ and $I/J$?

Answer 1

I think you have to understand the answer to the second question first.

Also, is there a difference between $(I+J)/J$ and $I/J$?

For arbitrary ideals $I,J$ of a ring $R$, "$I/J$" may not even be defined. That should only be written if $I$ contains $J$, which may not always be the case.

Of course, there is a simple way to move to the smallest ideal containing $I$ and $J$: it's called $I+J$. So it would always be possible to write "$(I+J)/J$".

If it happens that $I\supseteq J$, then $I+J=I$, so that both notations work and represent the same set.

Is it true that $I\frac{R}{J} \cong (I+J)/J$?

Yes, because $I\frac{R}{J} =(I+J)/J$. By definition the left side is $\{\sum i_k(r_k+J)\mid i_k\in I, r_k\in R\}$, which clearly contains $i+J$ for every $i\in I$ since $R$ has identity (ostensibly.) But it must also contain at a minimum $0+J=j+J$ for all $j\in J$.

You can write out the mutual containments that officially prove the equality.