why $\varphi(J)=(J+I)/I$ and not $J/I$ where $\varphi:R\longrightarrow R/I$?

Question

Let $R$ a ring and $I$ an ideal of $R$. I want to show that there is a bijection between the ideal of $R$ containing $I$ and the ideal of $R/I$. Let $\varphi: I\longrightarrow R/I$ defined by $\varphi(x)=x+I$. Let $J$ an ideal of $R$ containing $I$. Why $$\varphi(J)=(J+I)/I$$ and not $\varphi(J)=J/I$ ? Indeed, $$\varphi(J)=\{x+I\mid x\in J\}=J/I.$$

Answer 1

Since $J$ is an ideal containing $I$, $J+I=J$ anyway, because $I+J$ is the smallest ideal containing $I$ and $J$. So there is no difference.

Answer 2

You're probably being misled by the general statement that, for every ideal $J$ of $R$ (containing $I$ or not), $$ \varphi(J)=(J+I)/I $$ In the special case when $I\subseteq J$, obviously $J+I=J$.