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, $$\varphi(J)=\{x+I\mid x\in J\}=J/I.$$
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Yes, compare with the answers here. – Dietrich Burde Nov 01 '16 at 13:45
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Because $J/I$ means nothing if $I$ is not contained in $J$. – egreg Nov 01 '16 at 14:05
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@egreg: Like I said: $J$ is an ideal containing $I$. – user330587 Nov 01 '16 at 14:06
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And, in this case $J+I=J$, so where's the problem? You're probably confusing with the general statement that $\varphi(J)=(J+I)/I$ for all ideals of $R$, containing $I$ or not. – egreg Nov 01 '16 at 14:15
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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.
Dietrich Burde
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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 $J+I=J$.
egreg
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