# Subtlety in Correspondence Theorem for Rings

I have something of a subtle question about the correspondence theorem for rings. The theorem is typically stated like this: Let $A$ be a ring, and $I$ an ideal of $A$. There is a $1-1$, order-preserving correspondence between \begin{align*} \{\mbox{Ideals}\ J\ ,\ I\subseteq J \subseteq A\} &\leftrightarrow \{\mbox{Ideals}\ \bar{J}\subseteq A/I \} \\ \end{align*} I just finished reading a direct proof which constructed the forward direction, and inverse maps for the $1-1$ correspondence. For the forward direction, we consider the canonical epimorphism $\varphi : A \rightarrow A/I$ sending $a \mapsto a+ I$, and restrict to $J$. That is we consider the map $$\varphi |_{J} : J \rightarrow A/I$$ The proof proceeds to establish that $\mbox{ker}(\varphi |_{J}) = I$. For this step it matters that $I \subset J$, because otherwise we might have $$\mbox{ker}(\varphi |_{J}) = I \cap J \ne I$$ The isomorphism theorem for rings tells us that $$\mbox{Im}(\varphi |_{J}) \cong J / \mbox{ker}(\varphi |_{J}) = J/I$$ Then the argument concludes that because $J/I$ is an ideal of $A/I$, the image does in fact ''land'' in the correct set.
Basically I'm wondering where exactly it is necessary that $I \subseteq J$. It seems that the above argument could be sidestepped. Isn't it clear by the definitions (and isn't it true even if $I \nsubseteq J$) that $\mbox{Im}(\varphi |_{J}) = J/I$? I'm surely missing some detail. Thanks so much in advance for your help!

• If we don't have $I\subseteq J$, then $J$ won't consist of cosets of $I$, and $J/I$ begins to look very suspicious. What would it be? – Jyrki Lahtonen Aug 30 '14 at 20:17
• I'm not sure I understand what you mean when you say "then $J$ won't consist of cosets of $I$." $J$ is an ideal of $A$ in a coset free world. – aherring Aug 30 '14 at 20:27
• Sorry about not being clear. What I meant to say was: $J$ is not a union of cosets of $I$, so what would the elements of $J/I$ be? After all $J/I$ should be a quotient group (in addition to being an ideal of $A/I$). And to that end it is already necessary that $I$ is a subgroup of $J$. Hence $I$ should be a subset of $J$. – Jyrki Lahtonen Aug 30 '14 at 20:41

Indeed the projection map sends all ideals of $R$ to ideals of $R/I$. This is surjective, but not generally an injective map - two different ideals of $R$ can be sent to the same ideal of $R/I$. And when you go back by taking an ideal of $R/I$ and unioning the cosets of $I$ it contains to get an ideal of $R$ you will wind up with an ideal that contains $I$, so the inverse map takes ideals of $R/I$ and returns ideals of $R$ containing $I$.