Clarify what "inclusion preserving" means in lattice isomorphism theorem I'm working through Dummit and Foote right now. The lattice isomorphism theorem is stated as follows: "Let I be an ideal of a ring R. The correspondence $A \leftrightarrow A/I $ is an inclusion preserving bijection between the subrings A of R that contain I and the set of subrings of R/I."
First off, to make sure I understand: is "$A \leftrightarrow A/I$" the map $\phi(A) = A/I$? Second, what does it mean that it preserves inclusion?
 A: Let's be clear about what the bijection is between. There is a 1-1 correspondence
$$
\{\mbox{Ideals $J$ s.t. $I\subseteq J\subseteq A$}\}\leftrightarrow\{\mbox{Ideals in $A/I$}\}
$$
This bijection is induced by the canonical projection $\pi:A\to A/I$. 
Note that if $J$ is an ideal containing $I$, then $\pi(J)$ is an ideal in $A/I$. Also, if $K\lhd A/I$ is an ideal, then $\pi^{-1}(K)=\{x\in A\mid \pi(x)\in K\}$ is an ideal of $A$ containing $I$. 
To see that this gives you a bijection, it is enough to prove two things:


*

*If $K\lhd A/I$ is an ideal, then $\pi(\pi^{-1}(K))=K$. (This is by definition).

*If $J$ is an ideal containing $I$, then $\pi^{-1}(\pi(J))=J$. (It is clear that $J\subseteq\pi^{-1}(\pi(J))$. To get the reverse inclusion, suppose $a\in\pi^{-1}(\pi(J))$. Then $a+I=j+I$ for some $j\in J$. If follows that $a-j\in I\subset J$, forcing $a\in J$.)
Now, the next statement is that this bijection preserves inclusions. That is, if $I\subset J_1\subset J_2\subset A$ are two ideals, then $\pi(J_1)\subset\pi(J_2)$. 
This is called the Lattice Isomorphism Theorem because the ideal lattice for $A/I$ is the same as the portion of the ideal lattice for $A$ with minimal element $I$.
A: They do indeed mean the map $\phi: A \mapsto A/I$ with inverse $A/I \mapsto A$ when they talk about this correspondence. They're saying that it's a bijection.
The fact that it's inclusion preserving means that if $A \leq B$ in $R$ and both contain $I$, then $A/I \leq B/I$, and vice versa.
