What does the correspondence theorem (or 4th isomorphism theorem for rings) for rings mean and how is it used? That is, why do we care about it?
Edit:
My version of the correspondence theorem:
Let $R$ be a ring and $I$ be an ideal in $R$. Let $K$ be the set of ideals in $R$ containing $I$. Let $L$ be the set of ideals in $R/I$. Then there is bijection from $K$ to $L$ given by $\phi(I’)= \{ x+ I : x \in I’ \}$.
On
The correspondence says that if $R$ is a ring and $I$ is a two-sided ideal of $R$, that the following map is a bijection:
$$\{\mathrm{\ ideals \ of \ R \ containing \ I}\}\to \{\mathrm{ideals \ of \ the \ quotient \ ring \ R/I}\}$$ $$J \mapsto J/I = \pi(J)$$
where $\pi: R \to R/I$ is the natural quotient map.
We have a similar correspondence for subrings of the quotient ring.
Why do we care about it? Because it tells us that we know how ideals in the quotient ring look like and complete knowledge of ideals of $R$ gives us complete knowledge about the ideals in the quotient ring.
One of the most basic applications of this theorem is the result that a quotient ring $R/I$ is a field if and only if $I$ is a maximal ideal. (For me, all rings are commutative with unity.)
Every non-zero ring $R$ has at least two ideals, namely $\{0\}$ and $R$. And $R$ is a field if and only if these are the only two ideals in $R$.
If $I\subseteq R$ is a proper ideal, then there are at least two ideals in $R$ containing $I$, namely $I$ and $R$. And $I$ is maximal if and only if these are the only two ideals in $R$ containing $I$.
Now the set of ideals in $R$ containing $I$ is in bijection with the set of ideals of $R/I$. So $I$ is maximal if and only if there are exactly two ideals in $R$ containing $I$ if and only if there are exactly two ideals in $R/I$ if and only if $R/I$ is a field.