Prove a bijection of ideals exists.

Question

A problem from Intro to Abstract Algebra by Hungerford

If $f:R\rightarrow S$ is a surjective homomorphism of rings with kernel $K$, prove that there is a bjective function from the set of all ideals of $S$ to the set of all ideals of $R$ that contain $K$.

By the first isomorphism theorem we know that $R/K \cong S$. Also, we know that every ideal of $R/K$ is of the form $I/K$ (proven in part a of this problem). I tried proving that $\exists g$ st. $g:I\rightarrow I/K$ is bijective, since $I/K$ is isomorphic to the ideals of $S$ (is this correct?). But I haven't really gotten anywhere. Any help would be appreciated.

Answer 1

Prove that

$$\overline I\le R/K\implies I:=\{r\in R\;;\;r+K\in\overline I\}\le R\;\;\wedge\;\;K\le I$$

and now you have two mappings of ideals $\;R\to R/K\;,\;\;R/K\to R\;$ inverse to each other.

This is just the ring version of the very important Correspondence Theorem, also studied in group theory.