How do I prove the lattice theorem for rings? Let $ f: R \rightarrow S$ be an onto homomorphism from a ring $R$ to a ring $S$. Prove that there is a one-to-one, order-preserving correspondence between the ideals of $S$ and the ideals of $R$ that contain $ker(f)$. 
I am confused by this problem and I also need an example to help see what they're talking about.
 A: It is a general fact that there is a one-to-one correspondence between the ideals in a quotient ring $R/I$ ($I$ and ideal) and the ideals in $R$ containing $I$. This correspondence is given by the canonical map $\phi:R\rightarrow{R/I}$.
More specifically:
If $\overline{I}$ is an ideal in $R/I$, then $\phi^{-1}(\overline{I})$ is an ideal in $R$ (the inverse image of an ideal under a homomorphism is an ideal). Note that this ideal contains $I$. Then prove that the map taking $\overline{I}$ to $\phi^{-1}(\overline{I})$ is bijective.
In this specific situation we have a surjective ring homomorphism $f:R\rightarrow{S}$. Then it is a general fact that $R/ker(f)\cong{im(f)}$ (have you learnt this?) 
But $im(f)=S$, since $f$ is surjective, and so we may think of $S$ as $R/ker(f)$. Then what you ask follows directly from what I wrote above.
A: I'll restate the correspondence theorem for groups for you below.

Lemma. (Correspondence Theorem for Groups.) If $G$ is a group and $N\unlhd G$ a normal subgroup of $G$, then every subgroup of the quotient group $G/N$ has the form $H/N$, where $N\leqslant H \leqslant G$.  Conversely if $N\leqslant H \leqslant G$ then $H/N\leqslant G/N$.
Proof. Let $\overline{H}$ be a subgroup of $G/N$.  Let us look at the preimage of $H$ under the canonical homomorphism $x\mapsto Nx$, that is $H=\{g\in G : Ng \in \overline{H}\}$.  Since $N$ is the identity of $\overline{H}$, all $n\in N$ are in $H$.  Recall that $\mu(a)^{-1}=\mu(a^{-1})$ (a property of homomorphisms), so we have that $(Na)^{-1}=Na^{-1}\in \overline{H}$ for any $a\in H$, whence $a^{-1}\in H$.  Finally, we verify closure by noting that, for every $a,b\in H$, $(Na)(Nb)=N(ab)\in \overline{H}$ so $ab\in H$.  It follows that $H$ is a subgroup of $G$ that contains $N$.  The converse is verified similarly by looking at $\mu[H]$.

So with this, you can tell that there is a bijection between the additive subgroups of $R$ containing an ideal $I$ and the additive subgroups of $R/I$.  From there, verify that the associated subrings and ideals are also in one to one correspondence by checking to make sure the multiplicative groups still match up.  The proof is similar to the proof of the above lemma.
A: Denote the set of ideals of $R$ containing $\ker(f)$ by $X$ and the set of ideals of $S$ by $Y$. The question asks for an order-preserving bijection $X \to Y$.
Hint: Look at $$I \mapsto \{f(x) \mid x \in I\}.$$
