$\DeclareMathOperator{\Id}{Id}\DeclareMathOperator{\im}{Im}$Let $R\leq S$ be rings, and $\Id(R), \Id(S)$ be their ideal posets (in fact, lattices), ordered by inclusion.
Furthermore, if $A, B$ are subgroups of the additive group of a ring (in particular, subrings or ideals), let $AB := A\cdot B$ denote the subgroup generated by $\{ab\mid a\in A, b\in B\}$. We have two ways of translating between those, which “feel” inverse:
Although $λ$ and $ρ$ need not be precise inverses, we can see that they are reasonably well-behaved as follows:
Theorem Let $λ\colon \Id(R)\to \Id(S), ρ\colon \Id(S)\to\Id(R)$ as above.
Proof. $I\subseteq I^\prime$ implies $IS\subseteq I^\prime S$ by monotonicity of the “generated subgroup” hull operator, showing monotonicity of $λ$. $J\subseteq J^\prime$ clearly implies $J\cap R\subseteq J^\prime\cap R$, proving the other half of 1 (nota bene: intersecting with a fixed set is a kernel operator). For 2, we need to show that $I\subseteq (IS)\cap R$ for an $I\unlhd R$. Because $1\in S$, we see that $I\subseteq IS$. Since also $I\subseteq S$, we are done. Now for 3, let $J\unlhd S$. Clearly, $ρ(J) = J\cap R\subseteq J$. Since $ρ(J)S$ is the smallest ideal in $S$ containing $ρ(J)$, we must have $ρ(J)S\subseteq J$ as desired. Now consider a $ρ(J)\in \im(ρ)$ for some $J\unlhd S$. By monotonicity, we can apply $ρ$ to both sides of the intensivity relation $λ(ρ(J))\subseteq J$, resulting in $ρ(λ\circ ρ(J)))\subseteq ρ(J)$. On the other hand, extensivity of $ρ\circ λ$ implies that $ρ(J)\subseteq ρ\circ λ(ρ(J))$, proving equality. We have seen that Elements of $\im(ρ)$ are invariant under $ρ\circ λ$ – on the other hand, all such invariant elemetns clearly lie in $\im(ρ)$, proving the 4. Proving 5 works analogously: Applying $λ$ to the extensivity relation gives $λ(I)\subseteq λ(ρ\circ λ(I))$, intensivity gives $\cdots \subseteq λ(I)$, and we're done. Now, assume that $I\subseteq ρ(J)$. Applying $λ$ gives $λ(I)\subseteq λ(ρ(J))\subseteq J$. Conversely, if $λ(I)\subseteq J$, we get $ρ(J)\supseteq ρ(λ(I))\supseteq I$. This proves 6.
As you might have guessed it, we're dealing with a (monotone) galois connection.
It follows that $λ,ρ$ are mutual inverse isomorphisms when restricted to the sub-posets $\im(ρ)\leq \Id(R)$ and $\im(λ)\leq \Id(S)$, respectively.
Let $R\leq S$ be an integral extension of dedekind domains (e.g. implying that prime ideals are maximal). We can show that for prime ideals $\mathfrak p\unlhd R$, we have $λ(S)={\mathfrak p}S\neq S$. Hence, applyign $ρ\circ λ$ to $\mathfrak p$ can yield either $\mathfrak p$ or all of $R$. Since $\mathfrak p\subseteq ρ(λ(\mathfrak p)) = {\mathfrak p}S\cap R$, the latter being all of $R$ is equivalent to $1\in {\mathfrak p}S$, or equivalently, that ${\mathfrak p}S = S$, which cannot happen. Hence, $\mathrm{spec}{(R)}\subseteq \im(ρ)$ – in other words: $$ \mathfrak p = {\mathfrak p}S \cap R. $$
Please note that this is probably one of the more masochistic and indirect ways to prove this.
LEMMA Let $R\leq S$ be an extension of integrally closed (i.e., over their fraction fields) domains such that every element of $S$ is integral over $R$ (i.e., annihilated by a monic polynomial with coefficients in $R$. Let further $r\in R, s\in S$. If $rs\in R$, then $s\in R$ as well.
Proof. Let $K≤L$ denote the fraction fields of $R, S$, respectively, everything equipped with the canonical embeddings. Since $rs\in R\subseteq K$, we must have $s = r^{-1}rs\in K$ as well. Since $s$ is still integral over $R$, we have a monic $p\in R[X]$ such that $p(s) = 0$. Since $R$ is integrally closed over its fraction field $K$, this condition already ensures that $s$ must lie in $R$.
Counterexample. Consider $S=ℤ[1/3]$ over $R=ℤ$: choosing $r=3$ and $s=1/3$, we clearly have $rs=1\in R$. This is because $1/3$ bears witness to the fact that $ℤ[1/3]$ is not integral over $ℤ$ – evidently, its minimal polynomial over $ℤ$ is $3x-1$, which is of course not monic.
If we have a chain of such integral extensions $R\leq S\leq T$ of integrally closed fields, do we have transitivity for all three $ρ$s and $λ$, respectively?