Examples of rings whose polynomial rings have large dimension

Here is a classification of all integral domains $A$ with $\dim A=1$, $\dim A[X]=3$:

First, consider the map $\operatorname{Spec} A[X]\to\operatorname{Spec} A$ given by $\mathfrak{p}\to\mathfrak{p}\cap A$. The fiber over $P$ is isomorphic to $\operatorname{Spec} A_P / PA_P\otimes_A A[X] \cong (A_P / PA_P)[X]$, which is one-dimensional. It follows that a chain of three distinct primes in $A[X]$ cannot have the same intersection with $A$.

Suppose that there is a chain of four prime ideals in $A[X]$. By the previous paragraph, intersecting with $A$ must give us $(0)\subset (0) \subset P \subset P$ for some maximal $P\subset A$. It follows that the first three elements of the chain are $(0) \subsetneq \mathfrak{q} \subsetneq P[X]$, with $\mathfrak{q}\cap A = (0)$. The existence of such a $\mathfrak{q}$ is therefore equivalent to $\dim A[X] = 3$.

Then the map $A[X]\to K(A[X]/\mathfrak{q})$ restricts to an injection $A\to K(A[X]/\mathfrak{q})$, which induces an injection $K(A)\to K(A[X]/\mathfrak{q})$. So $\mathfrak{q}$ is the kernel of a map from $A[X]$ to a field extension of $K(A)$. Let $t$ be the image of $X$ under this map.

Since $\mathfrak{q}\neq (0)$, $t$ is algebraic over $K(A)$. Since $\mathfrak{q}\subset P[X]$, $t$ is transcendental over $K(A/P)$.

So all examples arise from a dimension $1$ integral domain $A$ with a maximal ideal $P$, such that there is some $t\in \overline{K(A)}$ that is transcendental over $K(A/P)$—i.e. every integral equation satisfied by $t$ has coefficients in $P$.

(Note: In the original example, $t$ satisfies the polynomial $p(T) = YT - tY$, so this is probably close to the simplest example of an element of $K(A)$ transcendental over $K(A/P)$.)

Conversely, such a $t$ gives rise to our desired prime ideal $\mathfrak{q}$.

Can the condition $t\in \overline{K(A)}$ be simplified to $t\in K(A)$? If so, the condition is much simpler: we must have some $t\in K(A)$ that can only be written as a fraction with both numerator and denominator in $P$.