If $A$ is a commutative ring with unity, then a fact proved in most commutative algebra textbooks is:

$$\dim A + 1\leq\dim A[X] \leq 2\dim A + 1$$

Idea of proof: each prime of $A$ in a chain can arise from at most two prime ideals of $A[X]$.

The left equality holds when $A$ is a noetherian ring, but this paper of Seidenberg shows that, for rings of fixed dimension, every value in the interval is achievable.

There seems to be a shortage of elementary references on these counterexamples, however, so I thought it might be good to collect some specific examples of commutative rings $A$ with $\dim A[X] > \dim A$, or knowledge about simple cases.

In particular, it would be good to have some examples of rings $A$ with $\dim A = 1$ and $\dim A[X]=3$.

Here is an example I just worked out, inspired by Seidenberg's paper.

Let $k$ be a field, with $t$ transcendental over $k$, and let $A = \{f\in k(t)[[Y]] \mid f(0) \in k\}$. Then $A$ has unique nonzero prime ideal $P = \{f\in k(t)[[Y]] \mid f(0) = 0\}$, with $A/P \cong k$, so $\dim A = 1$.

To see that $\dim A[X]=3$, it is enough to exhibit a prime ideal lying between $(0)$ and $P[X]$. Consider the map $A[X] \to k(t)[[Y]]$ given by $p(X) \mapsto p(t)$. The image is an integral domain, so the kernel $K$ is prime. But $(YX-tY)\in K$, while $Y\notin K$, so $(0)\subsetneq K \subsetneq P[X]$.