= Solution
{c}
For any <prime number> $p$, there is a <prime chain>
$$
(0)\subsetneq(p)\subsetneq(p,t)
$$
in $\mathbb Z[t]$. These are <prime ideals> because the respective <quotient rings> are $\mathbb Z[t]$, $\mathbb F_p[t]$, and $\mathbb F_p$, all <integral domains>. The inclusions are strict because $p\ne0$ and $t\notin(p)$. Hence $\dim\mathbb Z[t]\geq2$.
To prove the upper bound, suppose there were a <prime chain> of length at least three. Since $\mathbb Z[t]$ is an <integral domain>, we could insert $(0)$ at its beginning if necessary and select a chain
$$
(0)\subsetneq P_1\subsetneq P_2\subsetneq P_3.
$$
By <3/c>, $P_2\cap\mathbb Z\ne(0)$. This intersection is a proper <prime ideal> of $\mathbb Z$: if a product of integers lies in it, primality of $P_2$ puts one factor in it. Therefore $P_2\cap\mathbb Z=(p)$ for a <prime number> $p$, and $(p)\subseteq P_2$.
We need to check that $P_2\ne(p)$. Here <prime ideals below a principal prime> give the necessary restriction: no nonzero <prime ideal> can lie strictly below $(p)$. If $0\ne P\subsetneq(p)$, then $p\notin P$. Choose $0\ne f\in P$ and factor out the largest common power of $p$ from its integer coefficients:
$$
f=p^kg,\qquad g\notin(p).
$$
Such a finite $k$ exists because $f$ has at least one nonzero integer coefficient. Since $p\notin P$ and $P$ is a <prime ideal>, repeated primality gives $g\in P\subseteq(p)$, a contradiction. Applied to $P_1\subsetneq P_2$, this rules out $P_2=(p)$.
It follows that $P_2/(p)$ is a nonzero <prime ideal> of
$$
\mathbb Z[t]/(p)\cong\mathbb F_p[t].
$$
As in <3/c>, the <polynomial division algorithm> makes $\mathbb F_p[t]$ a <principal ideal domain>, so its nonzero <prime ideals> are maximal by <3/a>. Thus $P_2/(p)$ is maximal. The <ideal correspondence theorem> makes $P_2$ a <maximal ideal> of $\mathbb Z[t]$, contradicting $P_2\subsetneq P_3$. Consequently
$$
\boxed{\dim\mathbb Z[t]=2.}
$$
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