 Solution

For any prime number $p$, there is a prime chainin $\mathbb{Z}[t]$. 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 $p\ne 0$ and $t\notin (p)$. Hence $\dim \mathbb{Z}[t]\ge 2$.

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 chainBy 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 $0\ne P⊊(p)$, then $p\notin P$. Choose $0\ne f\in P$ and factor out the largest common power of $p$ from its integer coefficients: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 $P_1⊊P_2$, this rules out $P_2=(p)$.

It follows that $P_2/(p)$ is a nonzero prime ideal ofAs in c, the polynomial division algorithm makes ${\mathbb{F}}_p[t]$ a principal ideal domain, so its nonzero prime ideals are maximal by a. Thus $P_2/(p)$ is maximal. The ideal correspondence theorem makes $P_2$ a maximal ideal of $\mathbb{Z}[t]$, contradicting $P_2⊊P_3$. Consequently