= Solution
Take $I=(2,t)\subseteq\mathbb Z[t]$. The <ring homomorphism>
$$
\theta:\mathbb Z[t]\longrightarrow\mathbb F_2,
\qquad \theta(g)=g(0)\pmod2
$$
is surjective. Its <kernel of a ring homomorphism> is exactly $I$: a <polynomial> maps to zero precisely when its constant coefficient is even, in which case it has the form $2a+th(t)$ with $a\in\mathbb Z$ and $h\in\mathbb Z[t]$. By the <first isomorphism theorem for rings>,
$$
\mathbb Z[t]/I\cong\mathbb F_2.
$$
This <quotient ring> is a <field>, so $I$ is a <maximal ideal> and in particular a <prime ideal>.
Suppose $I=(h)$ were a <principal ideal>. Because $2\in I$, the nonzero <polynomial> $h$ divides the constant $2$, forcing its <polynomial degree> to be zero: $h=d$ for a nonzero <integer> $d$. But $t\in I$ also makes $d$ divide the coefficient $1$ of $t$, so $d=\pm1$. This would give $I=\mathbb Z[t]$, contradicting the properness of the <kernel of a ring homomorphism> $\ker\theta$. Therefore
$$
\boxed{(2,t)\text{ is a prime ideal of }\mathbb Z[t]
\text{ that is not principal}.}
$$
There is no conflict with part (c): $I\cap\mathbb Z=2\mathbb Z\ne\{0\}$. The zero-intersection hypothesis is exactly what excludes this example.
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