Past exam of the algebra 2 course of the University of Oxford / 2015 / 3 / d / Solution

Take $I=(2,t)\subseteq \mathbb{Z}[t]$. The ring homomorphismis 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,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 $d=\pm 1$. This would give $I=\mathbb{Z}[t]$, contradicting the properness of the kernel of a ring homomorphism $\ker \theta$. ThereforeThere 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.