Polynomials in Integers is not Principal Ideal Domain
Let $\Z \sqbrk X$ be the ring of polynomials in $X$ over $\Z$.
Then $\Z \sqbrk X$ is not a principal ideal domain.
From Polynomials in Integers with Even Constant Term forms Ideal, $J$ is indeed an ideal.
But $2 \in J$, and so $2$ is a multiple of $f$ in $\Z \sqbrk X$.
So $f = \pm 2$ or $f = \pm 1$.
But this contradicts the fact that $J = \ideal f$.
Hence the result by Proof by Contradiction.