Polynomials in Integers is not Principal Ideal Domain

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Let $\Z \sqbrk X$ be the ring of polynomials in $X$ over $\Z$.

Then $\Z \sqbrk X$ is not a principal ideal domain.


Let $J$ be the ideal formed from the set of polynomials over $\Z$ in $X$ which have a constant term which is even.

From Polynomials in Integers with Even Constant Term forms Ideal, $J$ is indeed an ideal.

Aiming for a contradiction, suppose $J$ is a principal ideal of $\Z \sqbrk X$ such that $J = \ideal f$.

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.