Polynomials in Integers is not Principal Ideal Domain

Theorem

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

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

Proof

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.

$\blacksquare$

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $9$: Rings: Exercise $21$