A non-unit which is neither irreducible, nor a product of irreducibles.

Question

I am having trouble with this exercise

Find a non-unit $d \in D = \{f \in \mathbb{Q}[x] \mid f(0) \in \mathbb{Z}\}$ where $d$ is neither irreducible, nor a product of irreducibles.

I find it contradictory that an element can be non-irreducible, yet not a product of irreducibles.

For those interested, this is exercise 32.10 from A First Course in Abstract Algebra (Rings, Groups and Fields) The Third Edition by Marlow Anderson and Todd Feil.

Answer 1

Consider the element $x \in D$. Notice that $x = 2(\frac{1}{2} x)$. Now $2$ is not a unit since $\frac{1}{2} \notin D$ and $x$ is clearly not a unit (it isn't a unit in $\mathbb{Q}[x]$). Hence $x$ is not irreducible.

Moreover a similar argument shows that $\alpha x$ is not irreducible for any $\alpha \in \mathbb{Q}^*$. It is not hard to see that if $f(x), g(x) \in D$ are such that $f(x)g(x) = x$ then one of $f(x)$ or $g(x)$ is of the form $\alpha x$.