Exhibit an integral domain $R$ and a non-zero non-unit element of $R$ that is not a product of irreducibles.

Exhibit an integral domain $R$ and a non-zero non-unit element of $R$ that is not a product of irreducibles.

My thoughts so far: I don't really have a clue. Could anyone direct me on how to think about this? I'm struggling to get my head round irreducibles.

Thanks.

Solution 1:

HINT $\ $ Consider a domain, not a field, that is closed under square-roots. It has no irreducibles since every element factors $\rm\ d\ =\ \sqrt{d}\ \sqrt{d}\:.\ $ For example, the domain of all algebraic integers.

Solution 2:

A standard example of a non-Noetherian domain is the ring $R=\{f(x) \in \mathbb{Q}[x]: f(0) \in \mathbb{Z} \}$; that is, the ring of rational polynomials with integer constant term. The units of $R$ are just $\pm 1$. The polynomial $x$ is reducible, since it factors as $2 \cdot \frac{1}{2}x$, but is not a product of irreducibles, since one of the factors would have to be $qx$ for some rational $q$, and again this factors as $2 \cdot \frac{q}{2}x$.