An integral domain may not contain any irreducible elements. One such example is the ring of all algebraic integers. Any non-zero non-unit $\vartheta$ of this ring satisfies an equation $$x^n\!+\!a_1x^{n-1}\!+\!\cdots\!+\!a_{n-1}x\!+\!a_n = 0$$ with integer coefficients $a_j$ , since it is an algebraic integer; moreover, we can assume that $a_n = \mbox{N}(\vartheta) \neq \pm 1$ (see norm and trace of algebraic number: theorem 2). The element $\vartheta$ has the decomposition $$\vartheta = \sqrt{\vartheta}\!\cdot\!\sqrt{\vartheta}.$$ Here, $\sqrt{\vartheta}$ belongs to the ring because it satisfies the equation $$x^{2n}\!+\!a_1x^{2n-2}\!+\!\cdots\!+\!a_{n-1}x^2\!+\!a_n = 0,$$ and it is no unit. Thus the element $\vartheta$ is not irreducible.