# ring without irreducibles

An integral domain may not any irreducible elements.  One such example is the ring of all algebraic integers.  Any nonzero non-unit $\vartheta$ of this ring satisfies an equation

 $x^{n}\!+\!a_{1}x^{n-1}\!+\!\cdots\!+\!a_{n-1}x\!+\!a_{n}=0$

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: 2).  The element $\vartheta$ has the

 $\vartheta=\sqrt{\vartheta}\!\cdot\!\sqrt{\vartheta}.$

Here, $\sqrt{\vartheta}$ belongs to the ring because it satisfies the equation

 $x^{2n}\!+\!a_{1}x^{2n-2}\!+\!\cdots\!+\!a_{n-1}x^{2}\!+\!a_{n}=0,$

and it is no unit.  Thus the element $\vartheta$ is not irreducible.

Title ring without irreducibles RingWithoutIrreducibles 2014-05-29 11:39:19 2014-05-29 11:39:19 pahio (2872) pahio (2872) 15 pahio (2872) Example msc 13G05 FieldOfAlgebraicNumbers