ring without irreduciblesRingWithoutIrreducibles2005-05-01 14:52:442006-09-24 02:52:05Exampleirreducible% this is the default PlanetMath preamble. as your knowledge
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\newtheorem*{thmplain}{Theorem}An integral domain may not \PMlinkescapetext{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: \PMlinkescapetext{theorem} 2).\, The element $\vartheta$ has the \PMlinkescapetext{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.