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Homering without irreducibles

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An integral domain may not contain 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: 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

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

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

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FieldOfAlgebraicNumbers

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## Comments

## no irreducibles

Even simpler I suppose is to just consider that a field

has all nonzero elements are units, so there are no irreducibles.

## Re: no irreducibles

True, but this may be too trivial =o)