# rational integers in ideals

Proof.  Let  $\mathfrak{a}\neq(0)$  be any ideal of $\mathcal{O}_{K}$.  Take a nonzero element $\alpha$ of $\mathfrak{a}$.  The norm (http://planetmath.org/NormInNumberField) of $\alpha$ is the product

 $\mbox{N}(\alpha)\;=\;\alpha^{(1)}\underbrace{\alpha^{(2)}\cdots\alpha^{(n)}}_{\gamma}$

where $n$ is the degree of the number field and $\alpha^{(1)},\,\alpha^{(2)},\,\cdots,\,\alpha^{(n)}$ is the set of the http://planetmath.org/node/12046$K$-conjugates  of  $\alpha=\alpha^{(1)}$.  The number

 $\gamma=\frac{\mbox{N}(\alpha)}{\alpha}$

belongs to the field $K$ and it is an algebraic integer  , since $\alpha^{(2)},\,\cdots,\,\alpha^{(n)}$ are, as algebraic conjugates of $\alpha$, also algebraic integers.  Thus  $\gamma\in\mathcal{O}_{K}$.  Consequently, the non-zero integer

 $\mbox{N}(\alpha)\;=\;\alpha\gamma$

belongs to the ideal $\mathfrak{a}$, and similarly its opposite number.  So, $\mathfrak{a}$ contains positive integers, in fact infinitely many.

Title rational integers in ideals RationalIntegersInIdeals 2013-03-22 19:08:47 2013-03-22 19:08:47 pahio (2872) pahio (2872) 6 pahio (2872) Theorem msc 12F05 msc 06B10 msc 11R04 CharacteristicPolynomialOfAlgebraicNumber IdealNorm