# number field that is not norm-Euclidean

Proposition.  The real quadratic field $\mathbb{Q}(\sqrt{14})$ is not norm-Euclidean.

Proof.  We take the number  $\gamma=\frac{1}{2}+\frac{1}{2}\sqrt{14}$  which is not integer of the field ($14\equiv 2\pmod{4}$).  Antithesis:  $\gamma=\varkappa+\delta$  where  $\varkappa=a+b\sqrt{14}$  is an integer of the field ($a,\,b\in\mathbb{Z}$) and

 $|\mbox{N}(\delta)|=\left|\left(\frac{1}{2}-a\right)^{2}-14\left(\frac{1}{2}-b% \right)^{2}\right|<1.$

Thus we would have

 $|\underbrace{(2a-1)^{2}-14(2b-1)^{2}}_{E}|<4.$

And since  $(2a-1)^{2}=4(a-1)a+1\equiv 1\pmod{8}$,  it follows  $E\equiv 1-14\cdot 1\equiv 3\pmod{8}$,  i.e.  $E=3$.  So we must have

 $\displaystyle(2a-1)^{2}\equiv(2a-1)^{2}-14(2b-1)^{2}\equiv 3\pmod{7}.$ (1)

But  $\{0,\,\pm 1,\,\pm 2,\,\pm 3\}$  is a complete residue system modulo 7, giving  the set  $\{1,\,2,\,4\}$  of possible quadratic residues modulo 7.  Therefore (1) is impossible.  The antithesis is wrong, whence the theorem 1 of the parent entry (http://planetmath.org/EuclideanNumberField) says that the number field is not norm-Euclidean.

Note.  The function N used in the proof is the usual

 $\mbox{N}:\,\,r\!+\!s\sqrt{14}\,\mapsto\,r^{2}\!-\!14s^{2}\quad(r,\,s\in\mathbb% {Q})$

defined in the field $\mathbb{Q}(\sqrt{14})$.  The notion of norm-Euclidean number field is based on the norm (http://planetmath.org/NormAndTraceOfAlgebraicNumber).  There exists a fainter function, the so-called Euclidean valuation, which can be defined in the maximal orders of some algebraic number fields (http://planetmath.org/NumberField); such a maximal order, i.e. the ring of integers of the number field, is then a Euclidean domain.  The existence of a Euclidean valuation guarantees that the maximal order is a UFD and thus a PID.  Recently it has been shown the existence of the Euclidean domain $\mathbb{Z}[\frac{1+\sqrt{69}}{2}]$ in the field $\mathbb{Q}(\sqrt{69})$ but the field is not norm-Euclidean.

The maximal order $\mathbb{Z}[\sqrt{14}]$ of $\mathbb{Q}(\sqrt{14})$ has also been proven to be a Euclidean domain (Malcolm Harper 2004 in Canadian Journal of Mathematics).

Title number field that is not norm-Euclidean NumberFieldThatIsNotNormEuclidean 2013-03-22 16:56:56 2013-03-22 16:56:56 pahio (2872) pahio (2872) 15 pahio (2872) Example msc 13F07 msc 11R21 msc 11R04 UniqueFactorizationAndIdealsInRingOfIntegers