1

Is the lattice [$2+\sqrt{11},3-2\sqrt{11}$] an ideal in $O_{11}$

$N(2+\sqrt{11})=4-11=-7$

$N( 3-2\sqrt{11} )=9-4*11=-35$

$Tr( \left(2+\sqrt{11})(3-2\sqrt{11} )\right)=56$

since $\left[\alpha,\beta \right] \implies N(\alpha)x^2+Tr(\alpha\beta')xy+N(\beta)y^2$

$(-7x^2+56xy-35y^2)\frac{1}{-7}=x^2-8xy+5y^2$

$\therefore [a,b+c\omega _7]=[1,-8+5\sqrt{11}] $

also $O_{11}=[1,\omega_{11}]=[1,\sqrt{11}]$ since $11\equiv 3\mod 4$

how should i conclude this is an ideal in $O_{11}$

Jonas Kgomo
  • 1,032
  • 9
  • 19

1 Answers1

2

There is a simple criterion for such. If module $\rm\:M = [a,b\!+\!\sqrt{d}] = a\,\Bbb Z + (b\!+\!\sqrt{d})\,\Bbb Z\:$ is an ideal then it contains the norm $\rm\: N(b\!+\!\sqrt{d}) = (b\!+\!\sqrt{d})(b\!-\!\sqrt{d}) = b^2\!-\!d,\ $ so $\rm\,\ a\mid b^2\!-\!d.\:$ This necessary condition is also sufficient. The proof is easy:

The module $\rm\:M = [a,b\!+\!\sqrt{d}]\:$ is an ideal of $\rm\,R = \Bbb Z[\sqrt{d}]\iff$ M is closed under multiplcation by elements of $\rm\,R\iff$ $\rm\sqrt{d}\ M \subseteq M\iff a\sqrt{d},\, (b\!+\!\sqrt{d})\sqrt{d} \in M.\:$ The first inclusion is clear since $\rm\:a\sqrt{d}\, =\, a(b\!+\!\sqrt{d})-ab\in M.\:$ For the second inclusion

$$\ \ \ \ \begin{eqnarray}\rm -(b\!+\!\sqrt{d})\sqrt{d} &=\,&\rm (b\!+\!\sqrt{d})(b\!-\!\sqrt{d}\, -\,\color{#c00}b)\\[.2em] &=\,&\rm\:\!\ b^2\!-d\, -\,\color{#c00}b(b\!+\!\sqrt{d})\end{eqnarray}$$

which lies in module $\rm\, M = [a,b\!+\!\sqrt{d}]\iff a\mid b^2\!-\!d = N(b\!+\!\sqrt{d})$.

The criterion generalizes to test idealness of the module $\rm\,[a,b\!+\!c\,\omega]\,$ in the ring of integers of a quadratic number field, e.g. see section 2.3 in Franz Lemmermeyer's notes.

This is a special case of module normal forms that generalize to higher degree number fields, e.g. see the discussion on Hermite and Smith normal forms in Henri Cohen's $ $ A Course in Computational Number Theory.

Bill Dubuque
  • 272,048