Thread: Norm Question....

Hi guys!

I've just looked at a past year's number theory questions and one is as follows:

Let R = Z[Sqrt(-6)] & K = Q[Sqrt(-6)]. Put a = 3+2Sqrt(-6) & b = 7-6Sqrt(-6). By choosing c close to b/a, or otherwise, find c in R such that:

Normk(b-ac) < Normk(a)

I have no ideas! :-s

Any help would be most appreciated! x

Originally Posted by AAM

Hi guys!

I've just looked at a past year's number theory questions and one is as follows:

Let R = Z[Sqrt(-6)] & K = Q[Sqrt(-6)]. Put a = 3+2Sqrt(-6) & b = 7-6Sqrt(-6). By choosing c close to b/a, or otherwise, find c in R such that:

Normk(b-ac) < Normk(a)

I have no ideas! :-s

Any help would be most appreciated! x

I assume $\displaystyle N(x+y\sqrt{-6}) = x^2 + 6y^2$. Form $\displaystyle \tfrac{b}{a} = p+q\sqrt{-6}$ where $\displaystyle p,q\in \mathbb{Q}$ by multiplying the numerator and denominator of $\displaystyle \tfrac{b}{a}$ by the conjugate. Next choose $\displaystyle p_0,q_0$, intergers, so that $\displaystyle p_0,q_0$ are the closest integers to $\displaystyle p,q$ respectively. Set $\displaystyle c = p_0 + q_0 \sqrt{-6}$ and see that $\displaystyle N(b-ac) < N(a)$.

Oh, fantastic! :-D Nice and simple! Just wasn't quite sure what the question was asking for. :-s But I completely get it now!

Thank you. :-)