Quadratic Number Field

Nov 2009
169
2
Let \(\displaystyle D \in \mathbb{Z}\) be square free. Consider \(\displaystyle K = \{ x| ax^2+bx+c = 0, a,b,c \in \mathbb{Z} \}\).

For which conditions on \(\displaystyle a,b,c\in \mathbb{Z}\) is \(\displaystyle K= \{ u + v\sqrt{D}: u,v \in \mathbb{Q} \}\)?
 
Dec 2009
411
131
With the abc-formula we have \(\displaystyle K= \left\{\frac{-b}{2a}\pm \frac{1}{2a}\sqrt{b^2-4ac}\right\}\)

Hence we must have \(\displaystyle kD=b^2-4ac\) for some square \(\displaystyle k\)
(\(\displaystyle \sqrt{k}\in \mathbb{Q}\))
 
Nov 2009
169
2
Could you enlarge on that, what are the conditions on a,b,c?
 
Dec 2009
411
131
Ok.

So we can choose \(\displaystyle k=p^2\in \mathbb{Z}\) free.

Then we choose \(\displaystyle b\in \mathbb{Z}\) such that \(\displaystyle ac=\frac{b^2-kD}{4}\in \mathbb{Z}\) is not prime.

Then \(\displaystyle a,c\) divide this number.
 
Nov 2009
169
2
So you say that we choose a and c arbitrarily, such that \(\displaystyle ac\) is not a prime, and then b needs to be chosen such that \(\displaystyle 4ac = b^2 -kD\), for some \(\displaystyle k=p^2\) \(\displaystyle (p \in \mathbb{Z})\)

Is it that what you mean? And why is this what we want??
 
Dec 2009
411
131
with the abc-formula we have that \(\displaystyle K\) consists of the numbers \(\displaystyle \frac{-b}{2a}\pm \frac{1}{2a}\sqrt{b^2-4ac}\)

To make sure \(\displaystyle K\) has the desired property we must have \(\displaystyle \sqrt{b^2-4ac}=p\sqrt{D}\) for some number \(\displaystyle p\in \mathbb{Z}\).

That is, \(\displaystyle b^2-4ac= p^2D\).

This number \(\displaystyle p^2\in \mathbb{Z}\) can ofcourse be freely chosen.

Then we must choose \(\displaystyle a,b,c\in\mathbb{Z}\) that satisfy this relation. Thus we must choose a \(\displaystyle b\in \mathbb{Z}\) such that

there exists a composite number \(\displaystyle ac\in\mathbb{Z}\) that satisfies \(\displaystyle b^2 = p^2D+4ac\)

wich is, the same as saying \(\displaystyle \frac{b^2-p^2D}{4}\) must be a composite number. Only then we can choose \(\displaystyle a,c\in \mathbb{Z}\) such that \(\displaystyle ac=\frac{b^2-p^2D}{4}\).

And we've found \(\displaystyle a,b,c\) such that \(\displaystyle K\) has the desired property.
 
Dec 2009
411
131
Why does ac need to be composite?
How else can we find \(\displaystyle a,c\in \mathbb{Z}\)?

If \(\displaystyle ac=p\) prime then \(\displaystyle a,c\) can not be integers.
 

chiph588@

MHF Hall of Honor
Sep 2008
1,163
429
Champaign, Illinois
How else can we find \(\displaystyle a,c\in \mathbb{Z}\)?

If \(\displaystyle ac=p\) prime then \(\displaystyle a,c\) can not be integers.
\(\displaystyle a=1,\; c=p \)
 
Dec 2009
411
131
Ah shoot, big blunder here. Shameful moment here xp

Thanks chiph

Forget everything about ac being composite. I'm sorry for this error ;p


The rest you can take for granted.