"Integers" in quadratic field

Apr 2011
108
2
Somwhere in cyberspace.
Greetings,

Is it true that the subring of quadratic field, given by \(\displaystyle \{a+b\sqrt{D}|a,b \in Z\}\) is contained in the subring \(\displaystyle \{a+b\frac{1+\sqrt{D}}{2}|a,b \in Z\}\), if D is equivalent to 1 modulo 4.

Dummit and Foote use the terminology "slightly larger one" for the second, which implies that it contains the first. I tried to prove it but could not, so I did some examples. I picked D=5, and a=b=1. So if the conjecture is true, there must be \(\displaystyle c,d \in Z\), such that
\(\displaystyle 1+\sqrt{5}=c+d\frac{1+\sqrt{5}}{2}}\Leftrightarrow 1+\sqrt{5}=\frac{2c+d(1+\sqrt{5})}{2} \Leftrightarrow 2(1+\sqrt{5})=2c+d(1+\sqrt{5})\Leftrightarrow(2-d)(1+\sqrt{5})=2c\)

Case 1: \(\displaystyle d\ne2 \Rightarrow \sqrt{5}=\frac{2c}{2-d}-1\) that on the right looks like rational number
Case 2: \(\displaystyle d=2 \Rightarrow 1+\sqrt{5}=c\) which also implies that \(\displaystyle \sqrt{5}\) is rational number (integer actually)

So, check is needed and If I'm correct, I'm interested what constitutes the intersection of these subrings ?
 
Last edited:
Jun 2013
1,127
601
Lebanon
\(\displaystyle a+b\sqrt{D}=(a-b)+2b\left(\frac{1+\sqrt{D}}{2}\right)\)

shows that it is a subring


\(\displaystyle \frac{1+\sqrt{D}}{2}\) cannot be written as \(\displaystyle a+b\sqrt{D}\)

shows it is a proper subring
 
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Apr 2011
108
2
Somwhere in cyberspace.
I'm yet to digest Idea's answer, but just noted the grave error in my original post.
Case 2 (when d=2) implies that c=0 which shows that c and d do in fact exist in this particular example.
The error exist, because I realized the division by zero problem only at the moment of writing the original post. Stupid.
 
Last edited:
Apr 2011
108
2
Somwhere in cyberspace.
I got the Idea's answer too. You show how \(\displaystyle a+b\sqrt(D)\) be rewritten as member of the larger ring. Then show a member of the large ring that is not member of the smaller. Thanks very much!