#### mrproper

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:

#### Idea

$$\displaystyle a+b\sqrt{D}=(a-b)+2b\left(\frac{1+\sqrt{D}}{2}\right)$$

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

1 person

#### mrproper

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:

#### mrproper

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!