Show that $\Bbb Z\Big(\frac{1+\sqrt{d}}{2}\Big)$ is closed under multiplication.

Question

Here, we have that $d\equiv 1\bmod4$ and $$\Bbb Z\Big(\frac{1+\sqrt{d}}{2}\Big):=\Big\{u+v\Big(\frac{1+\sqrt{d}}{2}\Big): u,v\in\Bbb Z\Big\}.$$

Here is my attempt so far:

Suppose $A:=u+v\Big(\frac{1+\sqrt{d}}{2}\Big)$ and $B:=w+x\Big(\frac{1+\sqrt{d}} {2}\Big)$. Now, we consider the product $$\Big(u+v\Big(\frac{1+\sqrt{d}}{2}\Big)\Big)\cdot \Big(w+x\Big(\frac{1+\sqrt{d}} {2}\Big)\Big)=uw+ux\Big(\frac{1+\sqrt{d}} {2}\Big)+\frac{vx}{4}+\frac{vxd}{4}+\frac{vx\sqrt{d}}{4}.$$

Now, write $d=4k+1$ for some $k\in\Bbb Z$, then

$$AB=uw+ux\Big(\frac{1+\sqrt{d}} {2}\Big)+\frac{vx}{4}+\frac{vx(4k+1)}{4}+\frac{vx\sqrt{d}}{4}\\ =uw+ux\Big(\frac{1+\sqrt{d}} {2}\Big)+\frac{vx}{4}+\frac{4vxk+vx}{4}+\frac{vx\sqrt{d}}{4}\\ =uw+ux\Big(\frac{1+\sqrt{d}} {2}\Big)+\frac{vx}{2}+vxk+\frac{vx\sqrt{d}}{4}.$$

But this is where I'm stuck. Pointer on where to carry on please?

Answer 1

Let $\omega = {1+\sqrt{d}\over 2}$ and $d=4k+1$. Then

$$\omega ^2 = {1+2\sqrt{d}+d\over 4} = {2\sqrt{d}+4k+2\over 4}={1+\sqrt{d}\over 2}+k=\omega +k$$

So $$ (a+b\omega)(c+d\omega) = ac+(bc+ad)\omega +bd\omega ^2 = (ac+bdk)+(bc+ad+bd)\omega$$

and we are done.

Answer 2

Hint: Let $\theta = \frac{1+\sqrt{d}}{2}$. It is enough to prove that $\theta^2 = u + v \theta$ for some $u,v\in\mathbb Z$.

Solution:

Write $d=4t+1$. Then $\theta^2 = t + \theta$.

Answer 3

The mistake here:

It should be $$\frac{vx(1+\sqrt{d})^2}{4}=\frac{vx(1+d+2\sqrt{d})}{4}=\frac{vx(1+\sqrt{d})}{2}+\frac{vx(d-1)}{4},$$ which ends the proof.

Answer 4

\begin{eqnarray*} \left(u+v \left( \frac{1+ \sqrt{d}}{2} \right) \right) \left(w+x \left( \frac{1+ \sqrt{d}}{2} \right) \right) = uw+(ux+vw) \left(\frac{1+ \sqrt{d}}{2}\right)+ vx \left(\frac{1+d +2\sqrt{d}}{4}\right) \end{eqnarray*} now use $d=4k+1$ \begin{eqnarray*} \left(u+v \left( \frac{1+ \sqrt{d}}{2} \right) \right) \left(w+x \left( \frac{1+ \sqrt{d}}{2} \right) \right) = uw+kvx +(ux+vw+vx) \left(\frac{1+ \sqrt{d}}{2}\right). \end{eqnarray*}