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Prove that if $d$ is a non-square integer with $d \equiv 1 \mod 4$ then the set $$\mathcal{O}_d := \left\{\frac{a + b\sqrt{d}}{2}:a,b \in \mathbb{Z}, a \equiv b \mod 2 \right\}$$ is a ring, and in particular a integral domain.

Little bit stuck here.

For $x,y \in \mathcal{O}_d,$ we define $+: \mathcal{O}_d \times \mathcal{O}_d \to \mathcal{O}_d$ by $(x + y) = \dfrac{a + b \sqrt{d}}{2} + \dfrac{a' + b'\sqrt{d}}{2} = \dfrac{(a+a') + (b+b')\sqrt{d}}{2} = \dfrac{a'+b'\sqrt{d}}{2} + \dfrac{a + b\sqrt{d}}{2} = (y + x) \in \mathcal{O}_d.$

We have that $0 = \dfrac{0 + 0\sqrt{d}}{2} \in \mathcal{O}_d$ and $\dfrac{a + b \sqrt{d}}{2} + \dfrac{-a + -b\sqrt{d}}{2} = \dfrac{(a-a) + (b-b)\sqrt{d}}{2} = 0 .$ Hence $-a \in \mathcal{O}_d.$ Also $$\begin{align} (x + y) + z &= \left(\dfrac{a + b\sqrt{d}}{2} + \dfrac{a' + b'\sqrt{d}}{2}\right) + \dfrac{a'' + b''\sqrt{d}}{2} \\&= \dfrac{(a + a') + (b + b')\sqrt{d}}{2} + \dfrac{a'' + b''\sqrt{d}}{2} \\&= \dfrac{(a + a' + a'') + (b + b' + b'')\sqrt{d}}{2} \\&= \dfrac{a + b\sqrt{d}}{2} + \dfrac{(a'+a'')+(b+b'')\sqrt{d}}{2} \\&= \dfrac{a + b\sqrt{d}}{2} + \left(\dfrac{a'+b'\sqrt{d}}{2} + \dfrac{a''+b''\sqrt{d}}{2}\right) \\&= x + (y+z) \end{align}$$

for all $x,y,z \in \mathcal{O}_d.$

Hence $\mathcal{O}_d$ is a group under addition.

I've had difficulty showing the multiplicative operation $\circ: \mathcal{O}_d \times \mathcal{O}_d \to \mathcal{O}_d$ stays in the set. How do we define the operation so that $x \circ y \in \mathcal{O}_d?$ When I multiply two elements from $\mathcal{O}_d$ I get $x \circ y = \left(\dfrac{a + b \sqrt{d}}{2} \right)\left(\dfrac{a' + b' \sqrt{d}}{2} \right) = \dfrac{(a + b \sqrt{d})(a' + b'\sqrt{d})}{4} = \dfrac{aa' + ab'\sqrt{d} +ba'\sqrt{d} + bb'd}{4}= \dfrac{aa' + bb'd + (ab' +ba')\sqrt{d}}{4}$ which doesn't seem to be an element of $\mathcal{O}_d.$

user26857
  • 52,094
St Vincent
  • 3,070

3 Answers3

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Hint $\ $ Show that $\,{\cal O}_d = \Bbb Z\left[w\right],\ w = (1\!+\!\sqrt d)/2,\,$ using $\ w^2\! -w + n = 0,\,\ n =(1\!-\!d)/4$

Bill Dubuque
  • 272,048
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Hint: $\mathcal O_d$ is a subset of real numbers, so it is enough to show that it is closed under $+,-,\cdot$ and that it contains $0$ and $1$. I'm pretty sure the operations are (implicitly) the standard addition and multiplication.

For the closure under multiplication, just consider the four cases (depending on the remainders of $a,a'$ modulo $2$).

tomasz
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$aa'+bb'd\equiv aa'+bb'\equiv aa'+aa'\equiv 0\bmod 2$, so $u=\dfrac{aa'+bb'd}{2}\in\mathbb Z$.

$ab'+ba'\equiv aa'+aa'\equiv 0\bmod 2$, so $v=\dfrac{ab'+ba'}{2}\in\mathbb Z$.

Now note that $x\circ y$ writes $\dfrac{u+v\sqrt d}{2}$ with $u,v\in\mathbb Z$.

In the following $a-b=2k$, $a'-b'=2k'$ and $d=1+4l$.

We have $u-v=\dfrac{aa'+bb'd}{2}-\dfrac{ab'+ba'}{2}=\dfrac{a'(a-b)+b'(bd-a)}{2}=\dfrac{2ka'+b'(-2k+4bl)}{2}=$ $ka'+b'(-k+2bl)=k(a'-b')+2bb'l=2kk'+2bb'l\equiv 0\bmod 2$, so $x\circ y\in\mathcal O_d$.

user26857
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