Number Theory theorem regarding ring of integers in $\mathbb{Q}[\sqrt{D}]$

Question

Here is the theorem that I need to prove

For $K = \mathbb{Q}[\sqrt{D}]$ we have

$$\begin{align}O_K = \begin{cases} \mathbb{Z}[\sqrt{D}] & D \equiv 2, 3 \mod 4\\ \mathbb{Z}\left[\frac{1 + \sqrt{D}}{2}\right] & D \equiv 1 \mod 4 \end{cases} \end{align}$$

The theorem we need to use is this one that can be found in any generic number theory textbook.

an element $\alpha\in K$ is an algebraic integer if and only if its minimal polynomial has coefficients in $\mathbb{Z}$.

I tried many avenues of attack but it is extremely hard to prove. How do I prove it?

Answer 1

$\bullet$ Let $d \equiv 1 \mod 4.$

Let $A=x+y\sqrt{d} \in \mathcal O_K$ be arbitrary. We have

$(A-x)^2 - dy^2 = 0$. Expanding,

$A^2 -2xA +x^2 - dy^2 = 0$.

Now we use the fact that $m_A(x)$, the minimal polynomial of $A$, must be in $\Bbb Z[x]$: we must have $x=\frac{r}{2}$ for some $r \in \Bbb Z$.

Since $x^2 - dy^2 \in \Bbb Z$ too, we must have $y = \frac{s}{2}$.

So $x^2 - dy^2 = \frac{r^2}{4} -d\frac{s^2}{4}$.

This implies $r^2 - ds^2 \equiv r^2 - s^2 \equiv 0 \mod 4$.

So $r \equiv s \mod 2$, i.e. $r=s+2t$.

Putting this together, $A=\frac{r}{2} + \frac{s}{2}\sqrt{d} = t+s\frac{1+\sqrt{d}}{2}$ as required, showing that $\mathcal O_K = \Bbb Z[\frac{1+\sqrt{d}}{2}]$.

$\bullet$ Now assume $d \equiv 2 \mod 4$.

We can repeat the above argument, arriving at

$r^2 +2s^2 \equiv 0 \mod 4$.

The only solution is $(r, s) = (0, 0) \mod 4$, so in fact $x, y \in \Bbb Z$ and $\mathcal O_K = \Bbb Z[\sqrt{d}]$.

$\bullet$ For $d \equiv 3 \mod 4$ we have to consider $r^2 + s^2 \equiv 0 \mod 4$ and we come to the same conclusion as for $d=2$.

Answer 2

There are lots of approaches depending on how much you assume. Assuming one only knows traces and norms and your theorem, this should be a 'low brow' approach that gets there without any other knowledge assumptions.

Let $K=\mathbb{Q}(\sqrt{d})$, where $d\neq 1$ is a squarefree integer. We want to find $\mathcal{O}_K$. If $\alpha=a+b\sqrt{d} \in \mathcal{O}_K$, where $a,b \in \mathbb{Q}$, we know that $\text{Nm}_{K/\mathbb{Q}}(\alpha),\text{Tr}_{K/Q}(\alpha) \in \mathbb{Z}$. Hence, $a^2-db^2,2a \in \mathbb{Z}$. Multiplying $a^2-db^2$ by 4, we obtain $(2a)^2-d(2b)^2,2a \in \mathbb{Z}$.

Therefore, $2\mathcal{O}_K \subseteq \mathbb{Z}[\sqrt{d}]=\{a+b\sqrt{d} \colon a,b \in \mathbb{Z}\}$. Then we have an inclusion of abelian groups $$ \mathbb{Z}[\sqrt{d}] \subseteq \mathcal{O}_K \subseteq \dfrac{1}{2}\, \mathbb{Z}[\sqrt{d}] $$ The quotient $\frac{1}{2}\mathbb{Z}[\sqrt{d}]/\mathbb{Z}[\sqrt{d}]$ is a group of order 4 with coset representatives: 0, $\frac{1}{2}$, $\frac{\sqrt{d}}{2}$, and $\frac{1+\sqrt{d}}{2}$.

In order to determine $\mathcal{O}_K$, we need to determine which of these representatives are algebraic integers. Clearly, $0 \in \mathcal{O}_K$ and $\frac{1}{2} \notin \mathcal{O}_K$. The minimal polynomial of $\frac{\sqrt{d}}{2}$ is $x^2 - \frac{d}{4}$---which is not in $\mathbb{Z}[x]$ as $d$ is square free. Hence, $\frac{\sqrt{d}}{4} \notin \mathcal{O}_K$. Finally, the minimal polynomial of $\frac{1+\sqrt{d}}{2}$ is $$ \left(x - \dfrac{1+\sqrt{d}}{2}\right) \left(x - \dfrac{1-\sqrt{d}}{2}\right)= x^2-x+ \dfrac{1-d}{4}. $$ Then $\frac{1+\sqrt{d}}{2}$ has minimal polynomial $p_\alpha(x) \in \mathbb{Z}[x]$. [That is, $\frac{1+\sqrt{d}}{2} \in \mathcal{O}_K$ if and only if $d \equiv 1 \mod 4$.] Therefore, $$ \mathcal{O}_K= \begin{cases} \mathbb{Z}[\sqrt{d}], & d \not\equiv 1 \mod 4 \\ \mathbb{Z}\left[\frac{1+\sqrt{d}}{2}\right], & d \equiv 1 \mod 4. \end{cases} $$

Answer 3

You can find a proof in many number theory textbooks. In Hardy and Wright's classic Introduction to the Theory of Numbers, for instance, its Theorem 238 on p.207.

Answer 4

Let $\alpha=p+q\sqrt D$ with $p,q\in\mathbb Q$. The other root of its minimal polynomial is $\bar \alpha=p-q\sqrt D.$

The minimal polynomial is $(x-\alpha)(x-\bar\alpha)=x^2-(\alpha+\bar\alpha)x+\alpha\bar\alpha=x^2-2px+(p^2-q^2D). $

For $-2p$ to be an integer, say $-m$, $p=m/2$, where $m \in \mathbb Z$.

For $p^2-q^2D$ to be an integer, say $n$, this means also $q=l/2$ for some $l \in \mathbb Z$.

So $p^2-q^2D=(\frac m 2)^2-(\frac l 2)^2D$ is an integer so $m^2-l^2D$ is a multiple of $4$.

Modulo $4$, $m^2$ and $l^2$ could be $\equiv0$ or $1$.

Can you take it from here?