Exercise 6.7

Proof. (We write $\bar{\mathbb{Z}}$ the ring of algebraic integers in $ℂ$, and ${\mathcal{}O}_K$ (or ${\mathbb{Z}}_K$) the ring of algebraic integers in the field $K$.)

If $D=1$, $\mathbb{Q}[\sqrt{D}]=\mathbb{Q}$. If $D\ne 1$, as $D$ is square-free, $D$ is not a square, so $\sqrt{D}$ is irrational.

Let $\gamma =r+s\sqrt{D}\in \mathbb{Q}[\sqrt{D}](r,s\in \mathbb{Q})$ an algebraic integer of $\mathbb{Q}[\sqrt{D}]$ ( $D\in \mathbb{Z},D$ square-free). ${(\gamma -r)}^2=s^{2}D$, so ${\gamma }^2-2\mathit{r\gamma }+r^2-D{s}^2=0$. $\gamma$ is a root of

If $s=0$, then the minimal polynomial of $\gamma$ is $x-r$. As $r=\gamma$ is an algebraic integer and $r\in \mathbb{Q}$, then $r\in \mathbb{Z}$. In this case $r\in \mathbb{Z}$ and $s=0$.

If $s\ne 0$, $\gamma \notin \mathbb{Q}$, so no polynom of degree $d\le 1$ has the root $\gamma$. Thus the minimal polynomial of $\gamma$ is $p(x)$. From Exercise 6.5, $p(x)\in \mathbb{Z}[x]$, so (in the two cases $s=0,s\ne 0$)

Conversely, if $2r\in \mathbb{Z},r^2-D{s}^2\in \mathbb{Z}$, then $p(x)\in \mathbb{Z}[x]$ and $p(\gamma )=0$, thus $\gamma$ is an algebraic integer.

If $r,s\in \mathbb{Q}$, $D\ne 1$ square-free,

Let $\gamma =r+s\sqrt{D}\in \bar{\mathbb{Z}}$. We can write

Then

As $D$ is square-free, $D\not\equiv 0(\mathit{mod}4)$.

$\text{∙}$

Case 1: $D\equiv 2,3(\mathit{mod}4)$.

$n^2-4m=\frac{4D{b}^2}{d^2}$, so $d\mid 2a,d^2\mid 4D{b}^2$.

If $2\mid d$, $4\mid a^2-D{b}^2$, $a^2\equiv D{b}^2(\mathit{mod}4)$. As $d\wedge a\wedge b=1$, and $2\mid d$, $a$ or $b$ is odd, and $a^2\equiv D{b}^2(\mathit{mod}4),D\not\equiv 0(mod4)$, implies that $a$ and $b$ are both odd. Then $a^2\equiv b^2\equiv 1(\mathit{mod}4)$, so $D\equiv 1(\mathit{mod}4)$ : this is in contradiction with the hypothesis $D\equiv 2,3(\mathit{mod}4)$. So $d$ is an odd number.

Consequently, $d\mid a,d^2\mid D{b}^2$. If $p\in \mathbb{N}$ is a prime factor of $d$, $p\mid d,p\mid a$, and $d\wedge a\wedge b=1$, thus $p\nmid b$, and since $p^2\mid D{b}^2$, $p^2\mid D$, in contradiction with $D$ square-free. So $d\ge 1$ has no prime factor : $d=1$ and $r=a,s=b\in \mathbb{Z}$. Conversely, any $\gamma =a+b\sqrt{D},a,b\in \mathbb{Z}$ is an algebraic integer, so

$${\mathcal{}O}_{\mathbb{Q}[\sqrt{D}]}=\bar{\mathbb{Z}}\cap \mathbb{Q}[\sqrt{D}]=\{a+b\sqrt{D}|a,b\in \mathbb{Z}\}.$$

$\text{∙}$

Case 2: $D\equiv 1(\mathit{mod}4)$.

Then $r=\frac{n}{2},n\in \mathbb{Z}$. Write $s=\frac{u}{v},u\wedge v=1,v\ge 1$.

$m=r^2-D{s}^2=\frac{n^2}{4}-D\frac{u^2}{v^2}\in \mathbb{Z}$, $4D\frac{u^2}{v^2}=n^2-4m\in \mathbb{Z}$, so $v^2\mid 4D{u}^2$. Since $u\wedge v=1,u^2\wedge v^2=1$, so $v^2\mid 4D$. As $D$ is square-free, $v$ has no odd prime factor, so $v=2^k$. Since $D$ is odd, $k\le 1$ and $v=1$ or $v=2$. So $r,s$ are both half-integers: $r=n∕2,s=n^{\prime }∕2,n,n^{\prime }\in \mathbb{Z}$.

$4m=n^2-D{n{}^{\prime }}^2$, thus $n^2\equiv {n{}^{\prime }}^2(\mathit{mod}4)$, so $n,n^{\prime }$ have the same parity. Let $a=\frac{n+n^{\prime }}{2}\in \mathbb{Z},b=n^{\prime }\in \mathbb{Z}$. Then $n=2a-b,n^{\prime }=b$ and $\gamma =\frac{n}{2}+\frac{n^{\prime }}{2}\sqrt{D}=a-\frac{b}{2}+\frac{b}{2}\sqrt{D}=a+b\left(\frac{-1+\sqrt{D}}{2}\right)$.

Conversely, $\frac{-1+\sqrt{D}}{2}$ is a root of $x^2+x+\frac{1-D^2}{4}\in \mathbb{Z}[x]$, so every $a+b\left(\frac{-1+\sqrt{D}}{2}\right)$ is an algebraic integer.

$${\mathcal{}O}_{\mathbb{Q}[\sqrt{D}]}=\bar{\mathbb{Z}}\cap \mathbb{Q}[\sqrt{D}]=\{a+b\left(\frac{-1+\sqrt{D}}{2}\right)|a,b\in \mathbb{Z}\}.$$