proof of prime ideal decomposition in quadratic extensions of $\mathbb{Q}$

Much of the proof of this theorem is given in Marcus’ Number Fields (http://planetmath.org/NumberField); however, all of the details will be filled in here, and some aspects of the proof here will differ from those of Marcus.

Note that $\gcd \mathit{}\mathrm{(}a\mathrm{,}b\mathrm{)}$ refers to the greatest common divisor in $\mathrm{Z}$ of $a$ and $b$ (which must necessarily be rational integers).

Proof.

Let $d$ be a squarefree integer with $d\ne 1$ and $K=\mathbb{Q}(\sqrt{d})$.

If $p$ is a rational prime that divides $d$, then

Note that $⟨p,\sqrt{d}⟩\ne {𝒪}_K$. (If they were equal, then ${⟨p,\sqrt{d}⟩}^2$ would equal ${𝒪}_K$.)

If $d\equiv 3\mathrm{mod}4$, then $\mathrm{disc}(K)=4d$. Note that $2$ divides $\mathrm{disc}(K)$. Thus, $2$ ramifies in ${𝒪}_K$. Therefore, $2{𝒪}_K=P^2$ for some prime ideal $P$ of ${𝒪}_K$. Moreover, $P$ is the unique ideal of ${𝒪}_K$ of norm (http://planetmath.org/IdealNorm) $2$. Since $\sqrt{d}\equiv -1\mathrm{mod}⟨2,1+\sqrt{d}⟩$, then

Since $⟨2,1+\sqrt{d}⟩$ has $2$, it follows that $P=⟨2,1+\sqrt{d}⟩$ and $2{𝒪}_K={⟨2,1+\sqrt{d}⟩}^2$.

If $d\equiv 1\mathrm{mod}8$, then $\mathrm{disc}(K)=d$. Note that $2$ does not divide $\mathrm{disc}(K)$. Thus, $2$ does not ramify in ${𝒪}_K$. Since

we have that $⟨2,{\displaystyle \frac{1+\sqrt{d}}{2}}⟩$ and $⟨2,{\displaystyle \frac{1-\sqrt{d}}{2}}⟩$ must be distinct. Proving that these ideals are indeed given below.

If $d\equiv 5\mathrm{mod}8$, then consider the minimal polynomial $f(x)\in \mathbb{Z}[x]$ for ${\displaystyle \frac{1+\sqrt{d}}{2}}$. Since ${\displaystyle \frac{1+\sqrt{d}}{2}}\notin \mathbb{Q}$, it must be the case that $\mathrm{deg}f\ge 2$.

Thus, $f(x)=x^2-x+{\displaystyle \frac{1-d}{4}}$.

Let $P$ be a lying over $2$ in ${𝒪}_K$. Note that $f(x)$ has a root (http://planetmath.org/Root) in ${𝒪}_K$ and thus in ${𝒪}_K/P$. On the other hand, since $f(x)\equiv x^2+x+1\mathrm{mod}2$, $f(x)$ considered as an element of ${𝔽}_2[x]$ has no root in ${𝔽}_2$. Thus, ${𝒪}_K/P$ and ${𝔽}_2$ are not isomorphic. Therefore, $[{𝒪}_K/P:{𝔽}_2]>1$. Since , we have that $f(P|2)=2$. Thus, $2$ is inert in ${𝒪}_K$. It follows that $2{𝒪}_K$ is in ${𝒪}_K$.

If $p$ is an odd prime (http://planetmath.org/Prime) that does not divide $d$ and $d\equiv n^2\mathrm{mod}p$, then $p$ does not divide $\mathrm{disc}(K)$ (which equals either $d$ or $4d$). Thus, $p$ does not ramify in ${𝒪}_K$. Also, $p$ does not divide $n$. Since

we have that $⟨p,n+\sqrt{d}⟩$ and $⟨p,n-\sqrt{d}⟩$ must be distinct. It will be proven that these ideals are indeed .

Let $\parallel I\parallel$ denote the norm of the ideal $I$ (http://planetmath.org/IdealNorm) of ${𝒪}_K$ and $\sigma \in \mathrm{Gal}(K/\mathbb{Q})$ with $\sigma (\sqrt{d})=-\sqrt{d}$. Then

Note that $p^2=\parallel p{𝒪}_K\parallel =\parallel ⟨p,n+\sqrt{d}⟩\parallel \parallel ⟨p,n-\sqrt{d}⟩\parallel ={\parallel ⟨p,n-\sqrt{d}⟩\parallel }^2$. Therefore, $\parallel ⟨p,n+\sqrt{d}⟩\parallel =\parallel ⟨p,n-\sqrt{d}⟩\parallel =p$. It follows that the indicated ideals are .

Finally, if $p$ is an odd prime that does not divide $d$ and $d$ is not a square $\mathrm{mod}p$, then consider the minimal polynomial $g(x)=x^2-d$ for $\sqrt{d}$ over $\mathbb{Q}$. Let $P$ be a lying over $p$ in ${𝒪}_K$. Note that $g(x)$ has a root in ${𝒪}_K$ and thus in ${𝒪}_K/P$. On the other hand, since $\mathrm{disc}g(x)=-4(-d)=4d$, which is not a square in ${𝔽}_p$, then $g(x)$ considered as an element of ${𝔽}_p[x]$ has no root in ${𝔽}_p$. Thus, ${𝒪}_K/P$ and ${𝔽}_p$ are not isomorphic. Therefore, $[{𝒪}_K/P:{𝔽}_p]>1$. Note that . Thus, $f(P|p)=2$. Therefore, $p$ is inert in ${𝒪}_K$. It follows that $p{𝒪}_K$ is in ${𝒪}_K$.∎