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

Much of the proof of this theorem is given in Marcus’ 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 $\gcd(a,b)$ refers to the greatest common divisor in $\mathbb{Z}$ of $a$ and $b$ (which must necessarily be rational integers).

###### Proof.

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

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

$\begin{array}[]{ll}\langle p,\sqrt{d}\rangle^{2}&=\langle p^{2},p\sqrt{d},d% \rangle\\ &=\langle\gcd(p^{2},d),p\sqrt{d}\rangle\\ &=\langle p,p\sqrt{d}\rangle\\ &=\langle p\rangle\\ &=p\mathcal{O}_{K}.\end{array}$

Note that $\langle p,\sqrt{d}\rangle\neq\mathcal{O}_{K}$. (If they were equal, then $\langle p,\sqrt{d}\rangle^{2}$ would equal $\mathcal{O}_{K}$.)

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

$\begin{array}[]{ll}\mathcal{O}_{K}/\langle 2,1+\sqrt{d}\rangle&=\{a+b\sqrt{d}+% \langle 2,1+\sqrt{d}\rangle:a,b\in\mathbb{Z}\}\\ &=\{a-b+\langle 2,1+\sqrt{d}\rangle:a,b\in\mathbb{Z}\}\\ &=\{0+\langle 2,1+\sqrt{d}\rangle,1+\langle 2,1+\sqrt{d}\rangle\}.\end{array}$

Since $\langle 2,1+\sqrt{d}\rangle$ has $2$, it follows that $P=\langle 2,1+\sqrt{d}\rangle$ and $2\mathcal{O}_{K}=\langle 2,1+\sqrt{d}\rangle^{2}$.

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

$\begin{array}[]{ll}\displaystyle\left\langle 2,\frac{1+\sqrt{d}}{2}\right% \rangle\left\langle 2,\frac{1-\sqrt{d}}{2}\right\rangle&\displaystyle=\left% \langle 4,1+\sqrt{d},1-\sqrt{d},\frac{1-d}{4}\right\rangle\\ &\displaystyle=\left\langle 4,2,2\left(\frac{1-\sqrt{d}}{2}\right),\frac{1-d}{% 4}\right\rangle\\ &=\langle 2\rangle\\ &=2\mathcal{O}_{K},\end{array}$

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

If $d\equiv 5\operatorname{mod}8$, 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 $\operatorname{deg}f\geq 2$.

$\begin{array}[]{rl}\alpha&\displaystyle=\frac{1+\sqrt{d}}{2}\\ &\\ 2\alpha-1&=\sqrt{d}\\ (2\alpha-1)^{2}&=d\\ 4\alpha^{2}-4\alpha+1&=d\\ 4\alpha^{2}-4\alpha+1-d&=0\\ &\\ \displaystyle\alpha^{2}-\alpha+\frac{1-d}{4}&=0\end{array}$

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

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

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

$\begin{array}[]{ll}\langle p,n+\sqrt{d}\rangle\langle p,n-\sqrt{d}\rangle&=% \langle p^{2},pn+p\sqrt{d},pn-p\sqrt{d},n^{2}-d\rangle\\ &=\langle p^{2},2pn,pn-p\sqrt{d},n^{2}-d\rangle\\ &=\langle\gcd(p^{2},2pn),pn-p\sqrt{d},n^{2}-d\rangle\\ &=\langle p,pn-p\sqrt{d},n^{2}-d\rangle\\ &=\langle p\rangle\\ &=p\mathcal{O}_{K},\end{array}$

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

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

$\begin{array}[]{ll}\|\langle p,n+\sqrt{d}\rangle\|&=\sigma\left(\|\langle p,n+% \sqrt{d}\rangle\|\right)\\ &=\left\|\sigma\left(\langle p,n+\sqrt{d}\rangle\right)\right\|\\ &=\|\langle\sigma(p),\sigma(n+\sqrt{d})\rangle\|\\ &=\|\langle p,n-\sqrt{d}\rangle\|.\end{array}$

Note that $p^{2}=\|p\mathcal{O}_{K}\|=\|\langle p,n+\sqrt{d}\rangle\|\,\|\langle p,n-% \sqrt{d}\rangle\|=\|\langle p,n-\sqrt{d}\rangle\|^{2}$. Therefore, $\|\langle p,n+\sqrt{d}\rangle\|=\|\langle p,n-\sqrt{d}\rangle\|=p$. 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 $\operatorname{mod}p$, then consider the minimal polynomial $g(x)=x^{2}-d$ for $\sqrt{d}$ over $\mathbb{Q}$. Let $P$ be a lying over $p$ in $\mathcal{O}_{K}$. Note that $g(x)$ has a root in $\mathcal{O}_{K}$ and thus in $\mathcal{O}_{K}/P$. On the other hand, since $\operatorname{disc}g(x)=-4(-d)=4d$, which is not a square in $\mathbb{F}_{p}$, then $g(x)$ considered as an element of $\mathbb{F}_{p}[x]$ has no root in $\mathbb{F}_{p}$. Thus, $\mathcal{O}_{K}/P$ and $\mathbb{F}_{p}$ are not isomorphic. Therefore, $[\mathcal{O}_{K}/P\!:\!\mathbb{F}_{p}]>1$. Note that $1<[\mathcal{O}_{K}/P\!:\!\mathbb{F}_{p}]=f(P|p)\leq[K\!:\!\mathbb{Q}]=2$. Thus, $f(P|p)=2$. Therefore, $p$ is inert in $\mathcal{O}_{K}$. It follows that $p\mathcal{O}_{K}$ is in $\mathcal{O}_{K}$. ∎

## References

• 1 Marcus, Daniel A. Number Fields. New York: Springer-Verlag, 1977.
Title proof of prime ideal decomposition in quadratic extensions of $\mathbb{Q}$ ProofOfPrimeIdealDecompositionInQuadraticExtensionsOfmathbbQ 2013-03-22 15:59:06 2013-03-22 15:59:06 Wkbj79 (1863) Wkbj79 (1863) 20 Wkbj79 (1863) Proof msc 11R11