# Determination of the prime ideals lying over an odd prime in a quadratic order

We need some notation before we state the problem. Let $K$ be a quadratic number field, $d$ its discriminant. Let $R$ be an order of $K$, i.e. a subring of $K$ which is a free $\mathbb{Z}$-module of rank $2$. Let $D$ its discriminant. Let $x_1,\cdots, x_n$ be a sequence of elements of $R$. We denote by $[x_1,\cdots,x_n]$ the $\mathbb{Z}$-submodule of $R$ generated by $x_1,\cdots, x_n$. By this question, $R = [1, \omega]$, where $\omega = \frac{(D + \sqrt D)}{2}$. Let $\sigma$ be the unique non-identity automorphism of $K/\mathbb{Q}$. We denote $\sigma(\alpha)$ by $\alpha'$ for $\alpha \in R$. We denote $\sigma(I)$ by $I'$ for an ideal $I$ of $R$. Let $f$ be the order of $\mathcal{O}_K/R$ as a $\mathbb{Z}$-module. Then $D = f^2d$ (see this question). Let $p$ be a prime number such that gcd$(p, f) = 1$. Then $pR$ is regular by this question. Hence $pR$ is uniquely decomposed as a product of regular prime ideals by this question. So it is natural to ask how it is decomposed. I came up with the following proposition.

Proposition Let $K, R$, etc. be as above. Let $p$ be an odd prime number such that gcd$(p, f) = 1$.

Case 1 $D$ is divisible by $p$.

$P = [p, \omega]$ is a prime ideal and $pR = P^2$. Moreover $P = P'$.

Case 2 gcd$(D, p) = 1$ and $D$ is quadratic residue modulo $p$.

Let $b, b'$ be solutions of $(2x + d)^2 \equiv d$ (mod $p$) such that $b - b'$ is not divisible by $p$. Then $P = [p, b + \omega]$ and $P' = [p, b' + \omega]$ are distinct prime ideals and $pR = PP'$.

Case 3 gcd$(D, p) = 1$ and $D$ is quadratic non-residue modulo $p$.

$pR$ is a prime ideal.

Method of my proof I used the result of this question.

My question How do you prove the proposition? I would like to know other proofs based on different ideas from mine. I welcome you to provide as many different proofs as possible. I wish the proofs would be detailed enough for people who have basic knowledge of introductory algebraic number theory to be able to understand.

Let $P$ be a prime ideal of $R$ lying over $p$. Since $P \cap \mathbb{Z} = p\mathbb{Z}$, by this question, there exist integers $b, c$ such that $P = [p, b + c\omega], c \gt 0, p \equiv 0$ (mod $c$), $b \equiv 0$ (mod $c$). Then $c = 1$ or $p$. Suppose $c = 1$. Then $N_{K/\mathbb{Q}}(b + \omega) \equiv 0$ (mod $p$)(see here). $N_{K/\mathbb{Q}}(b + \omega) = (b + \omega)(b + \omega') = \frac{2b + D + \sqrt D}{2}\frac{2b + D - \sqrt D}{2} = \frac{(2b + D)^2 - D}{4}$. Hence $(2b + D)^2 - D \equiv 0$ (mod $p$). Hence $(2b + D)^2 \equiv D$ (mod $p$).

Suppose $c = p$. Then $P = [p, p\omega] = pR$.

Case 1 $D$ is divisible by $p$.

Suppose $P = [p, b + \omega]$. Then by the above argument, $(2b + D)^2 \equiv D$ (mod $p$). Since $D \equiv 0$ (mod $p$), $4b^2 \equiv 0$ (mod $p$). Since $p$ is odd, $b \equiv 0$ (mod $p$). Hence $P = [p, \omega]$.

We first claim that $P = [p, \omega]$ is an ideal. By this question, It suffces to prove that $N_{K/\mathbb{Q}}(\omega) = D(D - 1)/4$ is divisible by $p$.

By this question, $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$). Suppose $D \equiv 0$ (mod $4$). Let $D = 4m$. Then $D(D - 1)/4 = m(D - 1)$. Since $D$ is divisible by $p$, $m$ is divisible by $p$. Hence $D(D - 1)/4$ is divisible by $p$.

Suppose $D \equiv 1$ (mod $4$). Then $D(D - 1)/4$ is divisible by $p$.

Hence $P$ is an ideal. Since $N(P) = p$, $P$ is a prime ideal.

Since $\omega + \omega' = D$, $\omega' \in P$. Hence $P = [p, \omega] = [p, \omega'] = P'$. Since gcd$(p, f) = 1$, $P$ is a regular ideal by this question. Hence $P^2 = PP' = pR$ by this question.

Case 2 gcd$(D, p) = 1$ and $D$ is quadratic residue modulo $p$.

As we see in the beginning, $N_{K/\mathbb{Q}}(b + \omega) = \frac{(2b + D)^2 - D}{4}$. Hence $4N(b + \omega) = (2b + D)^2 - D$. Since $(2b + D)^2 - D \equiv 0$ (mod $p$) and $p$ is odd, $N_{K/\mathbb{Q}}(b + \omega)$ is divisible by $p$. Hence $P = [p, b + \omega]$ is an ideal of $R$ by this question. Hence $P' = [p, b + \omega']$ is also an ideal of $R$. Let $\theta = b + \omega$. Then $PP' = (p, \theta)(p, \theta') = (p^2, p\theta, p\theta', \theta\theta') \subset pR$. Since $D = f^2d$ and gcd$(D, p) = 1$, gcd$(f, p) = 1$. Hence $P$ and $P'$ are regular by this question. Hence $N(PP') = N(P)N(P') = p^2$ by this question. Since $N(pR) = p^2$, $PP' = pR$.

We will show that $P \ne P'$. Since $\omega + \omega' = D$, $P' = [p, b + \omega'] = [p, b + D - \omega]$. Suppose $P = P'$. Then $(b + \omega) + (b + D - \omega) = 2b + D \in P$. Hence $2b + D \equiv 0$ (mod $p$). Since $(2b + D)^2 \equiv D$ (mod $p$), $D \equiv 0$ (mod $p$). This is a contradiction.

It remains to prove that $P' = [p, b' + \omega]$. $P' = [p, b + D - \omega] = [p, -b - D + \omega]$. $N(-b - D + \omega) = \frac{-2b - D + \sqrt D}{2}\frac{-2b - D - \sqrt D}{2} = \frac{(-2b - D)^2 - D}{4}$. Hence $\frac{(-2b - D)^2 - D}{4}$ is divisible by $p$ by this question. Since $p$ is odd, $(-2b - D)^2 \equiv D$ (mod $p$). Hence $-b - D$ is a solution of $(2x + D)^2 \equiv D$ (mod $p$). It suffices to prove that $b$ is not congruent to $-b - D$ modulo $p$. Suppose $b \equiv -b - D$ (mod $p$). Then $2b + D \equiv 0$ (mod $p$). Since $(2b + D)^2 \equiv D$ (mod p), $D \equiv 0$ (mod $p$). This is a contradiction.

Case 3 gcd$(D, p) = 1$ and $D$ is quadratic non-residue modulo $p$.

Let $P$ be a prime ideal lying over $p$. By the argument in the beginning, $P = [p, p\omega] = pR$.