Determination of the prime ideals lying over $2$ in a quadratic order

Question

Let $K$ be a quadratic number field. Let $R$ be an order of $K$, $D$ its discriminant. By this question, $1, \omega = \frac{(D + \sqrt D)}{2}$ is a basis of $R$ as a $\mathbb{Z}$-module.

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 $x_1,\cdots, x_n$.

My Question Is the following proposition correct? If yes, how do you prove it?

Proposition

Case 1 $D$ is even.

$P = [2, \omega]$ is a prime ideal and $2R = P^2$.

Case 2 $D \equiv 1$ (mod $8$).

$P = [p, \omega]$ and $P' = [p, 1 + \omega]$ are distinct prime ideals and $2R = PP'$. Moreover $P' = \sigma(P)$, where $\sigma$ is the unique non-identity automorphism of $K/\mathbb{Q}$.

Case 3 $D \equiv 5$ (mod $8$).

$2R$ is a prime ideal.

Answer 1

Since $\omega = (D + \sqrt{D})/2$ has minimal polynomial equal to $x^2 - D x + D(D-1)/4,$ we may write $R = \mathbb Z[x]/(x^2 - Dx - D(D-1)/4),$ and so $R/2R = \mathbb F_2[x]/(x^2 - D x - D(D-1)/4)$ (identify $\omega$ with the image of $x$).

If $D$ is even (and hence a multiple of $4$), the polynomial $x^2 - D x - D(D-1)/4 $ is congruent to $x^2 + 1 = (x+1)^2$ mod $2$.

If $D \equiv 5 \bmod 8$, then this polynomial is congruent to $x^2 + x + 1$ (an irreducible polynomial) mod $2$.

If $D \equiv 1 \bmod 8,$ then this polynomial is congruent to $x^2 + x = x(x+1)$ mod $2$.

In the first case we see that $(2) = (x,2)^2$, in the second case that $(2)$ is prime, and in third case that $(2) = (x,2)(x+1,2)$.

This proves the proposition. (Actually, the way the proposition is phrased, you have to check that the $R$-modules with the indicated generators, which is what I am writing, are the same at the $\mathbb Z$-modules with these generators. This is straightforward.)

Answer 2

I realized after I posted this question that the proposition is not correct. I will prove a corrected version of the proposition.

We need some notation. 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 $R$.

Proposition Let $K$ be a quadratic number field, $d$ its discriminant. Let $R$ be an order of $K$, $D$ its discriminant. By this question, $D \equiv 0$ or $1$ (mod $4$). By this question, $1, \omega = \frac{(D + \sqrt D)}{2}$ is a basis of $R$ as a $\mathbb{Z}$-module. Let $f$ be the order of $\mathcal{O}_K/R$ as a $\mathbb{Z}$-module. Then $D = f^2d$ by this question. We suppose gcd$(f, 2) = 1$.

Case 1 $D$ is even.

Since $D \equiv 0$ (mod $4$), $D \equiv 0, 4$ (mod $8$). If $D \equiv 0$ (mod $8$), let $P = [2, \omega]$. If $D \equiv 4$ (mod $8$), let $P = [2, 1 + \omega]$. Then $P$ is a prime ideal and $2R = P^2$. Moreover $P = P'$.

Case 2 $D \equiv 1$ (mod $8$).

$P = [p, \omega]$ and $P' = [p, 1 + \omega]$ are distinct prime ideals and $2R = PP'$.

Case 3 $D \equiv 5$ (mod $8$).

$2R$ is a prime ideal.

We need the following lemmas to prove the proposition.

Lemma 1 Let $K, R, D, \omega$ be as in the proposition. Let $P = [2, b + \omega]$, where $b$ is a rational integer. Then $P$ is an ideal if and only if $(2b + D)^2 - D \equiv 0$ (mod $8$). Moreover, if $P$ satisfies this condition, $P$ is a prime ideal.

Proof: By this question, $P = [2, b + \omega]$ is an ideal if and only if $N_{K/\mathbb{Q}}(b + \omega) \equiv 0$ (mod $2$). $N_{K/\mathbb{Q}}(b + \omega) = (b + \omega)(b + \omega') = \frac{2b + D + \sqrt D}{2}\frac{2b + D - \sqrt D}{2} = \frac{(2b + D)^2 - D}{4}$. Hence $P$ is an ideal if and only if $(2b + D)^2 - D \equiv 0$ (mod $8$) Since $N(P) = 2$, $P$ is a prime ideal.

Lemma 2 Let $K, R, D, \omega$ be as in the proposition. Suppose gcd$(f, 2) = 1$ and there exist no prime ideals of the form $P = [2, b + \omega]$, where $b$ is an integer. Then $2R$ is a prime ideal of $R$.

Proof: Let $P$ be a prime ideal of $R$ lying over $2$. Then $P \cap \mathbb{Z} = 2\mathbb{Z}$. By this question, there exist integers $b, c$ such that $P = [2, b + c\omega], c \gt 0, 2 \equiv 0$ (mod $c$), $b \equiv 0$ (mod $c$). Then $c = 1$ or $2$. By the assumption, $c = 2$. Hence $P = [2, 2\omega] = 2R$.

Proof of the proposition

Case 1 $D$ is even.

Let $P = [2, b + \omega]$, where $b$ is an integer. We may assume that $b = 0$ or $1$. By Lemma 1, $P$ is a prime ideal if and only if $(2b + D)^2 - D \equiv 0$ (mod $8$).

Suppose $D \equiv 0$ (mod $8$). Then $(2b + D)^2 - D \equiv 0$ (mod $8$) if and only if $b = 0$. Hence $P = [2, \omega]$ is an ideal of $R$. $P' = [2, \omega'] = [2, D - \omega] = [2, -\omega] = [2, \omega] = P$.

Suppose $D \equiv 4$ (mod $8$). Then $(2b + D)^2 - D \equiv 0$ (mod $8$) if and only if $b = 1$. Hence $P = [2, 1 + \omega]$ is an ideal of $R$. $P' = [2, 1 + \omega'] = [2, 1 + D - \omega] = [2, -1 - D + \omega] = [2, 1 + \omega] = P$.

Since gcd$(f, 2) = 1$, $P$ is regular by this quuestion. Hence $PP' = 2R$ by this question.

Case 2 $D \equiv 1$ (mod $8$).

If $b = 0, 1$, then $(2b + D)^2 - D \equiv (2b + 1)^2 - 1 \equiv 0$ (mod $8$). Hence, by Lemma 1, $P = [2, \omega]$ and $Q = [2, 1 + \omega]$ are prime ideals of $R$. $P' = [2, \omega'] = [2, D - \omega] = [2, - D + \omega] = [2, 1 + \omega] = Q$. Since gcd$(f, 2) = 1$, $P$ is regular by this question. Hence $PP' = 2R$ by this question.

Case 3 $D \equiv 5$ (mod $8$).

Consider the following congruence equation. $(2b + D)^2 - D \equiv (2b + 5)^2 - 5 \equiv 0$ (mod $8$). Since $b$ does not satisfy this congruence equation when $b = 0$ or $1$, there exist no ideals of the form $[2, b + \omega]$. Hence $2R$ is a prime ideal by Lemma 2.