Determination of the prime ideals lying over $2$ in a quadratic order 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.
 A: 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.)
A: 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.
