# 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.

• I would like to point out the following policy of StackExchange because it doesn't seem to be well-known and some users seem to dislike a question to which the poster already knows the answer. [It’s also perfectly fine to ask and answer your own question, as long as you pretend you’re on Jeopardy! ? phrase it in the form of a question. To be crystal clear, it is not merely OK to ask and answer your own question, it is explicitly encouraged.] blog.stackoverflow.com/2011/07/… – Makoto Kato Nov 7 '13 at 4:12

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.)

• [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 $2$.] Dear Matt E, you mean it is congruent to $x^2$ or $(x+1)^2$ mod $2$, right? – Makoto Kato Nov 7 '13 at 21:43
• @MakotoKato: Dear Makoto, Yes, that's right. Regards, – Matt E Nov 7 '13 at 21:49
• Dear Matt E, I'm afraid I don't understand why $(2) = (x,2)^2$ in the first case. It seems that $(2) = (x,2)^2$ or $(x + 1, 2)^2$ according $D/4$ is even or odd. – Makoto Kato Nov 7 '13 at 23:07
• @MakotoKato: Dear Makoto, I think that you're correct. E.g. if $D = -4$, then $\omega = -2 + i,$ and $(2) = (-1+i)^2 = (1 + \omega, 2)^2,$ while if $D = -8,$ then $\omega = -4 + \sqrt{-2},$ and so $(2,\omega)$ = (2,\sqrt{-2}) = (\sqrt{-2}),$and so in this case$(2,\omega)^2 = (2)$. So I think the statement when$D$is even has to be modified to take this into account. Regards, – Matt E Nov 8 '13 at 0:52 I realized after I posted this question that the proposition is not correct. I will prove a corrected version of the proposition. We need some notation. Let$\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.