I'm reading Marcus number field book and at page 57 he asks the following

We give some applications of Theorem 27. Taking $\alpha=\sqrt{m}$, we can re-obtain the results of Theorem 25 except when p = 2 and m $\equiv $ 1 (mod 4); in this exceptional case the result can be obtained by taking $\alpha=\frac{1+\sqrt{m}}{2}$.

Where the theorems are the following

Theorem 25 With notation as above, we have:

If p | m, then $$ pR=(p,\sqrt {m})^2.$$

If m is odd, then $$ 2R= \begin{cases} (2,1+\sqrt {m})^2&\text{if $m\equiv 3\pmod4$}\\ \left(2,\frac{1+\sqrt{m}}{2}\right)\left(2,\frac{1-\sqrt{m}}{2}\right) & \text{if $m\equiv 1\pmod8$}\\ \text{prime if $m\equiv 5\pmod8$.} \end{cases}$$

If p is odd, $p\not| m$ then $$ pR=\begin{cases} (p,n+\sqrt{m})(p,n-\sqrt{m})\; \text{if $m\equiv n^2 \pmod p$}\\ \text{prime if $m$ is not a square mod $p$} \end{cases}$$ where in all relevant cases the factors are distinct.


Theorem 27 Now let g be the monic irreducible polynomial for $\alpha$ over K. The coefficients of g are algebraic integers (since they can be expressed in terms of the conjugates of the algebraic integer $\alpha$), hence they are in $\mathbb{A}\cap K = R$.

Thus g $\in$ R[x] and we can consider $\overline{g}\in$ (R/P)[x].

$\overline{g}$ factors uniquely into monic irreducible factors in (R/P)[x], and we can write this factorization in the form $$\overline{g} =\overline{g}_1^{e_1}\dots \overline{g}_n^{e_n}$$ where the $\overline{g}_i$ are monic polynomials over R. It is assumed that the $\overline{g}_i$ are distinct.

Let everything be as above, and assume also that p does not divide |S/R[$\alpha$]|, where p is the prime of $\mathbb{Z}$ lying under P. Then the prime decomposition of PS is given by $$Q_1^{e^1}\dots Q_n^{e_n}$$ where $Q_i$ is the ideal (P, $g_i(\alpha$)) in S generated by P and $g_i(\alpha)$; in other words, Qi = PS + ($g_i(\alpha$)). Also, f ($Q_i$ |P) is equal to the degree of $g_i$ .

I tried doing it but I think I'm doing something wrong. How do I use the relations between p and m?

I always get that the minimal polynomial of $\sqrt{m}$ is $x^2-m=(x-m)(x+m)$ and so $Q_1=(P,2\sqrt{m})\wedge Q_2=(P,0)$ whose product is not equal, for example, to $(p,\sqrt{m})$.

Can you help me?


First of all, the factorization of $x^2-m$ (when it exists) is $(x-\sqrt{m})(x+\sqrt{m})$ not $(x-m)(x+m)$ as you wrote. Thus the key question is whether $\sqrt{m}$ exists in $\frac{R}{P}$.

An example to illustrate this : take $m=7,p=29$. Then $m$ is a square modulo $p$ (since $6^2\equiv m\ \mod p$), so in $\frac{\mathbb Z}{p{\mathbb Z}}$, $x^2-m$ factorizes $x^2-m=x^2-7=(x-6)(x+6)$ ; you have $\bar{g_1}=x-6,\bar{g_2}=x+6$. Accordingly, the ideal $(p)$ decomposes as $(p)=(p,\sqrt{m}-6)(p,\sqrt{m}+6)$.

If you want to "visualize" those ideals more, note that $(p)$ is the set of all $x+y\sqrt{m}$ such that $p$ divides both $x$ and $y$, $(p,\sqrt{m}-6)$ is the set of all $x+y\sqrt{m}$ such that $p$ divides $x-6y$, and $(p,\sqrt{m}+6)$ is the set of all $x+y\sqrt{m}$ such that $p$ divides $x+6y$.

| cite | improve this answer | |
  • $\begingroup$ Very nice indeed, I'll think about it as soon as I have time $\endgroup$ – Frankie123 May 20 at 20:51

We start with the general case picking $g(x)=x^2-m,\; \alpha=\sqrt {m},\; K=\mathbb{Q},\; L=\mathbb{Q}(\sqrt {m})$ $p\not|\left|\frac {S}{\mathbb{Z}[\sqrt{m}]}\right|$.

A problem arises when $m\equiv1\; (mod\; 4)$ and p=2, in that special case we pick $\alpha=\frac{1+\sqrt{2}}{2}$ $g(x)=x^2-x+\frac{1-m}{4}$.

In the general case,

  1. if $p|m|$, $x^2-m\equiv x^2\; (mod\; p)\mathbb{Z}[x]$; so $g_1(x)=g_2(x)=x$ and $pS=Q^2$ where $Q=(p,\sqrt{m})$;
  2. if m is a non zero square (mod p), $m\equiv n^2\; (mod\; p)$ we get $$ x^2-m\equiv(x-n)(x+n)\; (mod\; p)\mathbb{Z}[x]$$ so $g_1(x)=x-n$, $g_2(x)=x+n$ and $pS=Q_1Q_2$ where $Q_1=(p,\sqrt{m}-n)$ and $Q_2=(p,\sqrt{m}+n);$
  3. if m is not a square (mod p) then g(x) is irreducible (mod p) and $Q=(p,g(\alpha)=0)=pS$ is a prime with residue degree 2.

In the special case, there are two possibilities:

  1. if $m\equiv 1\; (mod\; 8)$ then g(x) has the zeroes $\alpha_1=0$ and $\alpha_2=1$ in $\mathbb{F}_2$ so $g_1(x)=x$ and $g_2(x)=x-1$ work; in this case $Q_1=(2,\alpha)=\left(2,\frac{1+\sqrt {m}}{2}\right)$ and $Q_2=(2,\alpha-1)=\left(2,\frac{1-\sqrt{m}}{2}\right)$;
  2. if $m\equiv 5\; (mod\; 8)$ then g(x) is irreducible (mod 2) so $Q=(2,g(\alpha)=0)=2S$ is a prime of residue degree 2.
| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.