# Application of the decomposition of prime ideals as $Q_q^{e_1}Q_2^{e_2}\dots Q_R^{e_r}$

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.

and

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 $$x+6y$$.

• Very nice indeed, I'll think about it as soon as I have time – 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.