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?
Best Answer
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 $x+6y$.