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\begin{document}

\centerline{\bf{Lecture 21}}

{\it Algorithm for finding the primes of an algebraic number field lying over a prime} $\mathfrak p$:
\par
Consider the following generic situation:
\\
$$\begin{array}{ccc}
L& & O_L\\
|& & \\
K& & O_K\end{array}$$

\par
Let $\alpha\in O_L$ be an element such that $O_L\otimes_{O_K}O_{K,\mathfrak p}=O_{K,\mathfrak p}[\alpha]$ (such an $\alpha$ always exists).
$$\begin{array}{ccccc}
L    & & O_L&          &            \\
\vert& &    &\diagdown &            \\
\vert& &    &          & O_K[\alpha]\\
\vert& &    &\diagup   &            \\
K    & & O_K&          &            \end{array}$$
Let $f(X)\in O_K[X]$ be the minimal polynomial for $\alpha$. Decompose $\bar{f}$ in $(O_K/\mathfrak p)[X]$:
\[
\bar{f}=\prod_{i=1}^g \varphi_i^{e_i}
\]

For each $i\in\{1,2,...\, g\}$ pick $g_i\in O_K[X]$ that maps modulo $\mathfrak p$ to $\varphi_i\in (O_K/\mathfrak p)[X]$. \\
Then the prime ideals in $L$ lying over $\mathfrak p$ are $\mathfrak {P}_i=(\mathfrak p,\ g_i(\alpha))$, for $i\in\{1,2,...\, g\}$.
\par

\begin{exa}
$K=\Q$, $L=\Q[\sqrt{5}]$. Then $O_K=\Z$. We shall determine $O_L$. Since $O_L$ is a subring of $L$, it consists of elements of the form
$a+b\sqrt{5}$, with $a,\ b\ \in\Q$. Let $\alpha=a+b\sqrt{5}\ \in L$. Then
\[
(X-\alpha)(X-\bar{\alpha})=X^2-\Tr(\alpha)X+\N(\alpha).
\]
\par
Note that if $\Tr(\alpha),\N(\alpha)\in \Z$ then $\alpha$ is an element $L$ wich is a root of a polynomial with coefficients in $\Z$, hence $\alpha\in O_L$.
\\
Conversely, if $\alpha\in O_L$ and $\alpha\notin \Z$ then the minimum polynomial of $\alpha$ over $\Q$ has degree 2, which means that $X^2-\Tr(\alpha)X+\N(\alpha)$ 
is the minimum polynomial of $\alpha$ over $\Q$. But $\alpha$ is an algebraic integer, hence its minimum polynomial over $\Q$ has actually coefficients in $\Z$.
This implies $\Tr(\alpha),\N(\alpha)\in \Z$. Consequently, $\alpha\in O_L$ iff $\Tr(\alpha),\N(\alpha)\in \Z$.
\\
$\Tr(\alpha)=2a\in \Z$ implies that $a=1/a_1$, with $a_1\in \Z$.
\\
$\N(\alpha)=a^2-5b^2\in \Z$ implies that $\frac{1}{4}a_1^2-5b^2\in \Z$ and $a_1^2-20b^2\in 4\Z$. Then $20b^2\in \Z$ which implies $v_p(20b^2)\geq 0$ so that $v_p(20)+2v_p(b)\geq 0$
hence $v_p(b)\geq 0$ for $p\neq 2,5$.
\\
$v_2(20)+2v_2(b)\geq 0\ \Rightarrow v_2(b)\geq -1$.
\\
$v_5(20)+2v_5(b)\geq 0\ \Rightarrow v_5(b)\geq -0.5 \Rightarrow v_5(b)\geq 0$.
\\ These inequalities show that $b=\frac{1}{2}b_1$, with $b_1\in \Z$.
\\
Hence $a_1^2-5b_1^2\in 4\Z \Rightarrow a_1,b_1$ have the same parity, i.e. $a_1=b_1+2c$, with $c\in \Z$.
\\
Then $a+b\sqrt{5}=\frac{1}{2}(a_1+b_1\sqrt{5})=\frac{1}{2}(b_1+2c+b_1\sqrt{5})=c+b_1\frac{1+\sqrt{5}}{2}$.
\\ We conclude that $O_L\subseteq \Z[\frac{1+\sqrt{5}}{2}]$.
\\ The other inclusion is obviously true (the minimum polinomial over $\Q$ of $\theta=\frac{1+\sqrt{5}}{2}$ is $X^2-X-1$).
\\ Consequently: 
\[
\displaystyle O_L=\Z[\frac{1+\sqrt{5}}{2}].
\]

\par The archimedian places of $\Q[\sqrt{5}]$: there are two embedings of $L$ in $\R$:
\[
v_1,v_2:\Q[\theta]\hookrightarrow \R
\]
given by $v_1(\theta)=\frac{1+\sqrt{5}}{2}(=1.61803)$ and $v_2(\theta)=\frac{1-\sqrt{5}}{2}(=-0.61803)$.
In this case $g=2$, and thus $e=f=1$. 

\par $p=2$. We have seen that $f(X)=X^2-X-1$ is the minimum polynomial of $\theta$ over $\Q$. By reducing the coefficients of $f$ modulo $p(=2)$
we get an irreducible polinomial $\bar{f}(X)=X^2-X-1\in (\Z/2)[X]$. Thus:
$$\begin{array}{c}
F_2(\theta)\\
|2 \\
F_2 \end{array}$$
hence $f=2$. There is a unique prime in $L$, $\mathfrak p_2=(2)=2O_L$, lying over 2. Clearly $g=e=1$.

\par $p=3$. By reducing the coefficients of $f$ modulo 3
we get an irreducible polinomial $\bar{f}(X)=X^2-X-1\in (\Z/3)[X]$. Thus:
$$\begin{array}{c}
F_3(\theta)\\
|2 \\
F_3 \end{array}$$
hence $f=2$. There is a unique prime in $L$, $\mathfrak p_3=(3)=3O_L$, lying over 3. Again, $g=e=1$.

\par $p=5$. Note that $\bar{f}(X)=X^2-X-1=(X+2)^2\in (\Z/5)[X]$. Then
$e=2, g=f=1$. There is a unique prime in $L$ lying over 5, $\mathfrak p_5=(5,\theta+2)$.
 
\par $p=11$. Note that $\bar{f}(X)=X^2-X-1=(X-4)(X-8)\in (\Z/5)[X]$. Then
$g=2, e=f=1$. There are two primes in $L$ lying over 11, $\mathfrak p'_11=(11,\theta-4)$, and $\mathfrak p"_11=(11,\theta-8)$.

In fact, here is the splitting of any rational prime $p$ in $L$:
\[
p=5 \Rightarrow e=2, f=g=1;
\]
\[
p=2,3\ (mod\ 5) \Rightarrow f=2, e=g=1;
\]
\[
p=1,4\ (mod\ 5) \Rightarrow g=2, e=f=1.
\]
The case $p=5$ has been discussed above.
\\
If $p=2,3\ (mod\ 5)$ then $p$ is not a square modulo 5 and by the quadratic reciprocity law 5 is not a square modulo $p$.
\\ Let $\mathfrak P$ be a prime in $L$ lying over $p$. The polynomial $X^2-5$ is irreducible over $F_p$. 
On the other hand, the polynomial $X^2-5$ is not irreducible over $O_L$ so it can be factored nontrivially over $O_L/\mathfrak{P}$.
That means that the degree of $O_L/\mathfrak{P}$ over $F_p$ is larger that 1, so $f=2$. This implies that $e=g=1$.\\

Finally, suppose that $p=1,4\ (mod\ 5)$, i.e. $p$ is a square modulo 5. Then 5 is a square modulo $p$, which implies that the 
polynomial $X^2-5$ can be factored nontrivially in $F_p[X]$. This means that $g=2$ and $e=f=1$.
\end{exa}

\vskip .05cm
\par {\bf Normalization of absolute values}
\begin{Def} \label{D:I.8.1}
Let $K$ be a number field. The {\it normalized} absolute values are defined as follows:
\\
$\norm x = |\imath(x)|_{\R}$, if $v$ is $\imath:K\hookrightarrow \R$ is a real embedding;
\\
$\norm x = |\imath(x)|^2_{\C}$, if $v$ is $\imath:K\hookrightarrow \C$ is a non-real embedding;
\\
$\norm x = q_v^{-v(x)}$, where $q_v=|O_K/p_v|$, if $v$ is a (non-archimedian) $p_v$-adic valuation.
\end{Def}
\par Let $M_K$ be the set of normalized absolute values of $K$. We can think of $M_K$ as the set of equivalence classes of absolute values ({\it places})
of $K$.
\\
For instance, Ostrowski's theorem assures that $M_{\Q}=\{\infty,2,3,5,...\}$. The normalized absolute values of $\Q$ are:
\[
\Vert x\Vert_{\infty}=|x|_{\R};
\] 
\[
\Vert x \Vert_p=(\frac{1}{p})^{v_p(x)}.
\] 

\end{document}