%! TEX root = DA.tex
% vim: tw=50
% 25/10/2024 12PM

$L / K$ extension of number fields, then $w \in
M_L$ lies above $v \in M_K$; in notation $w \mid
v$.

If both are embeddings and $w|_K = v$ or $w|_K =
\ol{v}$ or both are finite and $w$ lies over $v$
as prime ideals, i.e. $w \mid v \O_L$.

\begin{remark*}
  $\sum_{v \mid \infty} d_v = [K : \Qbb]$,
  $\sum_{v \mid p} d_V = [K : \Qbb]$.
\end{remark*}

\begin{proposition*}[Product formula]
  Let $K$ be a number field. Then for all $\alpha
  \in K \neq 0$, we have
  \[
    \prod_v |\alpha|_v^{d_v} = 1
  .\]
\end{proposition*}

\begin{proof}
  We compute $N(\alpha \O_K)$ in two ways.
  \[
    N(\alpha \O_K) = \prod_{v \in M_{K, f}}
    N(v)^{\ord_v(\alpha)} = \prod_{v \in M_{K, f}}
    p^{f_v \cdot \ord_v(\alpha)}
  ,\]
  where $p$ is the rational prime lying below $v$.

  Recall $|\alpha|_v = p^{-\ord_v(\alpha) / e_v} =
  p^{-\ord_v(\alpha) \cdot f_v / d_v}$. So
  \[
    N(\alpha \O_K) = \prod_{v \in M_{\alpha, f}}
    |\alpha|_v^{-d_v}
  .\]
  Also,
  \[
    N(\alpha \O_K) = |N(\alpha)| = \prod_{v \in
    M_{K, \infty}} |\alpha|_v^{d_v}
  .\]
  Dividing the equations gives the desired
  result.
\end{proof}

Now we define
\[
  H(\alpha) = \left(\prod_{v \in M_K} \max(1,
  |\alpha|_v) \right)^{\frac{1}{[K : \Qbb]}}
.\]
We will also use $h(\alpha) = \log H(\alpha)$. We won't
be using that much, but we mention it mostly
because it is used in the literature.

$H$ is known as ``multiplicative height'', while
$h$ is known as ``logarithmic / absolute / Weil
height''.

\begin{proposition}
  Let $L / K$ be an extension of number fields.
  Let $\alpha \in K$. Then $H(\alpha)$ as defined
  above is the same for $K$ and $L$.
\end{proposition}

\begin{proof}
  \textbf{Claim 1:} If $w \in M_L$, $v \in M_K$
  such that $w \mid v$ then $|\alpha|_w =
  |\alpha|_v$ for all $\alpha \in K$.

  \textbf{Claim 2:} $\sum_{w \mid v} d_w = [L : K]
  d_v$.

  Assuming these claims are true, then for $\alpha
  \in K$
  \[
    \prod_{w \mid v} \max(1, |\alpha|_w)^{d_w} =
    \max(1, |\alpha_v|)^{[L : K] d_v}
  \]
  Then
  \[
    \left(\prod_{w \mid v} \max(1, |\alpha|_w)^{d_w}
    \right)^{\frac{1}{[L : \Qbb]}} =
    \max(1, |\alpha_v|)^{\frac{d_v}{[K : \Qbb]}}
  \]
  Which implies the desired result.

  \textit{Proof of Claim 1:} Will show if $v, w$ are
  embeddings then
  \[
    |\alpha|_w = |w(\alpha)| = |v(\alpha)| =
    |\alpha|_v
  .\]
  If $w, v$ are prime ideals, then we need
  \[
    \frac{\ord_w(\alpha)}{e_w} =
    \frac{\ord_v(\alpha)}{e_v}
  .\]
  For this, note that for all ideals $I \subset
  \O_K$, we have
  \[
    \ord_w(I \cdot \O_L) = \ord_w(v \cdot \O_L)
    \cdot \ord_v(I)
  .\]
  Use this for $p\O_K$ and $\alpha\O_K$ in the
  role of $I$:
  \begin{align*}
    e_w
    &= \ord_w(v \cdot \O_L) e_v \\
    \ord_w(\alpha)
    &= \ord_w(v \cdot \O_L) \cdot \ord_v(\alpha)
  \end{align*}

  \textit{Proof of Claim 2:} Omitted.
\end{proof}

\begin{proposition*}
  Let $\alpha \in \ol{\Qbb}_{\neq 0}$. Then
  \[
    H(\alpha) =
    M(f_\alpha)^{\frac{1}{\deg(f_\alpha)}}
  .\]
\end{proposition*}

\begin{remark*}
  Recall $2^{-d} H(f_\alpha) \le H(\alpha)^d \le
  (1 + d) H(f_\alpha)$.
\end{remark*}

\begin{proof}
  Enough to prove
  \[
    |a_d|^{[K : \Qbb]} = \prod_{v \in M_{K, f}} \max
    (1, |x|_v)^{d_v}
  ,\]
  where $K$ is a number field with $\alpha \in K$.

  If $K = \Qbb(\alpha)$, then this is immediate
  from the definitions.
