The question is quite simple but I don't figure out why the following is true. In "Algebraic geometry" of Cornell/Silverman, chapter "The theory of height functions", he consider $x_0, \dots, x_n$ global section of the line bundle $O(1)$ of $\mathbb{P}^n$, and $P \in V(K)$, where $V$ is some variety embedded in $\mathbb{P}^n$ and $K$ is a global field of integers ring $R$. We can suppose $x_0(P) \neq 0$. The point where I'm stuck is the equality : $$ \left| N_{K/\mathbb{Q}}\left(\sum_{i=0}^n R(x_i/x_0)(P) \right)\right|^{-1} = \prod_{v \in M_{K}^0} \text{max}_i || (x_i/x_0)(P)||_v$$
In fact, I would have written :
$$\left| N_{K/\mathbb{Q}}\left(\sum_{i=0}^n R(x_i/x_0)(P) \right)\right|^{-1} = \prod_{v \in M_{K}^0} \left|\left| \sum_{i=0}^n (x_i/x_0)(P) \right|\right|_v \leq \prod_{v \in M_{K}^0} \text{max}_i \left|\left| (x_i/x_0)(P)\right|\right|_v $$
as we deal with non-archimedean absolute value, but how we have the equality ? (for a given $v \in M_{K}^0$, we could have $v(x_i(P))= v(x_0(P))$ for all $i$, and then how we can actually deduce the above equality ?)
Thank you !
The variety is irrelevant here, choosing a $K^{n+1}$ representative for $P$, the $a_i = x_i(P)$ are any elements of $R=O_K$ with $a_0\ne 0$.
$I=\sum_{i=1}^n a_i O_K$ is an ideal. Say it is non-zero. Factorize in prime ideals $$I= \prod_j P_j^{e_j}$$
Then $$e_j = \inf_i v_{P_j}(a_i)$$ so that the ideal norm is $$N(I)= \prod_j N(P_j)^{e_j}=\prod_P N(P)^{\inf_i v_{P}(a_i)}=\prod_v \sup_{i\ge 1} |a_i|_v^{-1}$$ where $\prod_v$ is over the non-archimedian absolute values normalized such that $|\pi_v|_v^{-1}$ is the cardinality of the residue field.
We get similarly that $$N(a_0 O_K)=\prod_v |a_0|_v^{-1}$$ and extending the ideal norm to fractional ideals through $N(a_0^{-1}I)= N(I) / N(a_0 O_K)$ we get $$N(\sum_{i=1}^n \frac{a_i}{a_0} O_K)^{-1}=N(a_0O_K)/N(I)=\prod_v \sup_{i\ge 1} |\frac{a_i}{a_0}|_v$$
For the case $I=\{0\}$: set $N(I)=|O_K|=\infty$.