height of a point on elliptic curve I m bit confused about the definition of Weil Height over number field. In our lecture, it is defined as $|x|_p=|O_k/(p^{-ord_p(x)})|$ where $O_K$ is the rings of integers for number field $K$ and $p$ is prime ideal of $O_K$， for finite places. If $K$ has $r_1$ real embeddings and $r_2$ complex embeddings we define $|x|_{\sigma_{i}}=|\sigma_i(x)|$ for $\sigma_i:K\rightarrow\mathbb{R}$ and $\sigma_j:K\rightarrow\mathbb{C}$ we have $|x|_{\sigma_j}=|\sigma_j(x)|^2$
Then we define for $P\in \mathbb{P}_K^n$ $$H(P)=(\prod_{v}\max\{|x_0|_v,...,|x_n|_v\})^{1/[K:Q]}$$Finally we define $H(x)=H(1:x)$ where $x\in K$.
This definition is bit confusing to me. So how do you compute $H(\frac{1+\sqrt{-5}}{2})$? You have to look at the ideals of $\mathbb{Z}(\frac{1+\sqrt{-5}}{2})$. Thanks
 A: If $\mathfrak p$ is an integral ideal of $K$, the absolute value of $x\in K$ at $\mathfrak p$ is given by $|x|_{\mathfrak p}=N(\mathfrak p)^{-ord_{\mathfrak p}(x)}$. Let $K=\mathbb Q(\sqrt{-5})$, so that $\mathcal O_K=\mathbb Z[\sqrt{-5}]$ and let $x=\frac{1+\sqrt{-5}}{2}$. Then, as your definition shows, we have that
$$H(x)=\left(\prod_v\max\{|1|_v,|x|_v\}\right)^{1/2}.$$
For every finite place $v$ not dividing $(x)$, we have that $|x|_v=|1|_v=1$. This shows that we only have to look at finite places dividing $(x)$ and at the complex place of $K$. Now you can check that $(x)=\mathfrak p_2^{-1}\mathfrak p_3$, where $\mathfrak p_2=(2,1+\sqrt{-5})$ and $\mathfrak p_3=(3,1+\sqrt{-5})$ are two integral prime ideals of norm $2,3$ respectively. Therefore $|x|_{\mathfrak p_2}=2^{-ord_{\mathfrak p_2}(x)}=2$, while $|x|_{\mathfrak p_3}=3^{-ord_{\mathfrak p_3}(x)}=1/3$. Thus the contribution to the height given by the finite places is $2$. If $|\cdot |_{\infty}$ is the unique Archimedean absolute value on $K$, it is easy to see that $|x|_{\infty}=\left(\frac{1+\sqrt{-5}}{2}\right)\left(\frac{1-\sqrt{-5}}{2}\right)=3/2>1$, so that all in all $H(x)=\sqrt{3}$.
