# Height of points over number fields

I'm beginning to read about height theory in this text by J.H. Silverman. In page $5$ he gives the following definition of height, considering a number field $K$:

The height of a point $P=[x_0:...:x_N]\in\mathbb{P}^N(\mathbb{Q})$ with $x_0,...,x_N\in\mathbb{Z}$ and $\gcd(x_0,...,x_N)=1$ is defined by:

$$H_K(P):=\prod_{v\in M_K}\max(||x_0||_v, ..., ||x_N||_v)$$

where $M_K$ is a "complete set of (normalized) absolute values on $K$". He also says that "$M_K$ contains an archimedean absolute value for each embedding of $K$ into $\mathbb{R}$ or $\mathbb{C}$ and a $p$-adic absolute value for each prime ideal in the ring of integers of $K$".

I've tried to look it up in the internet and found many definitions, but none of them clarified everything that he says above.

I know what absolute values are (archimedian, non-archimedian, $p$-adic etc), but what does he mean by "normalized" absolute value, and why did he use parenthesis for it? How does he know that $M_K$ has this archimedian absolute values for each embedding and a $p$-adic absolute value for each prime in the integers?

It feels like he is talking about something that I should already know, but I don't. Where can I read more about these things?

• For $K = \mathbb{Q}$, there is one Archimedian absolute value $|.|_\infty$ coming from the embedding $\mathbb{Q} \to \mathbb{R}$ and one non-Archimedian absolute value $|.|_p$ for each prime number $p$. You would say $|x|_p = p^{-v_p(x)}$ ? But $|x|_p = q^{-v_p(x)}$ is also an absolute value for any $q > 1$. Thus we are talking of equivalence classes of absolute values.
We choose $|x|_p = p^{-v_p(x)}$ because this way $\prod_p |x|_p^{-1} = |x|_\infty$ ie. $\prod_v |x|_v = 1$.
This is indeed what we need for $H(X) = \prod_{v} \max_n |X_n|_v$ to be well-defined for $X \in \mathbb{P}^N(\mathbb{Q})$ ie. $H(aX) = \prod_{v} \max_n |a X_n|_v= \prod_{v} |a|_v \max_n | X_n|_v = H(X)$
• For $K$ any number field, it works the same way. For $x \in \mathcal{O}_K$ write $(x) = \prod_i \mathfrak{p}_i^{v_{\mathfrak{p}_i}(x)}$ and set $|x|_{\mathfrak{p}_i} = N(\mathfrak{p}_i)^{-v_{\mathfrak{p}_i}(x)}$ where $N(I) = \# \mathcal{O}_K/I$ is the ideal norm, so that $\prod_i |x|_{\mathfrak{p}_i}^{-1}=\prod_i N(\mathfrak{p}_i^{v_{\mathfrak{p}_i}(x)})= N(\prod_i\mathfrak{p}_i^{v_{\mathfrak{p}_i}(x)})=N(x)=|N_{K/\mathbb{Q}}(x)| = \prod_\sigma |x|_\sigma$ where the last product is over all the complex embeddings $\sigma : K \to \mathbb{C}$ (not over equivalence classes of Archimedian absolute values) and $|x|_\sigma= |\sigma(x)|_\infty$.
• You need the Product Formula to hold in the field $K$, i.e. that for $z\in K$, the product of all the normalized absolute values of $z$ is $1$, because that makes the height independent of the particular $(n+1)$-tuple of elements of $K$ that you use to describe your point in projective space. – Lubin Oct 9 '17 at 1:43
• @Lubin Does $\prod_v |x|_v = 1$ hold in many other fields, for example finite extensions of $k(t)$ ? – reuns Oct 9 '17 at 1:58
• If $K$ is complete for $|.|$ then It doesn't hold. Are there other cases ? – reuns Oct 9 '17 at 2:05
• The product formula is characteristic of Global Fields: finite extensions of $\Bbb Q$, and finite extensions of $\Bbb F_p(t)$. Period (I think). – Lubin Oct 9 '17 at 14:19