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 $|x|_\sigma= |\sigma(x)|_\infty$.

  • $\begingroup$ 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. $\endgroup$ – Lubin Oct 9 '17 at 1:43
  • $\begingroup$ @Lubin Does $\prod_v |x|_v = 1$ hold in many other fields, for example finite extensions of $k(t)$ ? $\endgroup$ – reuns Oct 9 '17 at 1:58
  • $\begingroup$ If $K$ is complete for $|.|$ then It doesn't hold. Are there other cases ? $\endgroup$ – reuns Oct 9 '17 at 2:05
  • 1
    $\begingroup$ The product formula is characteristic of Global Fields: finite extensions of $\Bbb Q$, and finite extensions of $\Bbb F_p(t)$. Period (I think). $\endgroup$ – Lubin Oct 9 '17 at 14:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.