# Monday February 3rd

## Extending Height Functions to Separable Closure 

We started with a product formula field $(K, \Sigma_K, A)$, then took $L/K$ finite and separable and equipped it with a PFF structure:
$\Sigma_L$ is equal to the places extending $v\in \Sigma_K$, and for $v\in \Sigma_L$, we have $\abs{x}_w = \abs{L_w/K_v (x)  }_v^{1/d}$.

We defined a height function 
\begin{align*}
H: \PP^n(K) &\to [1,\infty) \\
[x_0, \cdots, x_n] &\mapsto \prod_{v\in \Sigma_K} \max(\abs{x_0}_v, \cdots)
.\end{align*}

If $L/K$ is separable of degree $d$, we can consider $p\in \PP^n(K) \subset \PP^n(L)$ and define $H_K, H_L$ respectively.
It is then a fact that $H_L(p) = H_K(p)$.
For $\alpha \in \AA^1$, we have 
$$
H_L(\alpha) = H_K(N_{L/K}(\alpha))^{1/d} = H_K(\alpha^d)^{1/d} = H(\alpha)
.$$

This allows us to extend height functions to the separable closure, $H: \PP^n(K\sep) \to \RR$.

Lemma
: If $p\in \PP^n(K\sep)$ and $\sigma \in g_K = \aut(K\sep/K)$, then $\sigma(p) \definedas [\sigma(x_0), \cdots, \sigma(x_n)]$ does not change the height.

Proof
: Exercise.
	Check that the norms of points are the same as the norms of their Galois conjugates.
	
So the height is Galois-equivariant.

For $v\in \Sigma_K$, let $\eps_v$ be the **Artin constant**: the smallest $\eps_v \in \RR$ such that $\abs{x + y}_v \leq \eps_v \max(\abs{x}_v, \abs{y}_v)$.

> Artin constant: measures how much better a norm does than the triangle inequality.
> Note that $\eps_v = 1$ for ?

Lemma
: Let $K$ be a PFF and $d\in \ZZ^+$.
	If there exists an $M(K, d) > 0$ such that for all separable 
	$$
	f(T) = T^d + a_{d-1}T^{d-1} + \cdots a_1 T + a_0 \in K[T]
	,$$ 
	i.e. $f(T) = \prod_{i=1}^\alpha (T-\alpha_i)$ with $\alpha_i \in K\sep$, then 
	$$
	H(a_0, \cdots, a_{n-1}, 1) \leq M(K, d) \prod_{i=1}^d H(\alpha_i)
	.$$

Proof
: **Step 1:**
	Suppose $\alpha_i \in K$ for all $i$.
	If so, we can take $M(K, d) = \prod_{v\in \Sigma_K} \eps_v^{d-1}$.

	Why?
	We need to show that for all $v\in \Sigma_K$, we have 
	$$
	\max \abs{a_i}_v \leq \eps_v^{d-1} \prod_{i=1}^d \max(\abs{\alpha_i}_v)
	.$$
	If so, take the product over all $v$.

	Proceeding by induction, where the $d=1$ is clear, suppose $d\geq 2$ so result holds for $d-1$.
	Choose $k$ such that $\abs{\alpha_k}_v \geq \abs{\alpha_i}_v$ for all $i$.
	We can then write 
	$$
	f(T) = (T-\alpha_k)( T^{d-1} + b_{d-1}T^{d-2} + \cdots + b_1 T + b_0 )
	$$ 
	where $b_{d-1} = 1$ and $b_{-1} = b_d = 0$.
	Then for all $i$, we have $\alpha_i b_{i-1} - \alpha_k b_j$.

	We then apply the triangle inequality:

	\begin{align*}
	\abs{a_i}_v 
	&= \max_i \abs{b_{i-1} - \alpha_j b_i} \\
	&\leq \eps_v  \max \abs{b_{i-1}}_v \abs{\alpha_k b_i}_v \\
	&\leq \eps_v (\max_n \abs{b_i}_v) \max(\abs{\alpha_k}_v, 1) \\
	&\leq_{IH} \eps_v^{d-1} \max_i \abs(\abs{a_i}_v, 1)
	.\end{align*}

	**Step 2:**
	Now suppose $\alpha, \cdots, \alpha_d \in K\sep$.

	Replace $K$ with $K(\alpha_1, \cdots, \alpha_d) \definedas L$.
	This is an extension of $K$ of degree at most $d!$.

	Note that if we define $\eps = \max \eps_v$, and $\abs{\Sigma_K^{Arch}} = A$, then 
	$$
	\prod_{v\in \Sigma_K} \eps_v^{d-1} \leq (\eps^A)^{d-1} = \eps^{A(d-1)}
	.$$
	We also have 
	$$
	\abs{\Sigma_L^{Arch}} \leq d! A
	$$ 
	and for the argument over $K$, we can take $M(K, d) = \eps^{A d!(d-1)}$.

Theorem (Northcott implies Strong Northcott)
: Let $K$ be a PFF satisfying the Northcott property.
	Then for all $d, n \in \ZZ^+$ and for all $R\in \RR$, the set $$\theset{ P \in \PP^n(K\sep) \suchthat [K(P): K] \leq d,~ H(P) \leq R  }$$ is finite.

I.e. passing to finite separable extensions preserves the Northcott property, although this statement is stronger, since this is a *uniform* bound (hence the term "strong").

Stated scheme-theoretically, we can write $P = [x_0, \cdots, x_n]$ where some $x_i = 1$ be rescaling.
Then $K(P) = K(x_0, \cdots, x_n)$ is a finite separable extension.

Proof
: Let $P \in \PP^n(K\sep)$ with $\deg P \leq d$, where $P = [x_0, \cdots, x_n]$ with one $x_i = 1$.

	\begin{align*}
	H(P) 
	&= \prod_{v\in \Sigma_K} \max_i \abs{x_i}_v \\
	&= \prod_{v\in \Sigma_K} \max_i (\abs{x_i}_v  , 1) \\
	&\geq \max_i \prod_{v\in \Sigma_K} \max(\abs{x_i}_v  , 1) \\
	&= \max_i H(x_i)
	.\end{align*}

	So if $H(P) \leq C$ and $[K(P):K] \leq d$, then $\max_i H(x_i) \leq C$ and $\max_i [K(x_i): K] \leq d$.

	Define $$X(K, C, d) = \theset{ x \in K\sep \suchthat H(x) \leq C,~ [K(x): K] \leq d  },$$ we'll show this is a finite set.
	Let $x \in X(K, C, d)$ and $d' \definedas [K(x): K] \leq d$, and $$f(T) \definedas \minpoly x = \prod_{i=1}^{d'} (T - x_i) = T^{d'} + \sum_{i=0}^{d'-1} a_i T^i.$$
	Take $M\definedas \max_{1\leq i \leq d} M(K, i)$ from the previous lemma.

	By the lemma, we have $$H(a_0, \cdots, a_{d'-1}, 1) \leq M \prod{i=1}^{d'} H(x_i).$$
	But Galois conjugate points have the same height, so this is equal to $M H(x)^{d'} \leq MC^d$.
	By the Northcott property, there exist only finitely many such tuples $(a_0, \cdots, a_{d'-1}, 1)$, and thus only finitely many possibilities for $x$.

## Key Theorem: Bounds on Heights of Morphisms for PFFs

Theorem (The Key Theorem)
: Let $K$ be a PFF, and $f: \PP^n \to \PP^m$ be a degree $d$ morphism over $K$ (given by a collection of homogeneous polynomials of degree at most $d$).
	Then there exist constant $C_1, C_2 > 0$ depending only on $K$ such that 

	\begin{align*}
	C_1 H(P)^d \leq H(f(P)) \leq C_2 H(P)^d
	.\end{align*}

Proof
: See Theorem 8.5.6 in Silverman I.