# Finiteness Theorems of ANT (Thursday, November 17)

:::{.remark}
Let $K\in \NF$ with $r$ real places and $s$ complex places with $[K: \QQ] = N = r+2s$.
The main theorems:

1. $\ZZ_K\in \zmod^\fg$, so $\ZZ_K \cong \ZZ^N \in \zmod$.
2. $\size \Pic \ZZ_K < \infty$, and in fact for any (not necessarily maximal) order $\OO$ in $K$, $\size \mathrm{ICM}(\OO) < \infty$.
3. Dirichlet's unit theorem: $\ZZ_K\units$ is finitely-generated with $\ZZ_K\units \isoas{\zmod} \ZZ^{r+s-1} \times \mu(K)$, where $\size \mu(K) < \infty$.
4. Hermite's theorems: 1) there are only finitely many number fields of bounded discriminant, and 2) given a finite set of primes that are allowed to ramify in a cover, there are only finitely many such covers.

Recall the **Minkowski bound**, which has many uses: for $K\in \NF$ of degree $N$, define the Minkowski constant as
\[
M(K) \da \qty{4\over \pi}^s {n!\over n^n}\abs{\delta_K}^{1\over 2}
,\]
then for every integral ideal $a\in \mathrm{Int}(\ZZ_K)$ there exists a nonzero $x\in a$ such that $\abs{ \Norm(x) }\da \abs{ \Norm_{K/\QQ}(x) }$ satisfies
\[
\abs{\Norm(x)} \leq M(K) \cdot \Norm(a)
.\]
:::

:::{.theorem title="Minkowski"}
Without loss of generality let $N\geq 2$, then

a. There is a lower bound
\[
\abs{\delta_K} \geq \qty{\pi\over 4}^{2s} {N^{2N} \over (N!)^2 } \geq {\pi\over 3} \qty{3\pi \over 4}^{N-1}
,\]
so the discriminant grows rapidly in the degree $N$.

b. In particular, $\abs{\delta_K} \neq 1$, so at least one prime ramifies.

:::

:::{.remark}
Note that in the relative case, one can have everywhere unramified extension fields or arbitrary degree.
:::

:::{.proof title="?"}
Muck about with the Minkowski bound: let $a = \ZZ_K$ and note $\Norm(a) = 1$ since it is the unit ideal.
So $\exists x\in \ZZ_K$ nonzero (an algebraic integer) such that $\abs{\Norm(x)} \leq \Norm(\ZZ_K) M(K) = M(K)$, but $\abs{\Norm(x)} = \size \ZZ_K/\gens{x} \in \ZZ_{\geq 0}$ and so we can conclude that $M(K) \geq \abs{\Norm(x)} \geq 1$.
Since $M(K)\geq 1$, applying the Minkowski bound yields
\[
\abs{\delta_K} \geq \qty{\pi\over 4}^s {N^{2N}\over (N!)^2 } \geq \qty{\pi\over 4}^{2n} {N^{2N} \over (N!)^2} \da a_N
.\]
One can check that $a_2 = { \pi^2 \over 4} > 2 > 1$ and
\[
{a_{N+1}\over a_N} = {\pi\over 4}\qty{1 + {1\over N}}^{2N} \geq {\pi\over 4} \qty{1 + {2N \over N}} = {3\pi\over 4} > 2 > 1
\]
by taking the two most obvious terms of the binomial expansion.
Thus for all $N\geq 2$ we have 
\[
\abs{\delta_K} \geq {\pi^2\over 4}\qty{3\pi\over 4}^{N-2} = {\pi\over 3}\qty{3\pi\over 4}^{N-1}
.\]
Note that Stirling's approximation $N!\sim \qty{N\over e}^N \sqrt{2\pi N}$, and so $M(K) \approx {1\over e^N}$.
:::

:::{.remark}
Note that $\abs{\delta_K} \geq \ceiling{\pi^2\over 4} = 3$, so in particular $\abs{\delta_K}\neq 2$ -- this would be interesting if it were not for the following:
:::

:::{.theorem title="Stickelberger"}
For any $K\in \NF$,
\[
\delta_K \equiv 0,1 \mod 4
.\]
:::

:::{.theorem title="Hermite 1"}
For all nonzero $d\in \ZZ$, there are only finitely many $K\in \NF$ with $\delta_K = d$.
:::

:::{.remark}
Choose an embedding $\QQ\embeds \CC$ and an algebraic closure $\QQbar \subseteq \CC$, so by convention we'll regard $\QQ \injects \QQbar \injects \CC$.
Note that finite degree number fields have a finite number of conjugate number fields in $\QQbar$, so we can either count fields up to isomorphism or just regard them all as subfields of $\QQbar$.
:::

:::{.proof title="of Hermite 1"}
By the last result, the discriminant tends to $\infty$ and so it's ETS finiteness for a fixed $N$ and hence a fixed $r$ and $s$.
Suppose $N\geq 2$ and let $K\in \NF$ with $\delta_K = d$.
Let $B \subseteq \RR^{r} \times \CC^n \cong \RR^N$ be the convex body defined in the following way:

- Case 1: if $r\geq 1$, so we have a real place, set $B = \ts{\tv{y_1,\cdots, y_r, z_1,\cdots, z_s}}$ such that 
  - $\abs{y_1} \leq 2^{N-1} \qty{\pi\over 2}^{-s} \abs{d}^{1\over 2}$.
  - For all $2\leq i\leq r$, $\abs{y_i} \leq {1\over 2}$,
  - For all $j$, $\abs{z_j}\leq {1\over 2}$.
- Case 2: if $r=0$, so we have no real places, take $B$ where
  - $\abs{z_1 - \bar z_1} \leq 2^{N-1} \qty{\pi\over 2}^{1-s} \abs{d}^{1\over 2}$ (a condition on $\Im z_1$),
  - $\abs{z_1 +\bar z_1} \leq {1\over 2}$
  - For all $2\leq j\leq s$, $\abs{z_j} \leq {1\over 2}$.

Where these conditions come from: reverse-engineering the smallest thing to which we can apply Minkowski's convex body theorem to guarantee a lattice point in $B$.

:::{.exercise title="?"}
Show $B$ is a compact convex body and show $\Vol B = 2^{N-s}\abs{d}^{1\over 2}$.
Compare this to the $B_t$ used to prove that theorem; this $B$ is simpler!
:::

One can compute $\covol \sigma(\ZZ_K) = 2^{-s}\abs{d}^{1\over 2}$ and check that $\Vol B = 2^N \covol \sigma(\ZZ_K)$ (recalling that $\sigma(\ZZ_K)$ is an embedded lattice).
Applying Minkowski's convex body theorem, one can find a nonzero $x\in \sigma(\ZZ_K) \intersect B$.


:::{.claim}
$K = \QQ[x]$ for this particular $x$.
:::

One then looks at both of the cases above, which use similar arguments:

1. If $\abs{\Norm(x)} = \abs{\prod_{i=1}^N \sigma_i(x)}\in \ZZ_{\geq 0}$ where $\abs{\sigma_i(x)}\leq {1\over 2}$ for all $i > 2$ and $\abs{\sigma_i(x)} > 1$, so $\sigma_1(x) \neq \sigma_i(x)$ for any such $i$ and thus $x$ does not lie in any proper subfield of $K$.
  This follows from an exercise in field theory.

2. Suppose $\abs{\sigma_1(x)} > 1$, which must be a non-real embedding by assumption, and note $\abs{\sigma_1(x)} = \abs{\overline{\sigma_1(x)}}$.
  If $\sigma(x) \in B$ then $\abs{\Re(\sigma_1(x))} \leq {1\over 4}$ and in particular $\sigma_1(x) \not\in \RR$ and thus $\sigma_1(x)\neq \overline{\sigma_1(x)}$ since it's not real.
  So $\sigma_1(x) \neq \sigma_j(x)$ for any $j\geq 3$, noting $\sigma_2 = \bar{\sigma_1}$, and as above $x$ is a primitive element for $K$.
  Note that the minimal polynomial of $x$ is $f(t) = \prod_{i=1}^N \prod (t-\sigma_i(x))$.
  Since the roots of $f$ are bounded by assumption, the coefficients (which are symmetric functions of the roots) are also bounded.
  Since the coefficients are integers, there are only finitely many choices for $f$ and thus only finitely many choices for $K$.

:::

:::{.proof title="of Hermite 2"}
Fix $N = S \da \ts{p_1 < \cdots < p_r}$ a finite set of primes.
Then
\[
\size \ts{K \subseteq \CC \st [K:\QQ] = N, \, K \text{ unramified for all }\ell \not\in S} < \infty
.\]
:::

:::{.remark}
Consider how this could fail: one would need an unbounded discriminant with an arbitrary large $p\dash$adic valuation for some $p\in S$, so this is all we need to rule out.
So we'll look for an upper bound on $v_p \delta_K$ depending only on $N$.
:::

:::{.lemma title="?"}
Let $K\in \NF$ of degree $N \geq 2$ and let $p$ be a prime, then
\[
v_p(\delta_K) \leq N \floor{\log_p N} + N - 1 \leq N \floor{\log_2 N} + N - 1
.\]
:::

:::{.remark}
Note that applying Hermite 1 and this lemma proves Hermite 2, since this yields only finitely many possibilities for the discriminant and for a given discriminant there are only finitely many number fields.
:::

:::{.proof title="of lemma"}
Recalling the *different*, we have
\[
v_p(\delta_K) = v_p(\Norm(\Delta_{K/\QQ})) = \sum_{P\divides p} f_P v_P(\Delta_K)
\]
where $f_p$ is the residual degree.
Writing $\Delta_K = \prod_{i=1}^r P_i^{a_i}$, we have $\Norm(\Delta_K) = \prod_{i=1}^r p^{f_{P_i} a_i} = p^{\sum_{i=1}^r f_i a_i}$. 
We can now use the fact that $v_P(\Delta) \leq e_p -1 + v_P(e_P)$, so $a_i = v_{P_i}(\Delta)$ and thus 
\[
v_{p}(\delta_K) \leq \sum_{i=1}^r f_i(e_i - 1 + v_{P_i}(e_i) )
.\]
:::