# Ch. 21: Consequences of Minkowski's Bound (Saturday, May 22)

:::{.remark}
We were discussing a lower bound on $\abs{\discriminant_K}$.
We have Minkowski's class number bound: every ideal class has a representative of norm at most $M_K$, where
\[
M_K \da \qty{4\over \pi}^{r_2} \qty{n! \over n^n}\sqrt{\abs{\discriminant_K}}
.\]
A consequence of the theorem is $M_K \geq 1$, and rearranging yields a bound
\[
\abs{\discriminant_K} \geq \qty{n^n \over n!}^2 \qty{\pi/4}^{2r)2}
.\]
What is this bound really telling us?
Let's bound the right-hand side from below.
Note that $\pi/4<1$, and since this is raised to a power, this might make things smaller.
The worst case, i.e. the smallest it could be, is when $2r_2$ is as large as possible, so using that $r_1 + r_2 = n$ we have
\[
\abs{\discriminant_K} \geq \qty{n^n\over n!}^2 \qty{\pi/4}^n \da B_n
.\]
How does $B_n$ grow?
We could use Stirling's formula, but we'll take a crude bound by looking at ratios:
\[
{B_{n+1} \over B_n} 
&= (1 + {1\over n})^{2n} (\pi/4)\\
&= \qty{ 1 + {2n \choose 1} {1\over n} + \cdots }(\pi/4) \\
&\geq 3\pi/4
.\]
Noting that $B_2 = \pi^2/4$, so by induction
\[
B_n \geq (3\pi/4)^{n-2} (\pi^2/4)
.\]
:::

:::{.remark}
Some consequences:

a. $\abs{\discriminant_K} > 1$ for all number fields $K\neq \QQ$.
b. $\abs{\discriminant_K}\to\infty$ as $[K:\QQ]\to\infty$.

The following says why (a) is important:
:::

:::{.theorem title="Dedekind"}
$p$ ramifies in $\ZZ_K \iff p \divides \discriminant_K$.
:::

:::{.proof title="?"}
Omitted, see book.
:::

:::{.remark}
So by (a), every number field $K\geq \QQ$ there is at least one ramified prime $p$.
:::

:::{.remark}
Note that if $\abs{\discriminant_K} = 1$, then $K=\QQ$, i.e. there is only one such number field.
What about for any fixed number $n$?
The next theorem says that there are only finitely many number fields occurring below a prescribed bound:
:::

:::{.theorem title="Hermite's theorem"}
For every $X>0$, there are finitely many number fields $K$ with $\abs{\discriminant_K} \leq X$.
:::

:::{.remark}
Since $\abs{\discriminant_K}\to\infty$ as $[K:\QQ]\to\infty$, it suffices to prove this theorem with a fixed $n\da [K: \QQ]$.
We'll make a simplifying assumption that $r_1 = n$ -- this doesn't simplify the proof so much, but rather simplifies the notation.
The full proof in the book is not so different.
:::

:::{.remark}
Define a region
\[
R \da \ts{\tv{\elts{x}{n}} \in \RR^n \st \abs{x_{i\leq n-1}} \leq {1 \over 2},\,\, \abs{x_n} \leq T } \subseteq \RR^n
,\]
where we'll specify $T$ in a moment.

Note that

- $R$ is centrally symmetric
- $R$ is convex, by the triangle inequality,
- The volume is easily computable: $\vol R = 1\cdot 1\cdot \ldots \cdot 1 \cdot 2T = 2T$.

Choose $T = 2^n \sqrt{X}$.
Suppose $K$ is totally real with \( \abs{ \discriminant_K } \leq X \).
Recall that 
\[
\covol \iota \ZZ_K = 2^{r_2} \sqrt{\abs{\discriminant_K}} 
\leq \sqrt{X}
.\]

Then 
\[
\vol R = 2^{n+1 } \sqrt{X} > 2^n \covol \iota \ZZ_K
,\]
so Minkowski's theorem can be applied: there is a nonzero $\alpha\in \ZZ_K$ with $\iota( \alpha) \in R$.
Then $\abs{\sigma_i( \alpha)} \leq 1/2$ for $i\leq n-1$ and $\abs{ \sigma_n ( \alpha)} \leq T$.
The claim is that there are only finitely many such \( \alpha \), since it was a root of a polynomial in a finite set $P_{n, T}$.
The claim is now that $K = \QQ( \alpha)$, so this $\alpha$ uniquely determines $K$.
Since there were finitely many \( \alpha \), there can only be finitely many such $K$.

The claim is that the size of $\sigma_n( \alpha)$ has to be big, say at least 1.
We have control over the product, since
\[
1\leq \abs{N \alpha} 
= \prod \abs{ \sigma_i( \alpha)} 
\leq {1\over 2^{n-1}} \abs{ \sigma_n( \alpha)}
,\]
since the first $n-1$ terms contribute at most $1/2$ each.
So $\abs{\sigma_n( \alpha)} \geq 2^{n-1} \geq 1$.
Suppose now that $\QQ(\alpha) < K$ is a proper subfield, so $[K : \QQ( \alpha)] = d > 1$.
Then every embedding $\QQ(\alpha) \embeds \CC$ extends to $d$ embeddings $K\embeds \CC$, but this means that for any element $x \in \QQ(\alpha)$, the images $\sigma_1(x),\cdots, \sigma_n(x)$ would have the same element repeated $d$ times.
But we know that $\sigma_n( \alpha)$ is different from all of the other $\sigma_i( \alpha)$, so this is a contradiction.
:::

:::{.remark}
Idea of proof: making sure some image of $\alpha$ under one embedding is different than all of the other images.
:::

:::{.remark}
Some remarks on modern research!
Let $N_n(X)$ be the set of number fields of degree $n$ where $\abs{\discriminant_K} \leq X$.
:::

:::{.conjecture}
For each fixed $n$,
\[
\lim_{X\to \infty} {N_n(X) \over X} = \delta_n > 0
,\]
where $\delta_n$ is some particular constant.

Some known results: 

- $n=2$: known to Gauss, since this is more or less equivalent to counting squarefree numbers.
- $n=3$: much harder, Davenport-Heilbronn.
- $n=4, 5$: Bhargava, part of what resulted in his 2014 Fields medal.

:::

:::{.remark}
One could restrict this problem, e.g. by prescribing a particular Galois group.
See Mahler's conjecture.
:::

# Chapter XYZ: Relative Extensions, Galois Theory, Prime Splitting

:::{.remark}
Up until now: we've compared extensions over $\QQ$:

\begin{tikzcd}
	K && {\ZZ_K} \\
	\\
	\QQ && \ZZ
	\arrow[no head, from=1-1, to=3-1]
	\arrow[no head, from=1-3, to=3-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJLIl0sWzAsMiwiXFxRUSJdLFsyLDAsIlxcWlpfSyJdLFsyLDIsIlxcWloiXSxbMCwxLCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzIsMywiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dXQ==)

We'll next talk about extensions between number fields:

\begin{tikzcd}
	L && {\ZZ_L} \\
	\\
	K && {\ZZ_K}
	\arrow[no head, from=1-1, to=3-1]
	\arrow[no head, from=1-3, to=3-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJMIl0sWzAsMiwiSyJdLFsyLDAsIlxcWlpfTCJdLFsyLDIsIlxcWlpfSyJdLFswLDEsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbMiwzLCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV1d)

We can ask the same sorts of questions about prime ideals factoring.
Note that if $I = \prod P_i^{e_i}$ with each $P_i \in \spec \ZZ_K$ , then extending to $\ZZ_L$ yields 
\[
I\ZZ_L = \prod (P_i \ZZ_L)^{e_i}
.\]
So we want to understand the following: given a prime ideal $P$ of $\ZZ_K$, how does $P\ZZ_L$ factor?
:::

:::{.definition title="lies above"}
Let $K\leq L$ be number fields, and suppose $Q\in \spec \ZZ_L, P\in \spec \ZZ_K$.
Then we say $Q$ **lies above** $P$ if $Q \supseteq P$, or equivalently $Q \divides P\ZZ_L$.
:::

:::{.proposition title="?"}
Every nonzero $Q\in \spec \ZZ_L$ lies above a unique nonzero $P \in \spec \ZZ_K$.
:::

:::{.proof title="?"}
Consider $P \da Q \intersect \ZZ_K \in \spec \ZZ_K$.
This is nonzero because taking the norm of any element of $Q$ yields a nonzero integer still in the ideal.
Then $Q$ lies above $P$ by definition.
Why is this unique?
Suppose $Q$ lies about $P'$, we'll show $P = P'$.
Since $Q$ lies above $P'$, $Q\supseteq P'$.
But $P'\normal \ZZ_K$, so $Q\intersect \ZZ_K \supseteq P'$, so $P \contains P'$.
We know $P'$ is maximal, since $\ZZ_K/P'$ is a finite domain and thus a field, so $P = P'$.
:::

:::{.definition title="Ramification and residue degrees"}
Let $P\in \spec \ZZ_K$, then write 
\[
P\ZZ_L = \prod Q_i^{e_i}
\]
with $Q_i \in \spec \ZZ_L$.
Then the $Q_i$ are the prime ideals of $\ZZ_L$ above $P$.
The exponent $e_i$ is called the **ramification degree**, usually denoted $e(Q_i/P)$.
We have the following picture:

\begin{tikzcd}
	L && {\ZZ_L} && Q \\
	\\
	K && {\ZZ_K} && P
	\arrow[no head, from=1-1, to=3-1]
	\arrow[no head, from=1-3, to=3-3]
	\arrow[no head, from=1-5, to=3-5]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMCwwLCJMIl0sWzAsMiwiSyJdLFsyLDAsIlxcWlpfTCJdLFsyLDIsIlxcWlpfSyJdLFs0LDIsIlAiXSxbNCwwLCJRIl0sWzAsMSwiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFsyLDMsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbNSw0LCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV1d)


The inclusion $\ZZ_K \injects \ZZ_L$ induces a ring morphism $\ZZ_K \to \ZZ_L/Q$, where the kernel is $\ZZ_K \intersect Q = P$.
Thus there is an injection $\ZZ_K/P \injects \ZZ_L/Q$, which is an inclusion of finite fields.
So we'll define 
\[
f(Q/P) \da [\ZZ_L/Q : \ZZ_K/P]
\]
to be the **residue degree** of $Q/P$.
Note that 
\[
\# \ZZ_L/\QQ = \qty{\# \ZZ_K/P}^{f(Q/P)}
\quad\text{i.e.}\quad 
N_L(Q) = N_K(P)^{f(Q/P)}
.\]

:::

:::{.theorem title="efg theorem"}
If $P \in \spec \ZZ_K$ is nonzero with $P\ZZ_L = \prod Q_i^{e_i}$, then 
\[
\sum_i e(Q_i/P) f(Q_i/P) = [L: K]
.\]
:::

:::{.lemma title="?"}
If $I\in \spec \ZZ_K$ is nonzero, then extend to $L$ to get $I\ZZ_L$.
Then
\[
N_L (I\ZZ_L) = \qty{ N_K(I)}^{[L : K]}
.\]
:::

:::{.proof title="?"}
Omitted.
Idea of why it's true: the norm of an ideal is supposed to be a "product of conjugates", although naive conjugates of an ideal might not remain an ideal in the field one starts with.
So interpret norms as products of images under all embeddings into $\CC$.
But then just interpret $[L: K]$ is the number of lifts of embeddings $K\embeds \CC$ to $L\embeds \CC$.
:::

:::{.proof title="of efg theorem"}
Take norms in $L$, then 
\[
N( P\ZZ_L) 
= \prod N(Q_i)^{e_i} \\
= N_K(P)^{e_1 \sum f(Q_i/P)}
.\]
On the other hand, the left-hand side is $N_K(P)^{[L: K]}$, so the exponents must be equal.
:::

:::{.remark}
What are the specific prime ideals involved in the factorization, i.e. is there a generalization of Dedekind-Kummer here?
:::

:::{.proposition title="Generalized Dedekind-Kummer theorem"}
Write $L = K(\theta)$ for some $\theta \in L \intersect \bar\ZZ = \ZZ_L$.
Let $m(x)$ be the minimal polynomial of $\theta$ over $K$, so $m\in \ZZ_K[x]$.
Let $P\in \spec \ZZ_K$ lying above $p\in \spec \ZZ$.
Then as long as $p\notdivides [\ZZ_L: \ZZ_K[\theta]]$, then the factorization of $P\ZZ_L$ mirrors the factorization of $m$ over the residue field $\ZZ_K/P$.
:::

:::{.proof title="?"}
Omitted, see "Number Rings" by Marcus.
Paul strongly recommends!

:::