# Thursday, November 03

:::{.remark}
Last time: applying the Dedekind-Kummer theorem to show that if $B/A, L/K$ is a NT1 square with $L/K$ degree $n$ and separable, if $\exists p\in \mspec A$ for which $\size\ts{P\mid p \st f(P\mid p) = 1} > \size(A/p)$ then $B$ is not monogenic, i.e. $\not\exists \alpha\in B$ such that $B = A[ \alpha ]$.
Note that necessarily $\size (A/p) < \infty$ for this to apply.
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:::{.remark}
NTII would show more: for all primes $p$ and all $n, r\in \ZZ_{\geq 0}$ with $1\leq r\leq n$ and $n\geq 2$, there exist infinitely many degree $n$ number fields $K$ in $\bar{K}$ such that $\ZZ_K$ has precisely $r$ degree one primes lying over $(p)$.
Thus if $n>p$, choosing $r>p$ yields a non-monogenic $\ZZ_K$.

> More generally, see Krasner's lemma.

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:::{.example title="?"}
Take a cubic field over which 2 completely splits, so $p=2$ and $n = 3 > p$.
This strategy couldn't possibly work for $n=2$ though, since all quadratic number fields are monogenic.
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:::{.example title="?"}
Let $L_1, L_2/ K$ be finite degree extensions and take the compositum $L_1 L_2$, i.e. the field they generate.
Consider the 3 corresponding Dedekind domains $B_1, B_2, B_{12}$, and let $A$ correspond to $K$ and $B$ to $\bar K$.
Fact: if $p\in \mspec A$ splits completely in $B_1$ and $B_2$, then it splits completely in $B$.

Take $p=2$ and a biquadratic field $K \da \QQ(\sqrt{d_1}, \sqrt{d_2})$ where $d_1, d_2, d_1/d_2\not\in \squares{\QQ}$ and $d_1, d_2$ squarefree.
Note that $2$ splits in $\ZZ_{\sqrt{d}} \iff d\equiv 1\mod 8$, which is proved by Dedekind-Kummer.
So take $d_1\equiv d_2 \equiv 1 \mod 8$, e.g. $d_1 = 17, d_2 = 33$.
Since 2 splits in the two intermediate fields, it splits in $K$.
Here there are 4 degree 1 primes lying over 2, and $\size \ZZ/2 = 2$ and $4>2$, so $\ZZ_K$ is not monogenic.
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:::{.exercise title="?"}
Run this example with $p=3$.
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:::{.remark}
Monogenic number fields is a topic of recent research interest.
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## Geometry of Numbers

:::{.remark}
Let $\Omega \subseteq \RR^n$ be a convex body (a bounded convex set with nonempty interior containing zero) and a $\ZZ\dash$lattice $\Lambda \subseteq \RR^n$.
Note that $\Lambda = \ZZ S$ for $S$ an $\RR\dash$basis of $\RR^n$.
The main questions concern how $\Omega$ and $\Lambda$ interact.
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:::{.question}
Is $(\Omega \intersect \Lambda)\smz$ nonempty?
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:::{.question}
Consider the dilate $\alpha \Omega \da \ts{\alpha x \st x\in \Omega}$.
How large must \( \alpha \) be in order for \( (\alpha \Omega \Lambda)\smz \) to be nonempty?

One can ask more refined questions, e.g. what are the successive minima (the smallest $\alpha_i$ such that $\alpha_i \Omega$ has $i$ linearly independent lattice vectors).
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:::{.remark}
Goal: Minkowski's convex body theorem, which says that if $\vol \Omega$ is large compared to $\covol \Lambda$, it will contain a lattice point.
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:::{.definition title="?"}
Let $\Omega$ be a bounded convex set (always).

- $\Omega$ is **symmetric** if $- \Omega = \Omega$, i.e. $x\in \Omega\implies -x\im \Omega$.
  Note that $\Omega$ necessarily contains zero if it is symmetric.
- For $x_1,\cdots, x_n \in \RR^N$, a **convex combination** is $\sum \lambda_i x_i$ where $\lambda_i \geq 0$ and $\sum \lambda_i = 1$.
  This generalizes $\lambda x_1 + (1- \lambda) x_2$ for $\lambda\in [0, 1]$ in 2 dimensions.

- For $S \subseteq \RR^N$, define the **convex hull** of $S$, $\Conv(S)$, to be all convex combinations of finite sequences of elements in $S$.
- A nonempty $S \subseteq \RR^N$ is **affinely linearly independent** if $\forall x_0 \in S$ the set $\ts{x-x_0\st x\in S\smts{x_0}}$ is $\RR\dash$linearly independent.
  Note that if this holds for one $x_0\in S$, it holds for all $s\in S$.
- If $S$ is affinely linearly independent in $\RR^k$ and $\size S = k$ then $S$ is a **simplex**.

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:::{.exercise title="?"}
Let $S \subseteq \RR^n$.

- Show there exists a unique smallest subset $\phi(S) \subseteq \RR^N$ such that $\phi(S)\contains S$ is convex and is minimal with respect to this property.
- Show that $\phi(S) = \Conv(S)$.

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:::{.exercise title="?"}
For $\Omega \subseteq \RR^N$, show that TFAE:

- $\Omega$ is convex.
- $\Omega^\circ$ (the interior) is convex.
- $\bar{\Omega}$ is convex.

Thus convexity is not at all affected by $\bd\Omega$.
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:::{.definition title="Jordan measurability"}
A subset $\Omega \subseteq \RR^N$ is **Jordan measurable** iff its characteristic function $\chi_{\Omega}: \RR^n\to \RR$ is Riemann integrable.
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:::{.remark}
By the Lebesgue criterion, a bounded function is Riemann integrable iff the set of discontinuities has Lebesgue measure zero.
Since characteristic function are only discontinuous at the boundary, a set is Jordan measurable iff $\bd \Omega$ has measure zero.
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:::{.theorem title="?"}
Any bounded convex set $\Omega \subseteq \RR^N$ is Jordan measurable iff it is Lebesgue measurable.
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## Lattices in $\EE^N$

:::{.definition title="Lattices"}
If $V\in\mods{\RR}^\fd$, a **lattice** $\Lambda \subseteq V$ is the $\ZZ\dash$span of an $\RR\dash$basis.
In particular, $\Lambda \cong \ZZ^n \in \zmod$.
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:::{.remark}
Note that the notion of covolume needs a bit more structure than just an abstract vector space (namely a notion of Haar measure).
One could compare this definition to lattices in $\qmod$ -- a subgroup \( \Lambda \subseteq V \) for $V\in\qmod, \dim V = N$ is a lattice iff $\Lambda \cong \ZZ^N \in \zmod$.
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:::{.question}
If $\Lambda \subseteq \RR^n$ is a subgroup with a $\ZZ\dash$basis $\ts{v_1,\cdots, v_n}$, must $\Lamba$ be a lattice?
I.e. is it true that $\spanof_\RR \gens{v_1, \cdots, v_n}$?
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:::{.answer}
No, unless $N=1$.
Consider $\ZZ[\sqrt 2] \cong \ZZ\gens{1, \sqrt 2}$ which is free of rank 2 as a $\ZZ\dash$module, but is contained in $\RR\times \ts{0} \subsetneq \RR^2$, so $\ZZ\dash$linear independence does not imply $\RR\dash$linear independence.
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:::{.remark}
Thus one needs extra conditions on subgroups $\Lambda \subseteq \RR^N$ to determine if $\Lambda \cong \ZZ^N\in \zmod$ is a lattice.
Long story short: it's the topological condition that $\Lambda$ is *discrete* -- every point admits a neighborhood whose only point is that point.

> Recall that lattices over a PID are parameterized by $\GL_n(K)/\GL_n(\RR)$, and that works verbatim here.
  Todo: missed the precise statement here.

Note $\GL_n(\RR)$ acts on subgroups of $\RR^n$ by homeomorphisms and is transitive on lattices since it acts simply transitively on bases for lattices.
Thus if $\Lambda \subseteq \RR^N$ is a $\ZZ\dash$lattice then there is some $M\in \GL_N(\RR)$ with $M\ZZ^N = \Lambda$, and since $\ZZ^N$ is discrete, $\Lambda$ is discrete (and hence closed since $\RR^n$ is Hausdorff).
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