# Ch. 18: Dirichlet's Units Theorem Part I (Friday, May 21)

:::{.remark}
Some notation introduced last time: for $K$ a number field, $[K: \QQ] = n$ the degree, set $\sigma_{1}, \cdots, \sigma_{r_1}$ to be the real embeddings, and $\sigma_{r_1 + 1}, \cdots, \sigma_{r_1 + r_2}$ (along with their complex conjugates) to be the nonreal embeddings into $\CC$.
Let $\mu_K \da \ts{ z\in K \st z^n = 1, n\in \NN }$ be the $n$th roots of unity
Recall Dirichlet's unit theorem: there are $\elts{\eps}{r_1+r_2-1} \in \ZZ_K\units$ of infinite order such that
\[
\ZZ_K\units = \mu_K \times \prod_{i\leq r_1 + r_2 -1} \gens{ \mu_i } 
.\]
Moreover, $\# \mu_K < \infty$.
:::

:::{.remark title="An outline of the proof"}
Set $d = r_1 + r_2$.
We define a homomorphism
\[
\Log: K\units &\to \RR^{d} \\
\alpha &\mapsto \tv{ 
\log\abs{\sigma_i(\alpha)}, 
\cdots,
\log\abs{\sigma_{r_1}(\alpha)}, 
2 \log\abs{\sigma_{r_1+1}(\alpha)}, 
\cdots,
2 \log\abs{\sigma_{r_1+r_2}(\alpha)}, 
}
.\]
Note that we only used half of the non-real embeddings, but made up for it by including a 2!
We'll mostly be concerned with its restriction to $\ZZ_K\units$.


1. Show that the restriction has finite kernel $\mu_k$ -- we've completed this.

2. Show $\Log(\ZZ_K\units)$ is discrete as a subgroup of $\RR^{d}$

3. Use the fact that every discrete subgroup of $\RR^d$ is a lattice, the $\ZZ\dash$span of linearly independent vectors.


4. Prove the units theorem with $d-1$ replaced by $g = \rank \Log(\ZZ_K\units)$.

5. Deferred to next chapter: show $g = r_1 + r_2 - 1$.

:::

## Step 2: Discreteness

:::{.definition title="Discrete lattices"}
Let $d\in \ZZ^{\geq 0}$ and let $\Lambda \leq \RR^d$ be a subgroup.
Then $\Lambda$ is **discrete** iff for every $R>0$, $\Lambda \intersect B_R(0)$ is a finite set, where $B_R(0)$ is a ball of radius $R$ about 0.
:::

:::{.proposition title="?"}
$\Log(\ZZ_K\units)$ is discrete in $\RR^d$ for $d=r_1 + r_2$.
:::

:::{.proof title="?"}
Let $R>0$, we'll show that there are only finitely many $\alpha \in \ZZ_K\units$ for which $\Log(\alpha) \in B_R(0)$.
This forces $\log\abs{\sigma(\alpha)} \leq R$ for all $\sigma: K\embeds \CC$.
Thus $\abs{\sigma(\alpha)} \leq e^R$ for all such embeddings.
By a previous lemma, this makes $\alpha$ a root of a polynomial in some finite set $P_{n,e^R}$ -- what this set is isn't important, just recall that once you have a bound on the embeddings, this gives a bound on the coefficients of the field polynomial.
:::

## Step 3

:::{.theorem title="?"}
Every discrete subgroup of $\RR^d$ is a lattice.
:::

:::{.proof title="?"}
See the book!
:::

## Step 3

:::{.theorem title="Weak units theorem"}
Let $g$ be the rank of $\Log\ZZ_K\units \subseteq \RR^{r_1 + r_2}$.
Then there are fundamental units \( \elts{\eps}{g} \in \ZZ_K\units \) of infinite order such that
\[
\ZZ_K\units = \mu_K \times \prod_{i\leq g} \gens{ \eps_i } 
.\]
Note that this is the same theorem from before, just with $d$ replaced by $g$.

:::

:::{.proof title="?"}
Using that $g$ is by definition the rank of the lattice $\Log\ZZ$, choose $\elts{\eps}{g}\in \ZZ_K\units$ such that the $\Log(\eps_i)$ are $\RR\dash$linearly independent and their $\ZZ\dash$span is the lattice.
Let $\eps\in \ZZ_K\units$ be any unit, then $\Log(\eps) \in \Log\ZZ_K\units$, so there are integers $\elts{n}{g}\in \ZZ$ such that 
\[
\Log(\eps) = \sum n_i \Log(\eps_i)
.\]
Hence
\[
\Log\qty{ \eps / \prod_{i\leq g} \eps_i^{n_i} } = \vector 0
.\]
So it's in the kernel and equal to some $\zeta\in \mu_K$, so we can write 
$\eps = \zeta \prod_{i\leq g} \eps_i^{n_i}$.
This representation of $\eps$ is unique, in the sense that $\zeta$ and the $n_i$ are uniquely determined.
Why?
For any other representation, applying $\Log$ would kill the $\zeta$ part and write $\Log(\eps) = \sum n_{j} \eps_j$ in the basis $\eps_j$.
So the $n_i$ are unique, and $\zeta$ is just obtained by dividing, so it can be recovered uniquely as well.
:::

:::{.remark}
Our goal is to now show $g=r_1 + r_2-1$, which is hard.
It's easier to obtain a bound instead.
:::

:::{.proposition title="Lattice rank bound"}
Let $g = \rank \Log(\ZZ_K\units)$, then
\[
g\leq r_1 + r_2 - 1
.\]
:::

:::{.lemma title="?"}
Let $\Lambda \subseteq \RR^d$ be a lattice, then $\rank \Lambda = \dim_\RR V$ where $V$ is the smallest subspace of $\RR^d$ containing $\Lambda$.
:::

:::{.proof title="?"}
Let $g\da \rank \Lambda$, and write \( \Lambda= \sum_{i\leq g} \ZZ \vector v_i \) where the $\vector v_i$ are $\RR\dash$linearly independent. 
Then $V = \sum_{i\leq g} \RR \vector v_i$, and $\dim_\RR V = g = \rank \Lambda$.
:::

:::{.proof title="of proposition, lattice rank bound"}
Let $\vector w = \tv{1, 1, \cdots, 1}\in \RR^{r_1 + r_2}$.
If \( \alpha\in \ZZ_K\units \), then dotting sums the components, so
\[
\vector w \cdot \Log \alpha = \sum_{i\leq r_1} \log \abs{\sigma_i( \alpha)}
+ 2\sum_{r_1 + 1 \leq i \leq r_1 + r_2} \log \abs{\sigma_i( \alpha) }
.\]
Note that $\abs{\sigma_i(\alpha)} = \abs{\bar{\sigma( \alpha)}}$, so because we have a 2 here, we could have written this as
\[
\vector w \cdot \Log \alpha 
&= \sum_{\sigma:K\injects \CC} \log \abs{\sigma(\alpha)}\\
&= \log\prod_{\sigma: K\embeds \CC} \abs{\sigma( \alpha)} \\
&= \log \abs{ N ( \alpha)} \\
&= 0
,\]
since \( \alpha\in \ZZ_K\units \implies N( \alpha) = \pm 1 \).
This says that $\Log(\ZZ_K\units) \subseteq \vector w^\perp$, which has codimension $1$ and thus dimension $r_1 + r_2 - 1$.
Now $V \subseteq \vector w ^\perp$. so its rank is bounded by this dimension.
So $g\leq r_1 + r_2 - 1$.
:::

# Ch. 20: Unit Theorem, Part II

:::{.remark}
We still need to prove that this is an equality.
We'll start with some geometric preliminaries.
:::

:::{.definition title="The Minkowski Embedding"}
The **Minkowski embedding** of $K$ is the map
\[
\iota: K &\to \RR^{r_1} \times \CC^{r_2} \\
\alpha &\mapsto \tv{ 
\sigma_1(\alpha), \cdots, \sigma_{r_1}( \alpha),
\sigma_{r_1 + 1}( \alpha), \cdots, \sigma_{r_1 + r_2}( \alpha)
}
.\]
Identify $\CC \cong \RR^2$ using $a+bi \mapsto \tv{a, b}$, which identifies the codomain above with $\RR^{r_1} \times \RR^{2r_2} = \RR^n$.
So we view $\iota:K\to \RR^n$.
:::

:::{.remark}
Note $\iota$ is injective and $\QQ\dash$linear.
:::

:::{.proposition title="Extremely important"}
$\iota(\ZZ_K)$ is a lattice of full rank in $\RR^n$.
:::

:::{.proof title="?"}
Why is this a lattice?
We'll show it's a discrete subgroup.

Let $R>0$, then if \( \alpha\in \ZZ_K \) and $\iota(\alpha) \in B_R(0)$, then $\abs{\sigma( \alpha)} \leq R$ for all \( \sigma: K\embeds \CC \).
This forces \( \alpha \) to be a root of one of finitely many polynomials in $P_{n, R}$.

Now let $g\da \rank \iota(\ZZ_K)$ as a lattice.
This is a free abelian group of rank $g$, so isomorphic to $\ZZ^g$ since it's the $\ZZ\dash$span of $g$ linearly independent elements.
On the other hand, by the integral basis theorem, $\ZZ_K$ is a free abelian group of rank $n$.
Since $\iota$ is injective, $\iota(\ZZ_K)$ is free of rank $n$, and a fact from algebra implies $n=g$.
:::

:::{.remark}
Since $\iota(\ZZ_K$ is a full rank lattice, $\covol \iota(\ZZ_K)$ is well-defined, where the covolume is the volume of the fundamental parallelepiped spanned by any generating set, and is measured by the absolute value of the determinant of the matrix of basis elements.
It turns out that the covolume is the discriminant from before, but we don't need the exact value at the moment.
:::

:::{.remark}
We'll now make some further reductions.
Recall that $g\da \rank \Log(\ZZ_K\units) = \dim_\RR V$, where $V$ is the smallest subspace of $\RR^{r_1 + r_2}$ containing $\Log(\ZZ_K\units)$.
Recall the rank-nullity theorem:
\[
\dim_\RR V + \dim_\RR V^\perp = r_1 + r_2
\implies g = \dim_\RR V = r_1 + r_2 - \dim_\RR V^\perp
.\]
the goal is to show $\dim_\RR V^\perp = 1$.
We know $\vector w \da \tv{1, 1, \cdots, 1} \in V^\perp$, so it suffices to show every element of $V^\perp$ is a multiple of $\vector w$.
Suppose not, then there's an element of $V^\perp$ of the form \( \tv{c_1, c_2 \cdots, c_j, 0} \) by subtracting a suitable multiple of $\vector w$.
Not all of the $c_i$ are zero, since this would mean the original was a multiple of $\vector w$.
Define a map
\[
F: K\units&\to \RR \\
\alpha&\mapsto \tv{c_1, c_2, \cdots c_{r_1 + r_2 - 1}, 0 } \cdot \Log( \alpha)
.\]
Now note that \( F(\ZZ_K\units) = 0 \), since $V \supseteq \Log(\ZZ_K)$.

:::

:::{.proposition title="?"}
Let $c_1, c_2, \cdots, c_{r_1 + r_2 - 1} \in \RR$ not all 0.
Then if $F$ is defined as above, 
\( F(\ZZ_K\units) \neq 0 \).
:::

:::{.remark}
This requires some delicate inequalities, and an application of Minkowski's convex body theorem.
:::

:::{.lemma title="Key lemma"}
There is a sequence of nonzero $\alpha\in K$ such that 

- $\abs{F(\alpha)} \to \infty$, but
- $\abs{N( \alpha)} < C_1 = C_1(K, c_1, \cdots, c_{r_1 + r_2 - 1} )$ not depending on $\alpha$.
:::

:::{.proof title="of proposition, assuming the key lemma"}
Note that there are only finitely many ideals of $\ZZ_K$ of norm bounded by any constant $C_1$.
We saw this for quadratic fields: if $N(J) = m$, which by Lagrange's theorem yields $1+1+\cdots 1 = m \equiv 0 \mod J$.
So $m\in J$ and thus $J\divides m$.
But by unique factorization, any ideal only has finitely many divisors, so there are only finitely many possible values of $m$.

Note also that $\abs{N( \alpha )} = N(\gens{ \alpha } ) < C_1$, so there are only finitely many such ideals.
So given an infinite sequence of $\alpha$s, there are only finitely many different principal ideals they generate.
So by the pigeonhole principle, we can pass to a subsequence such that \( \gens{ a }  \) is constant.
Note that $F$ can't be constant on this sequence, since $F(\alpha) \to \infty$, so we can choose
\( \alpha_1, \alpha_2 \) such that 
\[
\gens{ \alpha_1 }= \gens{ \alpha_2 } \quad\text{but}\quad F(\alpha_1) \neq F(\alpha_2)  
.\]
If this happens, then \( \alpha_1/ \alpha_2 \in \ZZ_K\units \), and 
\[
F(\alpha_1/ \alpha_2) = F( \alpha_1)- F(\alpha_2) \neq 0
.\]
So we have $F(\ZZ_K\units)\neq 0$, since we've found a unit that maps to a nonzero element.
:::