# Units in $\ZZ_K$ (Lec. 9, Monday, February 15)

## Review

:::{.remark}
Today: chapter 8.
We'll continue with the statements from last time:
:::

:::{.proposition title="Subgroups of $\RR$ are either discrete or infinite cyclic"}
A discrete subgroup \( \Lambda \leq \RR \) is either $0$ or infinite cyclic, where *discrete* means having finite intersection with every interval $[-x, x]$.
:::

:::{.proof title="of proposition"}
Suppose \( \Lambda \neq 0 \), then we can choose a smallest positive element \( \alpha\in \Lambda \).
Why does this exist? 
There are only finitely many elements in $[0, \alpha]$, so there is a smallest, and we could replace \( \alpha \) with it.
The claim is that \( \Lambda = \ZZ \alpha \).
The reverse containment is clear because the RHS is necessarily a subgroup.
Toward a contradiction, suppose there is some \( \beta\in \Lambda\sm \ZZ \alpha \) with \( n \alpha < \beta < (n+1) \alpha \) for some $n\in \ZZ$.
This can't happen: subtracting $n$ from both sides yields 
\[
0 < \beta - n \alpha < \alpha
,\]
where the middle term is necessarily in \( \Lambda \), contradicting minimality of \( \alpha \).

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:::

:::{.remark}
Recall that to show the theorem we wanted, it was enough to show $\log U(\ZZ_K)^+$ is an infinite cyclic subgroup of $\GG_a(\RR)$.
We proved that this was a discrete subgroup.
If this were just the zero element, the only possible units would be $\pm 1$, so it suffices to find a unit \( \epsilon \in U(\ZZ_K) \) with \( \epsilon>0 \) and \( \epsilon\neq 1 \).
:::

## An Aside: Diophantine approximation

:::{.remark}
Let \( \alpha\in \RR \) and let \( Q \in \ZZ^+ \).
How well can we approximate \( \alpha \) with a fraction with denominator bounded by $Q$?
:::

:::{.theorem title="Dirichlet's Approximation Theorem"}
There is a $q \leq Q \in \ZZ^+$ with 
\[ 
\norm{q \alpha } \leq {1 \over Q+1} 
,\] 
where \( \norm{\wait} \) denotes the distance to the nearest integer.
:::

:::{.remark}
The way to think about this inequality: if the LHS is close to an integer $p$, then \( \alpha \) is close to \( p/q \).

:::

:::{.proof title="of Dirichlet's theorem"}
Chop the interval into $Q+1$ pieces, and think of the inequality as a condition on the fractional part of \( \alpha \), denoted \( \fractional{ qa } \da qa - \floor{ qa }\in [0, 1) \).
Note that if \( \fractional{ qa } \in [0, 1/Q+1)\) or \( [Q/Q+1, Q) \) for some $q$, then we are done.
If not, it must land in one of the $q-1$ middle intervals 
\[
[1/Q+1, 2/Q+1), 
[2/Q+1, 3/Q+1), 
\cdots,
[Q-1/Q+1, Q/Q+1)
\]
for all all $q\leq Q$.
But we have $Q$ choices for $q$ and only $Q-1$ intervals, so there are two values of $q$ with fractional part in the same interval.
So choose these, say $q_1<q_2 \leq Q$, and consider $q \da q_2 - q_1$.
Since \( \fractional{q_1 \alpha}, \fractional{q_2 \alpha} \) are in the same interval, we have \( \fractional{q \alpha} \in [0, 1/Q+1) \), putting it close to an integer.
:::

:::{.corollary title="Infinitude of elements of bounded norm"}
There are infinitely many pairs of positive integers $(p, q)$ such that 
\[
\abs{p^2 - dq^2}\leq 1 + 2 \sqrt{d}  
,\]
where $d$ was the squarefree integer for which $K = \QQ( \sqrt{d} )$.
:::

:::{.remark}
Note that the RHS does not depend on $p$ or $q$, and only depends on the field.
Moreover, this proof is also true with the 1 removed.
:::

:::{.proof title="of corollary"}
Using Dirichlet's approximation theorem, choose \( Q \in \ZZ^+ \) and \( 1\leq q\leq Q \) 
such that 
\[
\norm{ q \sqrt{d} } \leq {1 \over Q+1}
,\]
then there is a \( p \in \ZZ \) such that
\[
\abs{ p - q \sqrt{d} } \leq {1 \over Q+1}
.\]
We know $q$ is positive by Dirichlet's theorem, and $p$ is positive since $q\sqrt{d} \geq \sqrt{d} \geq 1$, and the distance from $p$ to $q$ is at most $1/2$.
We can now check
\[
\abs{p^2 - dq^2} 
&= \abs{p - q \sqrt{d}} \abs{p + q \sqrt{d}}    \\
&= \abs{p - q \sqrt{d}} \abs{(p- q \sqrt{d} ) + 2q \sqrt{d} }    \\
&\leq \abs{p - q \sqrt{d}} \abs{p- q \sqrt{d}} + \abs{2q \sqrt{d} }    \\
&\leq {1\over Q+1} \qty{ {1\over Q+1} + 2Q \sqrt{d} }\\
&= \qty{1\over Q+1}^2 + {2Q \over Q+1} \sqrt{d} \\
&< 1 + 2\sqrt{d}
,\]
where we've applied the triangle inequality and used the bound twice.
How do we know that this results in infinitely many distinct pairs?
Things could also go wrong if the same pairs resulted from all but finitely many choices of $Q$.
However, the bound from Dirichlet's theorem prevents this: any pair $(p, q)$ can arise for ay most *finitely* many starting values for $Q$.
Pick a $Q$, then produce $q$ satisfying the bound.
Then $\norm{ q \sqrt{d} } \neq 0$ since \( \sqrt{d} \) is irrational, and thus the LHS is some positive irrational number.
For a fixed $q$, choosing $Q'$ big enough can make the RHS smaller than the LHS, meaning that $q$ can not occur for that value of $Q'$ or anything larger.
In other words, we're using
\[
\norm{ q \sqrt{d} } \leq {1\over Q+1} \converges{Q \to \infty } \to 0
,\]
and there can't be any infinite sequences of $Q_i$ yielding the same fixed $q$, since the RHS would go to zero while the LHS does not.
:::

:::{.remark}
Choosing a pair $(p, q)$ as above, we'll have $p + q \sqrt{d} \in \ZZ_K$ and
\[
\abs{ N( p + q \sqrt{d} ) } 
= \abs{ p^2 - d q^2 } \\
< 1 + 2 \sqrt{d} 
.\]
So we have many elements in $\ZZ_K$ whose norm is bounded, which will force the existence of a nontrivial unit.
:::

:::{.lemma title="Finitely many ideals of bounded norm"}
For all real $x> 0$ there are finitely many nonzero ideals $I\normal \ZZ_K$ with $N(I) \da \abs{ \ZZ_K/I } \leq x$.
:::

:::{.proof title="of lemma"}
Suppose $N(I) \da m \leq x$ with \( m \in \ZZ^+ \); it's enough to show that for each $m$ there are at most finitely many $I$, since there are only finitely many values of $m \leq x$.
View $\ZZ_K / I$ as a group under addition, so by Lagrange every element has order dividing $m$.
We can check $m = 1 + 1 + \cdots + 1$, which must be the identity in $\ZZ_K/I$.
So $m\in I$, and since to contain is to divide, $I \divides \gens{ m }$. 
But \( \gens{ m }  \) has only finitely many ideal divisors.
Why? 
This is because there is unique prime factorization, and just like $n = \prod p_i^{n_i}$ in the integers, $n$ has $\sum n_i < \infty$ possible divisors.
:::

:::{.proof title="There exists a nontrivial unit"}
We now want to show that there exists a unit \( \epsilon>0 \) that is not equal to 1.
Consider all ideals \( I_{p, q} \da \gens{ p + q \sqrt{d} }  \)  where \( (p, q) \) is a pair of positive integers such that \( \abs{p^2 - dq^2} < 1 + 2 \sqrt{d}  \).
Taking norms amounts to taking absolute values of generators, so 
\[
N(I_{p, q} ) < 1 + 2 \sqrt{d}
\]
for all $p, q$.
By the last lemma, this means there are only finitely many different ideals.
On the other hand, there are infinitely many such pairs, so infinitely many pairs give rise to the same ideal.
Pick two pairs $(p, q)$ and $(p', q')$ such that \( \gens{ p + q \sqrt{d} } = \gens{ p' + q' \sqrt{d} } \).
If two ideals are equal, the generators differ by a unit, and so 
\[
(p + q \sqrt{d} ) = \eps (p' + q' \sqrt{d} ), && \eps \in U(\ZZ_K)
.\]
Everything in sight is positive, so solving for \( \eps \) yields \( \eps > 0 \).
But \( \eps \neq 1 \), since the pairs would have to have been the same by comparing coefficients in the expression above.
:::

:::{.remark}
This gives us the fundamental unit.
How do we actually find it?
See the book -- use continued fractions!
It's not surprising they'd come up, since they provide a more constructive proof of Dirichlet's approximation theorem.
:::

:::{.example title="of the fundamental unit"}
Take $d=2$, what is \( \eps_0 \)?
We have \( U(\ZZ_K) = \ts{ \pm \eps_0 ^k \st k\in \ZZ  } \), and so if we just look at positive units, the smallest power such that \( \eps_0^k > 1 \) will just be equal to \( \eps_0 \).
So we're really looking for the smallest unit greater than 1.
We proved that if \( \eps_0 = u + v \sqrt{d} \), then $u, v \geq 0$, and if \( \eps_0 > 1 \) is strict then $u, v > 0$ is strict as well.
We also know that $u, v \geq 1$, using that $\ZZ_K = \ZZ[\sqrt{2} ]$.
Luckily enough, $1 + \sqrt{2}$ is a unit, and so $\eps_0 = 1 + \sqrt{2}$.
:::

## Class Groups and the Class Number

:::{.remark}
This is now chapter 9.
Let $K$ be a quadratic field.
:::

:::{.definition title="Dilation Equivalence"}
If $I, J$ are nonzero ideals of $\ZZ_K$, we say $I, J$ are **dilation equivalent**
if there exists a \( \lambda\in K\units \) such that $I = \lambda J$.
:::

:::{.remark}
It's easy to check that this is an equivalence relation, so we'll use $I \approx J$.
:::

:::{.definition title="Class Group"}
The **class group** of $\ZZ_K$ is defined as 
\[
\Cl(\ZZ_K) \da \Id(\ZZ_K)/\approx
.\]

:::

:::{.remark}
A priori this is just a set, but we can descent the monoid structure to define a group multiplication.
We define $[I] [J] = [IJ]$, and it's easy to check that this is well-defined on equivalence classes.
The identity is \( [ \gens{ 1 }]  \), and for inverses we can use the fact that \( [I \bar{I}] = [\gens{ N(I) }] = N(I) [ \gens{ 1 } ]  \).
In fact, any $J$ for which $IJ$ is principal serves as an inverse for $I$.
So the inverses come from the *Principal Multiple Lemma*, and a similar story will go through for general number fields.
:::

:::{.remark}
This is an abelian group, wouldn't it be nice if it were finite?
This is one of the big theorems of number theory: $\Cl(\ZZ_K)$ is finite.
We can thus define the following:
:::

:::{.definition title="Class Number"}
The **class number** of $K$ is defined as:
\[
h_k \da \abs{ \Cl(\ZZ_K) }
.\]

:::

:::{.lemma title="Comparison bound between element norm and ideal norm"}
There is a constant $C$ depending on $K$ such that for every $I \in \Id(\ZZ_K)$ there is a nonzero \( \alpha\in I \) such that 
\[
\abs{N \alpha}\leq C N(I) 
.\]
In fact, one can take 
\[
C \da 1 + \Tr(\tau) + \abs{N(\tau)} 
.\]
:::

:::{.remark}
The norm of $I$ is a natural thing to compare $N \alpha$ to, since $I \divides \gens{ \alpha }$ and thus $N(I) \divides N(\gens{ \alpha })$, so there's no hope of the LHS being smaller than $N(I)$.
:::

:::{.proof title="?"}
Look at all elements \( a + b \tau \in \ZZ_K \) such that $0 \leq a,b, \sqrt{ N(I) }$.
How many elements does this yield?
Precisely $\qty{ \floor{ \sqrt{N(I)} } +1 }^2$.
Note that this is strictly larger than $N(I)$, using \( \floor{x} > x-1 \) for any $x$.
Then going to the quotient by $I$, there are exactly $N(I)$ elements, two elements reduce to the same element of $\ZZ_K/I$.
So their difference is in $I$, so we get something of the form \( a' + b' \tau \) where \( a;, b; \in \ZZ \) (where they could now be negative), but are bounded by 
\[
- \sqrt{N(I)}
\leq a', b' \leq 
 \sqrt{N(I)} 
.\]

The claim is now that the given value of $C$ in the theorem works:
\[
\abs{N( \alpha)}
&=
\abs{N( a' + b' \tau)} \\
&=
\abs{(a' + b' \tau) (a' + b' \bar \tau)} \\
&= \abs{(a')^2 + a'b' \Tr(\tau) + (b')^2 N \tau} \\
&\leq 
\abs{a'}^2 + \abs{a'} \abs{b'} \Tr(\tau) + \abs{b'}^2 N(\tau) \\
&\leq C N(I)
,\]
where we've used \( a', b' \leq \sqrt{N(I)} \) and collected terms in the last step.
:::

:::{.proposition title="Class representatives of small norm"}
Every ideal class contains an ideal $I$ of norm $N(I) \leq C$.
:::

:::{.corollary title="Class numbers are finite"}
$h_K < \infty$.
:::

:::{.remark}
Why is this true?
There are only finitely many ideals with this norm bound, and this says every ideal class belongs to this finite set.
:::

:::{.proof title="of proposition"}
Since we're working with a group, it suffices to work with inverses, since these still run over all elements.
It's enough to show that for every $I \in \Id(\ZZ_K)$, we can write $[I]\inv = [J]$ for some $J$ satisfying $N(J) \leq C$.
Choose a nonzero \( \alpha\in I \) with \( \abs{N( \alpha )}\leq C N(I)  \).
Since \( \alpha\in I \) we know that \( I \divides \gens{ \alpha }  \), so we can write \( \gens{ \alpha } = IJ  \) for some ideal $J$.
We have \( [I] [J] = [IJ] = [ \gens{ \alpha } ] = [ \gens{ 1 } ] \), since all principal ideals are dilation-equivalent to \( \gens{ 1 }  \).
This means that $[J] = [I] \inv$, and our hope is that it has small norm.
Taking norms in \( \gens{ \alpha } = IJ  \) yields
\[
\abs{N \alpha} 
&= N(I) N(J) \\
\implies N(J) 
&= { \abs{N \alpha} \over N(I) } \\
&\leq { C N(I) \over N(I)} \\
&= C
.\]
:::

:::{.example title="?"}
What we'll look at next: $\Cl( \ZZ [ \sqrt{-5} ])$.
We know this does not have unique factorization, and the claim is that the class group is nontrivial.
If it were, every ideal would be dilation-equivalent to \( \gens{ 1 }  \), making every ideal principal, and every PID is a UFD.
Here we'll have $C=6$.

\

One could try to write down all ideals of norm bounded by 6, but instead lets consider how they factor into primes.
Every ideal of norm at most 6 factors into prime ideals, whose norm is also bounded by 6.
So this factors into prime ideals lying above $2,3,5$, since any ideal lying above a prime $p$ has norm $p$ or $p^2$, and we need $p, p^2 < 6$ here.
We've worked out all such primes before, coming from the *prime factor mirroring theorem*:

- \( \gens{ 2 }= P_1^2,\, P_1 \da \gens{ 2, 1 + \sqrt{-5} } \) 
- \( \gens{ 3 } = P_2 P_3, P_2 \da \gens{ 3, 1 - \sqrt{-5} }, P_3 \da \gens{ 1 + \sqrt{-5} }  \) 
- \( \gens{ 5 }= P_4^2, P_4 \da \gens{ \sqrt{-5} }   \)  

This allows us to conclude that 
\[
\Cl(\ZZ[ \sqrt{-5} ]) = \gens{ [P_1], [P_2], [P_3], [P_4] }
.\]
In fact, since $P_4$ is principal we can leave it out.

:::