# Ch. 21: Applications of Minkowski's Theorem (Friday, May 21)

:::{.remark}
Recall that we're considering the Minkowski embedding $\iota: K\to\RR^n$.
We saw that $\iota(\ZZ_K)\subseteq \RR^n$ is a full rank lattice whose covolume is the following:
\[
\covol \iota \ZZ_K = 2^{r_2} \sqrt{ \abs{ \discriminant_K}}
.\]
We wanted to make this work for arbitrary ideals $I\normal \ZZ_K$, and the proposition was that
\[
\covol \iota I = 2^{r_2} \sqrt{ \abs{ \discriminant_K}} N(I)
.\]
:::

:::{.proof title="?"}
We know $I$ is free abelian of rank $n$, so choose a $\ZZ\dash$basis $\elts{\omega}{n}$ for $\ZZ_K$ and $\elts{\theta}{n}$ for $I$.
There is some $A\in \Mat(n\times n; \ZZ)$ such that
\[
\tv{\elts{\theta}{n}} = \tv{\elts{\omega}{n}}A
.\]
The index-determinant theorem tells us that $\# \ZZ_K/I = \abs{\det A}$, and this is equal to $N(I)$.
Now apply $\iota$ to both sides above to obtain $n\times n$ matrix with the $\iota(\theta_i)$ and $\iota(\omega_i)$ as column vectors:
\[
\tv{\elts{ \iota \theta^t }{n} } 
= 
\tv{\elts{\iota \omega^r}{n}}A
.\]
Taking $\abs{\det{\wait}}$ on both sides, we get
\[
\covol \iota(I) 
= \covol \iota(\ZZ_K) \abs{\det A} 
= \covol \iota(\ZZ_K) N(I)
.\]
:::

## Minkowski's Class Group Bound

:::{.remark}
We've proved that $\# \Cl(\ZZ_K) < \infty$ for a general number field, which is in the book and uses Dirichlet's approximation theorem and goes similarly to how it did for quadratic fields.
That proof isn't so good if you want to concretely compute the class group: what you'd like would be an upper bound on the smallest ideal in any ideal class.
This would allow computing all ideals up to that bound and discerning the structure based on these finitely many ideals.
:::

:::{.proposition title="Minkowski's Bound"}
Every ideal class contains a representative $I$ with $N(I) \leq M_K$, where 
\[
M_K \da \qty{4\over \pi}^{r_2} {n! \over n^n} \sqrt{\abs{\discriminant_K}}
.\]
:::

:::{.remark}
The precise form of this bound isn't important yet.
We can prove using this bound that some specific number fields have class number 1.
How do we prove this?
:::

:::{.lemma title="?"}
Let $B>0$, then the following are equivalent: 

1. Every ideal class contains a representative of norm at most $B$,

2. Every nonzero ideal $I\normal \ZZ_K$ contains a nonzero \( \alpha \) with $\abs{N \alpha} \leq N N(I)$.

:::

:::{.proof title="?"}
$2\implies 1$: 
done for quadratic fields.

$1\implies 2$:
Start with $I\normal \ZZ_K$ nonzero.
Pick $J \in [I]\inv$ with $N(J) \leq B$ using (1).
Then $IJ$ is principal, so write $IJ = \gens{ \alpha }$. 
This shows that $I \divides \gens{ \alpha }$, so $\alpha\in I$ since to divide is to contain. 
Moreover
\[
\abs{N ( \alpha)} 
&= N( \gens{ \alpha } ) \\
&= N(I) N(J) \\
&\leq B N(I)
.\]
:::

:::{.remark}
So we've reduced the problem, and it suffices to prove (2) in the above lemma taking $B = M_K$ from above.
Idea: we'll introduce a region $R \subseteq \RR^n$ which is centrally symmetric, convex, and $\vol(R) > 2^n \covol \iota(I)$ big enough.
Then we'll be guaranteed an $\alpha \in I$ nonzero with $\iota( \alpha)\in R$, coming from Minkowski's theorem.

What does this tell us?
The components of $\iota( \alpha)$ are the images of $\alpha$ under embeddings $\sigma_i$.
If we know these images, we can recover the norm as the product, so one ought to be to rig the region $R$ in order to control the size of $\alpha$.
So we'll choose $R$ such that 
\[
\abs{ N( \alpha)} \leq M_K N(I)
.\]
:::

:::{.question}
What kinds of regions $R$ correspond to $\abs{N( \alpha)} \leq X$ for an arbitrary $X$?
:::

:::{.answer}
It's precisely the following region:
\[
\ts{ \tv{x}{r_1 + r_2} \in \RR^{r_1} \oplus \CC^{r_2} \st \prod_{i\leq r_1} \abs{x_i} \prod_{j\leq r_2} \abs{x_{r_1 + j}}^2 \leq X } 
.\]
Note that this is centrally symmetric, but the convexity may be a problem.
:::

:::{.example title="?"}
Let $K$ be a real quadratic field, so $r_1 = 2, r_2 = 0$.
Then the region above is $R_0 = \ts{ (x_1, x_2) \st \abs{x_1 x_2} \leq X }$, which looks like the following:

![image_2021-05-22-17-53-50](figures/image_2021-05-22-17-53-50.png)

The solution: choose a convex region inside $R_0$ and apply Minkowski's theorem, noting that this maintains the bound.
For example, one can take
\[
R = \ts{ (x_1, x_2) \st \abs x_1 + \abs x_2 \leq 2x^{1/2} } 
,\]
which is a diamond lying inside $R_0$ due to the AM-GM inequality.
:::

:::{.proposition title="AM-GM Inequality"}
Let $\elts{t}{m} \geq 0$, then 
\[
{1\over m} \sum_{i\leq m} t_i \geq \qty{ \prod_{i\leq m} t_i }^{1/m}
.\]
:::

:::{.remark}
So we'll take the region
\[
R \da \ts{ \tv{\elts{x}{r_1 + r_2}} \in \RR^{r_1} \oplus \CC^{r_2} \st 
{1\over n} \qty{ \sum_{i\leq r} \abs{x_i} + 2\sum_{r_1 + 1 \leq j \leq r_1 + r_2 } \abs{x_j} } \leq X^{1/n}
} 
.\]
Then by AM-GM, $R \subseteq R_0$ and $n= r_1 + r_2$.
This is still centrally symmetric, and now convex using the triangle inequality.
When is $\vol R > 2^n \covol \iota(I)$?
We'll need to compute the volume of $R$, which is an involved exercises in multivariable calculus.
This is done in the book, and it turns out that
\[
\vol(R) = 2^{r_1} \qty{\pi \over 2}^{r_2} {n^n \over n!} X
.\]
Recall that
\[
2^n \covol \iota(I) = 2^n 2^{-r_2} \sqrt{\abs{\discriminant_K}}
,\]
and solving this linear equality for $X$ yields $X > M_K N(I)$ as defined before.

Now apply Minkowski's theorem: for any $X > M_K N(I)$ there is a nonzero \( \alpha\in I \), there is a nonzero \( \alpha\in I\) with $\iota( \alpha)\in R$, and hence $\abs{N( \alpha)} \leq X$.
Note that the inequalities don't quite match up as-is, since we can't take $X = M_K N(I)$.
Does this imply that we can find an $\alpha$ with $\abs{N( \alpha)} \leq X$?
The answer is yes, because we can choose $X$ with $\floor{X} = \floor{M_K M(I)}$.
Then $\abs{N( \alpha)} \leq X \implies \abs{N( \alpha)} \leq \floor{X} = \floor{M_K N(I)}$, since the left-hand side is an integer.
:::

## Example: Showing Number Fields are PIDs using Dedekind-Kummer and the Minkowski Bound

:::{.example title="?"}
Let $K = \QQ( \sqrt[3]{3})$.
Then $n=3$, and $r_1 = 1, r_2 = 1$ since embeddings permute the roots of $x^3-3$, which has exactly one real root.
We need the discriminant, so we need an integral basis, in which case it helps to know $\ZZ_K$.
By a homework problem, if $K = \QQ(\sqrt[3]{d})$ with $d$ squarefree and $d\not\equiv \pm 1\mod 9$, then $\ZZ_K$ is what you guess it'd be!
So here $\ZZ_K = \ZZ[\sqrt[3]{3}]$, and thus
\[
\discriminant_K = \Delta(1, 3^{1/3}, 3^{2/3})
= \discriminant (x^2 - 3) = -3^5
.\]

Thus
\[
M_K = \qty{4\over \pi} {3! \over 3^3} \sqrt{3^5} \approx 4.42
,\]
and thus we know every ideal class contains a representative of norm at most 4, using that $N(I)$ is an integer.
Any such $I$ is a product of ideals of norms $2,3,4$.
Prime ideals of norm 2 or 4 lie above the prime 2, and of norm 3 lie above 3.
If we can show that every prime ideal with $N(I) = 2,3,4$ is principal, then $I$ will be a product of principal ideals and thus principal.
Then since every class contains such an $I$, $\Cl(\ZZ_K)$ is trivial.
We look at all of the primes above 2 and 3 using Dedekind-Kummer, which says the factorization of \( \gens{ p }  \) mirrors the factorization of $x^2 - 3 \mod p$.

Note that if $\ZZ(\sqrt[3]{3}) \neq \ZZ_K$, there'd be some exceptional primes $p$ to worry about, but since these are equal here this is literally true for all $p$.

So we factor polynomials.
First mod 2:
\[
x^2 - 3 = x^3 - 1 = (x-1) (x^2 + x + 1)
,\]

where the second term is an irreducible quadratic with no roots in $\FF_2$.
This yields
\[
\gens{ 2 }= \gens{ 2, \sqrt[3]{3} - 1 } \gens{ 2, \qty{\sqrt[3]{3} }^2 + \sqrt[3]{3} + 1 }   
.\]
Next mod 2:
\[
x^3 -3 = x^3
,\]
which yields
\[
\gens{ 3 }= \gens{ 3, \sqrt[3]{3} }^3  
.\]

Although we have explicit factorizations, but it may not obvious whether or not they're principal.
It's easy to see that \( \gens{ 3, 3^{1/3} }  \) is principal since $3^{1/3} \divides 3$ and the first generator is redundant
For $\gens{2}$, it's less clear, but
\[
(3^{1/3} - 1) (3^{2/3} + 3^{1/3} + 1) = 2
,\]
and so the 2s are redundant generators in both terms. 
So every ideal above 2 and 3 is principal, so $I$ is principal.
Every ideal class contains such an $I$, so $\Cl(\ZZ_K) = 0$ and $\ZZ_K$ is a PID.
:::

:::{.remark}
Asking whether or not this ring is a PID is an undergraduate-level question, but it's not clear how one would determine this without the theory developed in this class.
:::

:::{.example title="?"}
Let $K = \QQ(\theta)$ where $\theta$ is a root of $f(x) = x^5-x^3 + 1$.
This is irreducible over $\QQ$ and has one real root.
Then $n=5, r_1 = 1, r_2 = 2$, but we need the discriminant to apply the Minkowski bound.
If we could prove $\ZZ_K = \ZZ(\theta)$, $\discriminant_K = \discriminant(f)$.
If it were squarefree, it'd correspond to an integral basis, so we compute
\[
\discriminant(f) = \discriminant(1, \theta, \theta^2, \theta^3, \theta^4) =3017
,\]
using determinant formulas in the book.
This is squarefree, so the power basis is an integral basis, so $\ZZ_K = \ZZ[\theta]$ and $\discriminant_K = 3017$.
Then
\[
M_k = \qty{4\over \pi}^2 {5! \over 5^5} \sqrt{3017} \approx 3.41
.\]
So every ideal class is represented by an ideal of norm at most $3$.

- Norm 1: Unit ideal and thus principal,
- Norm 2: Primes above 2
- Norm 3: Primes above 3

We'll apply Dedekind-Kummer.
First mod 2: $f(x)$ is irreducible mod 2, so $\gens{2}$ is inert and prime.
So there are no ideals of norm 2, since it would have to factor as a product of primes above 2, but the only possible factor is 2 which has norm $2^5 =32$.

Mod 3:
$f$ is no longer irreducible, but factors as $f=pq$ with $\deg p = 2,\deg q = 3$.
So there are no ideals of norm 3, since all prime ideals above 3 would have to have norms of $3^2$ corresponding to $p$ or $3^3$ corresponding to $q$.
Thus $\Cl(\ZZ_K) = 0$, since every $I$ is represented by an ideal of norm 1 and hence is principal.
:::

:::{.remark}
Next up we'll talk about a lower bound for $\abs{\discriminant_K}$
Obviously $M_K \geq 1$, since the theorem states that every ideal class has a representative of norm at most $M_K$, and $M_K< 1$ wouldn't make sense.
Using the formula, rearranging yields a bound
\[
\abs{\discriminant_K} \geq \qty{n^n \over n!}^2 \qty{\pi \over 4}^{2r_2}
.\]

:::