# Ideal Theory in Number Fields Continued (Lec. 16, Tuesday, March 30)

## Setting up the Theory

:::{.remark}
We want to develop theorems of ideal theory for $\ZZ_K$ for $K$ a general number field, i.e. factorization into prime ideals and the finiteness of the class group.
The strategy:

- Prove $\Cl(\ZZ_K)$ is a finite monoid,
- Prove $\Cl(\ZZ_K)$ has inverses and is thus a group, i.e. every nonzero ideal can be multiplied by another ideal to become principal (principal multiple lemma),
- Run proofs/corollaries as before.

Last time, we proved the first one.
:::

:::{.lemma title="When ideals are left identities under multiplication"}
Let $I, J \in \Id(\ZZ_K)$, then if $IJ = J$ then $I = \gens{ 1 }$. 
:::

:::{.remark}
Note that this is a special case of cancellation.
To prove this, we'll use that $\ZZ_K$ is Noetherian, i.e. every ideal is finitely generated as a \(\ZZ_K\dash\)module.
In fact, $\ZZ_K \cong \ZZ^n$, so any ideal is free of rank $\leq n$ as a \(\ZZ\dash\)module, hence finitely generated as a \(\ZZ\dash\)module, hence finitely-generated as a $\ZZ_K\dash$module since one can use the same generators.
:::

:::{.proof title="of lemma"}
Let $J = \gens{ \beta_1, \cdots, \beta_n }$, then since $IJ = J$, for every $j$ we can write \( \beta_j = \sum_{i=1}^m A_{ij} \beta_i  \).
This means that there is some matrix $A\in \Mat(m\times m, I)$ with entries $A_{ij} \in I$ such that
\[
\tv{\beta_1, \cdots, \beta_m}
=
\tv{\beta_1, \cdots, \beta_m}A
.\]
Then $A-\one \vector{\beta} = 0$, making $A-\one$ singular since not all of the $\beta_i$ were zero since they were generators of a nonzero ideal.
Now take the determinant mod $I$, which yields
\[
0 \equiv \det(A-\one) \equiv \det(-\one) \equiv \pm 1\mod I
,\]
but this can only occur if $1\in I$, making \( \gens{ 1 }= I  \).
:::

:::{.lemma title="Right-cancellation when principal ideals are involved"}
Let $I, J\normal \ZZ_K$, then if $IJ = \beta J$ with $\beta \in \ZZ_K \smz$, we have $I = \gens{ \beta }$. 
:::

:::{.remark}
Note that the previous lemma is a special case of this where $\beta = 1$.
One can then bootstrap the previous lemma to get this, see the book.
:::

:::{.lemma title="Principal Multiple Lemma"}
For all $I\in \Id(\ZZ_K)$ there is a $J \in \Id(\ZZ_K)$ such that $IJ$ is principal.
:::

:::{.proof title="of principal multiple lemma"}
Consider $[I], [I]^2, \cdots \in \Cl(\ZZ_K)$.
By the pigeonhole principal, there is some $k, \ell$ such that $[I^k] = [I]^\ell$, so $I^k = \lambda I^\ell$ for some $\lambda \in K\units$.
Note that any nonzero element of $K$ can be written as $k/n$ for $k\in K$ and $n\in \ZZ_K$.
So we can scale $\lambda$ to put it in $\ZZ_K$, yielding $\lambda = \alpha/m$ where \( \alpha\in \ZZ_K \) and $m\in \ZZ\units$.
We then have
\[
mI^k = \alpha I^\ell = (\alpha I^{\ell-k})I^k
.\]
We have enough to cancel the $I^k$, and so \( \gens{ m }= \alpha I^{\ell - k}  \).
Dilating both sides by \( \alpha\inv \) yields 
\( \gens{ m/\alpha } = I^{\ell-k}  \).
But this is a power of $I$ that is principal, so we can take $J \da I^{\ell-k-1}$.
:::

:::{.remark}
Note that the logical order in which these theorems are proved is slightly reversed.
:::

:::{.corollary title="Class groups are finite and $\Id$ is a cancellative monoid"}
\envlist

a. $\Cl(\ZZ_K)$ is a group, and thus a finite abelian group.

b. $\Id(\ZZ_K)$ is cancellative.
  Just show one can cancel principal ideals (by dilation), and then in general you cancel by multiplying both sides principal and cancelling that principal ideal.

To show unique factorization, we before showed factorization into irreducibles first, then uniqueness as a consequence of Euclid's lemma.

:::

:::{.lemma title="The monoid $\Id$ is atomic"}
$\Id(\ZZ_K)$ is atomic, i.e. every element factors into irreducibles.
:::

:::{.proof title="of lemma"}
We'll proceed by induction on $N(I)$, using that $N(AB) \leq N(A)$ for any $A$ and so $I\divides J \implies N(I) < N(J)$.
Before we used that $N(AB) = N(A) N(B)$, but we haven't proved that here yet.
We also don't know that "to divide is to contain" here, but since $I\divides J$ and $I\neq J$, we do obtain $J \subsetneq I$.
Hence there is a surjection
\[
\ZZ_K/J \to \ZZ_K/I
.\]
This has nontrivial kernel since $I\sm J \neq \emptyset$, so \( \abs{\ZZ_K/J}> \abs{\ZZ_K/I} \).
:::

:::{.lemma title="Analog of Euclid's Lemma"}
Irreducibles in $\Id(\ZZ_K)$ are prime.
:::

:::{.proof title="of lemma"}
Same as before!
Literally use the exact same words, we've set it up this way.
:::

:::{.theorem title="The monoid $\Id$ is a unique factorization monoid"}
$\Id(\ZZ_K)$ is a UFM, or equivalently every nonzero ideal factors uniquely as a product of prime ideals.
:::

## Modern Approach

:::{.question}
What is the widest class of domains for which the previous theorem holds?
:::

:::{.definition title="Dedekind Domains"}
Let $R$ be a domain that is not a field (since ideals in fields are uninteresting).
Then $R$ is a **Dedekind domain** if and only if 

a. $R$ is Noetherian,

b. $R$ is integrally closed,
  so if $K = \ff(R)$, then if $\alpha\in K$ is a root of a monic polynomial in $R[x]$ we have \( \alpha\in R \).[^rational_root_comparison]

[^rational_root_comparison]: 
Compare to the classical rational root theorem.

c. Every nonzero prime ideal is maximal.

:::

:::{.theorem title="Noether"}
TFAE:

1. $R$ is a Dedekind domain,

2. Every nonzero ideal of $R$ factors into prime ideals (not necessarily uniquely).

3. (2) along with uniqueness.
:::

:::{.proof title="of Noether's theorem"}
Omitted, this is an exercise in commutative algebra.
:::

:::{.proposition title="Rings of integers are Dedekind domains"}
For any number field $K$, $\ZZ_K$ is a Dedekind domain.
:::

:::{.proof title="of proposition"}
We can check the definitions directly:

a. This is a consequence of the integral basis theorem.

b. Suppose \( \alpha \in K \) and is a root of a monic polynomial in $\ZZ_K[x]$, we then want to show \( \alpha\in \ZZ_K \).
  Then \( \alpha \) is a root of a monic polynomial in $\bar{\ZZ}[x]$, and by a previous proof, any monic polynomial with $\bar{\ZZ}$ coefficients is itself in $\bar{\ZZ}$.
  Since \( \alpha \in K \) as well, we have \( \alpha\in \bar{\ZZ_K} \intersect K \da \ZZ_K \).

c. Let $P \normal \ZZ_K$ be nonzero.
  Then $\ZZ_K/P$ is a domain, but any finite integral domain is a field and we know this is finite since $N(P) < \infty$.
:::

## Norms Revisited

:::{.remark}
We left one theorem hanging when we discussed norms.
We proved that norms of ideals are finite, and $N(P)$ for $P$ principal is equal to $N(a)$ for $a$ any generator.
We haven't yet proved the following:
:::

:::{.theorem title="The norm is multiplicative"}
\[
N(IJ) = N(I) N(J) && \forall I, J \in \Id(\ZZ_K)
.\]
:::

:::{.remark}
If $I, J\in \Id(\ZZ_K)$, we have $\gcd(I, J) = I+J$, since this is the smallest ideal such that any $P\divides I, P\divides J$ must satisfy $P\divides \gcd(I, J)$.
:::

:::{.proof title="that the norm is multiplicative"}
It's enough to show $N(IP) = N(I) N(P)$ for $P$ prime, since every $J$ factors into primes and we can apply this result recursively.
Now
\[
N(IP) 
&= [\ZZ_K, IP] \\
&= [\ZZ_K: I] [I : IP] \\
&= N(I) [I: IP]
,\]
so it suffices to show $N(P) = [I: IP]$.

:::{.claim}
This is true because $\ZZ_K \cong I/IP$ are isomorphic as $\ZZ\dash$modules. 
:::

Choose \( \beta \in I\sm IP \), using that $I$ *properly* divides $IP$ when it is properly contained.
Define a map 
\[
\psi: \ZZ_K/p &\to I/IP \\
\alpha \mod p  &\mapsto \alpha \beta \mod IP
.\]
This is well-defined since for any two elements which differ by a multiple of $P$, multiplying by $\beta \in I$ lands in $IP$.

:::{.exercise title="?"}
Check that this is a well-defined group morphism.
:::

\

**Injectivity**:
Suppose that \( \alpha\mod P \in \ker \psi \), so \( \alpha \beta\in IP \) and $IP \divides \gens{ \alpha } \gens{ \beta }$.
Note that without the \( \alpha \) this would be false, so we're critically using that \( \beta \) is in $I$ but not $IP$: $IP\not\divides \gens{ \beta }$ since \( \beta \not\in IP \). 
So $IP$ divides this product but not $\beta$ while $I$ *does* divide \( \beta \), this forces $P\divides m \gens{ \alpha }$.
Then \( \alpha\in P \) and \( \alpha\mod P = 0 \), so $\ker \psi = 0$.

\

**Surjectivity:**
We might want to write \( \Im(\psi) = \gens{ \beta } / IP  \), but this doesn't quite make sense since $IP$ may not be a subgroup.
This can be fixed, $\im \psi = ( \gens{ \beta } + IP) / IP$. 
But this equals \( \gcd( \gens{ \beta }, IP ) / IP  \), and this numerator is $I$ since \( \beta\in I \) and \( \beta\not\in IP \).
So we have 
\[
\im(\psi) 
&= {\gens{ \beta } + IP \over  IP } \\
&= { \gcd( \gens{ \beta }, IP ) \over IP } \\
&= I/IP
.\]

:::

:::{.remark}
For quadratic fields, we could compute ideal norms by multiplying an ideal $I$ by its conjugate $\bar I$ to get a principal ideal generated by $N(I)$.
Here we don't know what conjugates mean yet for a general number field -- one could try applying all of the embeddings into $\CC$ and taking a product, but this may not yield an ideal in the same ring again.
In particular, if $K$ isn't Galois, the embedding can land outside of $K$ in $\CC$.
:::

:::{.definition title="Extending Ideals"}
For $R \subseteq S$ and $I\normal R$, define $IS$ to be the smallest ideal of $S$ containing $I$ (i.e. take all intersections), or equivalently take all finite $S\dash$linear combinations of elements from $I$.
:::

:::{.exercise title="Arithmetic of ideals"}
Check that 

- $(IJ)S = (IS)(JS)$,
- $(\alpha R)S = \alpha S$.
:::

:::{.theorem title="Norm is generated by product of conjugates"}
Let $I\in \Id(\ZZ_K)$ and let $L$ be the Galois closure of $K/\QQ$.
For each \( \sigma: K\embeds \CC \), the image \( \sigma(I) \) is an ideal of $\ZZ_{\sigma(K)} \subseteq \ZZ_L$.
Then 
\[
\prod_{\sigma: K\embeds \CC} \sigma(I) \ZZ_L = N(I) \ZZ_L
.\]
:::

:::{.remark}
This shows why the norm is multiplicative, and why \( N( \gens{ \alpha }) = |N( \alpha ) |  \).
:::

## Applications of Finiteness of Class Group

:::{.question}
Where do ideals come from?
:::

:::{.remark}
They're meant to generalize multiples of an integer in $\ZZ$, but not all ideals in a general number field are principal.
However, there is a way in which this is true for $\ZZ_K$ even when it's not a PID.
:::

:::{.theorem title="Dedekind's theorem on the actuality of ideals"}
Let $K$ be a number field and $I\normal \Id(\ZZ_K)$.
Then there is a \( \beta\in \bar{\ZZ} \) such that $I = \beta \bar{\ZZ} \intersect K$, or equivalently \( I = \beta \bar{\ZZ} \intersect \ZZ_K \).
:::

:::{.remark}
Example of a non-principal ideal: in $\ZZ[\sqrt{-5}]$, the ideal \( I \da \gens{ 2, 1 + \sqrt{-5} }  \) is not principal, i.e. not all such elements are given by multiples of some element in $\ZZ[ \sqrt{-5} ]$.
It turns out that instead this is all multiples (in $\bar{\ZZ})$ of $\sqrt{2}$.
So anything that's a multiple of $\sqrt{2}$ *and* an algebraic integer that's in $\ZZ[\sqrt{-5}]$ will be in $I$ and vice-versa.
So ideals are multiples of a single element, provided you allow that element to be outside of $\ZZ_K$ and in $\bar{\ZZ}$ instead.
:::

:::{.lemma title="Ideals become principal after extending"}
Let $K$ be a number field and $I\in \Id(\ZZ_K)$.
Then there is a finite extension $L/K$ in which $I\ZZ_L$ is principal.
So any ideal can be made principal after passing to some finite extension.
:::

:::{.proof title="of lemma"}
Let $m \da \# \Cl(\ZZ_K)$.
Then $I^m = \alpha\ZZ_K$ is principal since $m$ is the order of this group.
Let \( \beta\da \sqrt[m]{\alpha} \in \CC \) and let $L \da K( \beta)$.
Here \( \beta \) is an algebraic integer since it's an $m$th root of an algebraic integer.
The claim is that $I\ZZ_L$ is principal.
We have 
\[
(I \ZZ_L)^m
&= I^m \ZZ_L \\
&= ( \alpha\ZZ_K) \ZZ_L \\
&= \alpha\ZZ_L \\
&= \beta^m \ZZ_L \\
&= (\beta \ZZ_L)^m
.\]
But how can two ideals have the same $m$th power?
By unique factorization, they must be the same, so $I\ZZ_L = \beta \ZZ_L$.

> To be continued.

:::