# Ch. 16, Continued (Thursday, May 13)

## Actuality of Ideals

:::{.theorem title="Dedekind, Actuality of ideals"}
If $I$ is a nonzero ideal of $\ZZ_K$ for $K$ a number field, then there is an $\alpha\in \bar{\ZZ}$ such that 
\[
I = (\alpha \bar\ZZ) \intersect \ZZ_K
.\]

:::

:::{.lemma title="?"}
Let $I$ be a nonzero ideal of $\ZZ_K$, then there is an extension of number fields $L/K$ such that $I \ZZ_L$ is principal ,say $I \ZZ_L = \alpha\ZZ_L$.
:::

:::{.proof title="Idea"}
We've already proved this, but the idea was that we can take $m$ to be the order of the class group to obtain $I^m = \beta \ZZ_K$.
Then if $L = K( \beta^{1/m} )$ we will have $I\ZZ_L = \beta^{1/m} \ZZ_L$.
:::

:::{.remark}
How we'll use this to prove the theorem: the lemma shows that after passing to a suitable extension we can find $\alpha$, and the claim is that this $\alpha$ works.
We'll need one more result:
:::

:::{.lemma title="Paul's up-down lemma"}
Let $K$ be a number field and $I \normal \ZZ_K$ be a nonzero ideal.
Let $R \leq \bar\ZZ$ be a subring containing $\ZZ_K$, then 
\[
IR \intersect \ZZ_K = I \normal \ZZ_K
.\]

The picture: we can lift ideals from $\ZZ_K$ to $R$ and intersect them with $\ZZ_K$ to come back down, and the claim is that this lands where you started:

\begin{tikzcd}
	{\bar{\ZZ}} \\
	\\
	R && IR \\
	\\
	{\ZZ_K} && I && {IR \intersect \ZZ_K}
	\arrow[hook, from=5-1, to=3-1]
	\arrow[hook, from=3-1, to=1-1]
	\arrow["{(\wait) \cdot R}", from=5-3, to=3-3]
	\arrow["{\wait \intersect \ZZ_K}", from=3-3, to=5-5]
	\arrow[color={rgb,255:red,92;green,214;blue,214}, Rightarrow, no head, from=5-3, to=5-5]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMCw0LCJcXFpaX0siXSxbMCwyLCJSIl0sWzAsMCwiXFxiYXJ7XFxaWn0iXSxbMiw0LCJJIl0sWzIsMiwiSVIiXSxbNCw0LCJJUiBcXGludGVyc2VjdCBcXFpaX0siXSxbMCwxLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFsxLDIsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzMsNCwiKFxcd2FpdCkgXFxjZG90IFIiXSxbNCw1LCJcXHdhaXQgXFxpbnRlcnNlY3QgXFxaWl9LIl0sWzMsNSwiIiwyLHsibGV2ZWwiOjIsImNvbG91ciI6WzE4MCw2MCw2MF0sInN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XV0=)

:::

:::{.proof title="?"}
Follow your nose!
The containment $I \subseteq IR \intersect \ZZ_K$ is clear, since $I$ is contained in both terms.
For the reverse containment, we'll first do it for $I$ principal, and then reduce to that case.
So suppose $I = \alpha \ZZ_K$, then $IR = \alpha R$.
Take $\beta \in IR \intersect \ZZ_K$, then \( \beta\over \alpha \in R \intersect K \subseteq \bar\ZZ \intersect K = \ZZ_K \).
Then $\beta \in \alpha \ZZ_K$, which is what we wanted to show.
\

We'll reduce to the principal case by using a familiar trick.
Pick $m\in \ZZ^{\geq 0}$ so that $I^m$ is principal, then $I^m R \intersect \ZZ_K = I^m$ by the previous case.
Taking \( \gamma\in IR \intersect \ZZ_K \), we'll show it must be in $I$.
We know \( \gamma^m \in (IR)^m = I^m R \), and thus we have \( \gamma^m \in I^m R \intersect \ZZ_K = I^m \).
Now using unique factorization and to contain is to divide, this implies that $I^m \divides \gens{ \gamma ^m }= \gens{ \gamma }^m$.
So $I \divides \gens{ \gamma }$ and thus \( \gamma \in I \). 
:::

:::{.proof title="of theorem"}
Choose an extension $L/K$ with $I\ZZ_L = \alpha\ZZ_M$ principal.
consider $I\bar{\ZZ}$.
We have $I\bar\ZZ = (I\ZZ_L)\bar\ZZ$ -- this just says that extending $I$ to $\bar{\ZZ}$ can be done in two steps, first extending to $\ZZ_L$.
Concretely, these extensions are linear combinations of elements in $I$ with coefficients in $\ZZ_L$, so extending in stages yields the same thing.
We have
\[
I\bar\ZZ
&= (I\ZZ_L) \bar\ZZ \\
&= ( \alpha \ZZ_L) \bar\ZZ \\
&= \alpha\bar\ZZ
.\]
Now apply the up-down lemma to $R = \bar\ZZ$ to obtain
\[
I = I\bar\ZZ \intersect \ZZ_K
,\]
which is equal to \( \alpha\bar\ZZ \intersect \ZZ_K \).
:::

:::{.remark}
Is is true that in $\bar\ZZ$ that every ideal is principal?
The answer is no, since this would force it to be a UFD, but $\bar\ZZ$ has irreducibles at all since there are no irreducibles.
One *can* prove that every finitely generated ideal of $\bar\ZZ$ is principal.
One can also prove a variant of Gauss' lemma for $\bar\ZZ$: one can take the greatest common divisor to be the generator of a principal ideal.
:::

# Ch. 17: Prime Decomposition and General Number Rings

:::{.definition title="lies above"}
If $P\normal \ZZ_K$ is nonzero, then $P$ **lies above** a rational prime $p\in \ZZ$ if $p\in P$.
We also say $p$ **lies below** $P$.

The picture:

\begin{tikzcd}
	K && {\ZZ_K} && P \\
	\\
	\QQ && \ZZ && p
	\arrow[no head, from=1-1, to=3-1]
	\arrow[no head, from=1-3, to=3-3]
	\arrow[no head, from=1-5, to=3-5]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMCwwLCJLIl0sWzAsMiwiXFxRUSJdLFsyLDAsIlxcWlpfSyJdLFsyLDIsIlxcWloiXSxbNCwwLCJQIl0sWzQsMiwicCJdLFswLDEsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbMiwzLCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzQsNSwiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dXQ==)

:::

:::{.remark}
As in the quadratic field setting, every $P$ lies above some uniquely determined $p$ in the sense that $P \intersect \ZZ = p\ZZ$.
The proof here goes through in the same way.
Note that since $p\in P \implies \gens{ p } \in P$, we have \( \gens{ p }\divides P  \).
How do rational primes decomposes in $\ZZ_K$?
I.e., how does \( \gens{ p }  \) factor into prime ideals?
:::

:::{.remark}
There are two main theorems: one describes how these prime ideals factor, and the second (Dedekind-Kummer) will produce the factorization.
:::

:::{.theorem title="efg theorem"}
Let $K$ be a number field of degree $[K: \QQ] = n$ and let $p$ be a rational prime.
Factor 
\[
\gens{ p } = \prod_{i=1}^g P_ji{e_i}
,\]
where the $P_i$ are distinct primes ideals and $e_i > 0$.
Then for each $i$ we have $N(P_i) = p^{f_i}$ for some $f_i > 0$, and 
\[
\sum_{i=1}^g e_i f_i = n
.\]

Before the proof, it's helpful to introduce some notation.
:::

:::{.definition title="Residual degree and ramification index"}
In the setup above, call $f_i = f(P_i / p)$ the **residual degree** of $P_i/p$ and $e_i = e(P_i/p)$ the **ramification index**.
:::

:::{.proof title="of theorem"}
Consider the homomorphisms 
\[
\phi: \ZZ &\to \ZZ_K/P_i \\
a &\mapsto a \mod{P_i}
.\]
We have $\ker \varphi = P_i \intersect \ZZ$, and since $P$ lies over $p$, this equals $p\ZZ$.
So there is an induced injection $\FF_p \da \ZZ/p\ZZ \injects \ZZ_K/ P_i$, where $a\mod p \mapsto a \mod P_i$.
Thus $\ZZ_K/P_i \in \Vect_{\FF_p}$ is an $\FF_p\dash$vector space.
We want to show the size of the right-hand side, which is the size of $P_i$, is a power of $p$.
But any finite-dimensional vector space over $\FF_p$ has dimension $p^\ell$ for some $\ell$.
Let $f_i \da \dim_{\FF_p} (\ZZ_K/P_i)$, then $N(P_i) = \# \ZZ_K/P_i = p^{f_i}$.
\

Now proving that $\sum e_i f_i = n$ is an easy corollary: go back to the factorization and take norms of both sides.
Noting $N( \gens{ p } ) = p^n$ since $\ZZ_K \in \Ab_n^{\Free}$, we have
\[
p^n 
&= N( \gens{ p }  ) \\
&= \prod_{i=1}^g  N(P_i)^{\ell_i} \\
&= \prod_{i=1}^g p^{e_i f_i} \\
&= p^{\sum_i e_i f_i}
,\]
and the result follows by comparing exponents.
:::

:::{.observation}
Each $\ZZ_K/P_i$ is a field.
Since $P_i \in \spec \ZZ_K$, so $\ZZ_K/P_i$ is an integral domain.
It's finite by the above argument, using that norms are finite, so it's in fact a field.
It contains $\ZZ/p\ZZ$, and is thus an extension over it, and the extension degree $[\ZZ_K/P_i: \ZZ/p\ZZ] = f_i$.
So $f_i$ are called the residual degrees since they're the degrees of extensions of residue fields.
:::

:::{.definition title="Inert, split, ramified"}
More terminology: 

- If \( \gens{ p }  \) is prime, the $p$ is **inert**.
- If $g=n$, which forces $e_i = f_i = 1$ for all $i$, then $p$ **splits completely**.
- If $e_i > 1$ for any $i$, then $p$ **ramifies**.
:::

:::{.remark}
How do we actually determine the prime factors?
:::

:::{.theorem title="Dedekind-Kummer"}
Let $K$ be a number field of degree $n$ and suppose $K = \QQ(\alpha)$ where \( \alpha\in \bar\ZZ \).
This can be done by using the primitive element and scaling by an integer.
Let $p$ be a rational prime, and suppose $p\notdivides [ \ZZ_K: \ZZ[\alpha] ]$.
Then the prime ideal factorization of \( \gens{ p }  \) mirrors the factorization of $\min_{\alpha}(x) \mod p$.
\

More precisely, supposing 
\[
\min_{ \alpha}(x) = \prod_{i=1}^g p_i(x)^{e_i} \mod p && p_i \in \ZZ[x]
,\]
where the $p_i$ are the irreducible factors of $\min_{ \alpha}(x)$ in $\FF_p[x]$.
Then 
\[
\gens{ p } = \prod_{i=1}^g P_i ^{e_i} && P_i \da \gens{ p, p_i( \alpha) } 
.\]
These $P_i \in \spec \ZZ_K$ are distinct, and the residual degrees are given by $f_i = \deg(p_i)$.
:::

:::{.remark}
So as long as this condition holds, we factor $P$ by factoring a minimal polynomial mod $p$.
Note that if one can find an \( \alpha\in \ZZ_K \), the condition holds vacuously since the index is 1.
This always holds for quadratic fields, since $\ZZ_K = \ZZ[\tau]$, so this recovers the factorization statement in that setting.
In an annoying twist of fate, not every number field can be written as $\ZZ[ \alpha]$ for some single \( \alpha \), and so this theorem necessarily excludes some primes.
Is there some easy way to check the divisibility condition?
The answer is yes, coming from the discriminant.
:::

:::{.proposition title="?"}
Suppose $K = \QQ( \alpha)$ where \( \alpha\in \bar\ZZ \).
If $p^2 \notdivides \discriminant(\min_{\alpha} (x) )$, then $p\notdivides [\ZZ_K: \ZZ[\alpha]]$.
:::

:::{.remark}
A while ago we showed that 
\[
\discriminant( \min_{ \alpha}(x)) = \discriminant(1, \alpha, \alpha^2, \cdots , \alpha^{n-1}) = \discriminant_K \cdot [\ZZ_K: \ZZ[ \alpha] ]^2
.\]
Note that we're done if this is true: if $p$ divides the index, $p^2$ divides the last term, meaning $p^2$ divides the first.
:::

:::{.example title="?"}
Let $K = \QQ( \alpha)$ where \( \alpha \) is a root of the irreducible polynomial $f(x) = x^3 - x^2 - 2x + 8$.
Recall that we used this example when looking at integral bases.
By a computation, $\discriminant(f(x)) = -4 \cdot 503$, and the only squared prime dividing this is $p=2$.
We can thus apply Dedekind-Kummer for all $p\neq 2$.

- Mod 3, $f$ is irreducible.
  So \( \gens{ 3 }  \) is prime and $3$ is inert.

- Mod 5, $f(x) \equiv (x+1)(x^2 - 2)$.
  So \( \gens{ 5 }= P_1 P_2  \) where \( P_1 = \gens{ 5, \alpha+ 1 }  \) and \( P_2 = \gens{ 5, \alpha^2 - 2 }  \).
  Thus $f_1 = \deg(x+1) = 1, f_2 = \deg(x^2 - 2) = 2$.

- Mod 59, $f(x) \equiv (x+11)(x+20)(x+25)$.
  So \( \gens{ 59 } = P_1 P_2 P_3  \) where e.g. $P_1 = \gens{ 5, \alpha+ 11 }$ and so on.
  Note that $g=3=n$, so $59$ splits completely.

- Mod 503, $f(x) \equiv (x+259)(x+354)^2$.
  So \( \gens{ 503 }= P_1 P_2^2  \), and thus $503$ ramifies since an exponent is larger than 1.

Note that $\discriminant(f(x)) = 0 \mod 503$, so $f\in \FF_{503}[x]$ has a repeated root in an extension.
We can run this backward to show that $503$ is the only odd prime that ramifies: Dedekind-Kummer says this can only happen if $f$ has a repeated root mod $p$.
In this case, $\discriminant(f) \mod p = 0$, so $p\divides 4\cdot 503$.
Note that if you try to apply this theorem mod 2, this results in the wrong answer!
:::

:::{.lemma title="?"}
Let $K = \QQ( \alpha)$ be a number field where \( \alpha \in \bar\ZZ \).
Suppose $p\notdivides [\ZZ_K: \ZZ[ \alpha] ]$.
Then the inclusion $\iota: \ZZ[\alpha] \injects \ZZ_K$ induces an isomorphism
\[
\bar\iota: {\ZZ[\alpha] \over p \ZZ[ \alpha] } \mapsvia{\sim} \ZZ_K/p\ZZ_K
.\]
:::

:::{.proof title="?"}
Start by showing $p \ZZ_K \intersect \ZZ[ \alpha] = p \ZZ[ \alpha ]$.
The reverse containment is clear, so let \( \beta\in p \ZZ_K \intersect \ZZ[ \alpha ] \).
Then \( \beta/p \in \ZZ_K \), so look at its image in $\ZZ_K/ \ZZ[ \alpha]$.
The order of the image divides $p$, , since $p (\beta/p) = \beta \in \ZZ[ \alpha]$.
But we assumed that $p$ doesn't divide the size of this quotient, forcing the order to be 1 and thus $\beta/p\in \ZZ[ \alpha] \implies \beta \in p\ZZ[ \alpha]$.
:::

:::{.remark}
The map $\bar\iota$ is well-defined because $\iota( p \ZZ[ \alpha]) \subseteq p\ZZ_K$, and is obviously a homomorphism.
We want it to be an isomorphism, so what's the kernel?
We've just shown that $\ker \bar\iota = 0$, and comparing cardinality makes it a surjection.
Both have cardinality $p^n$, since both are free abelian groups of rank $n$.
:::

:::{.remark}
This is useful because the Dedekind-Kummer theorem describes the left-hand side, but we want to study the right-hand side instead.
The theorem says they're isomorphic!
:::