# Ch. XYZ: April 6 

## Multiplicativity in Towers

:::{.remark}
An important topic not in the book: relative extensions of number fields, vs absolute extensions over $\QQ$.

\begin{tikzcd}
	L && {\ZZ_?} && Q \\
	\\
	K && {\ZZ_K} && {P = Q \intersect \ZZ_K}
	\arrow[no head, from=1-5, to=3-5]
	\arrow[no head, from=1-3, to=3-3]
	\arrow[no head, from=1-1, to=3-1]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMCwwLCJMIl0sWzAsMiwiSyJdLFsyLDAsIlxcWlpfPyJdLFsyLDIsIlxcWlpfSyJdLFs0LDAsIlEiXSxbNCwyLCJQID0gUSBcXGludGVyc2VjdCBcXFpaX0siXSxbNCw1LCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzIsMywiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFswLDEsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XV0=)

Note that if $Q\contains P$, then $Q$ will divide the extension of $P$ to $\ZZ_L$, if $Q$ will show up in the prime factorization of $P\ZZ_L$.
We defined $e(Q/P)$ to be the exponent of $Q$ in the factorization of $P\ZZ_L$, and $f(Q/P)$ to be the degree of the field extension $\ZZ_L/Q$ over $\ZZ_K/P$.
:::

:::{.remark}
Note that "lying above" is transitive, and the following situation makes sense:

\begin{tikzcd}
	M && {P''} \\
	\\
	L && {P'} \\
	\\
	K && P
	\arrow[no head, from=3-1, to=5-1]
	\arrow[no head, from=1-1, to=3-1]
	\arrow[no head, from=1-3, to=3-3]
	\arrow[no head, from=3-3, to=5-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMCwyLCJMIl0sWzAsNCwiSyJdLFswLDAsIk0iXSxbMiw0LCJQIl0sWzIsMiwiUCciXSxbMiwwLCJQJyciXSxbMCwxLCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzIsMCwiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFs1LDQsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbNCwzLCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV1d)

How are the various $e$ and $f$ related?
It turns out they work similarly to degrees of field extensions:
:::

:::{.theorem title="Multiplicativity in towers"}
\[
e(P''/P) &= e(P''/P') \cdot e(P'/P) \\
f(P''/P) &= f(P''/P') \cdot f(P'/P) 
.\]
:::

:::{.proof title="?"}
Note that $f$ is the degree of a field extension where we take $\ZZ_K/P \embeds \ZZ_L$ and consider the induced quotient map to $\ZZ_L/Q$.
By composing inclusions we have a commutative diagram:

\begin{tikzcd}
	{\ZZ_K/P} && {\ZZ_L/P'} && {\ZZ_M/P''}
	\arrow[hook, from=1-1, to=1-3]
	\arrow[hook, from=1-3, to=1-5]
	\arrow[curve={height=30pt}, hook, from=1-1, to=1-5]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwwLCJcXFpaX0svUCJdLFsyLDAsIlxcWlpfTC9QJyJdLFs0LDAsIlxcWlpfTS9QJyciXSxbMCwxLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFsxLDIsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzAsMiwiIiwyLHsiY3VydmUiOjUsInN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV1d)

So for $f$, this reduces to the multiplicativity of degrees in towers of field extensions.


For $e$, write $P\ZZ_L = P'^{e(P'/P)} I$ where $I\in \Id(\ZZ_L)$ is nonzero and $P'\notdivides I$.
Similarly we can write $P' \ZZ_< = P''^{e(P'/P')} J$ where $J\in \Id(\ZZ_L)$ and $P''\notdivides J$.
Using general facts about ideal extensions and substituting yields
\[
P\ZZ_M 
&= (P\ZZ_L)\ZZ_M \\
&= (P'\ZZ_M)^{e(P'/P)} I\ZZ_M \\
&= P''^{e(P'/P) e(P'/P')} J^{e(P'/P)} (I\ZZ_M) \\
.\]

We know that $P''\notdivides J$, so it doesn't divide any power of $J$ either.
We can check that $P''\notdivides I\ZZ_M$: otherwise, if it did then $P''\supseteq I\ZZ_M$ and 
\[
P'' \intersect \ZZ_L \supseteq (I\ZZ_M) \intersect \ZZ_L \supset I
\]
by intersecting both sides with $\ZZ_L$.
But we know $P'' \intersect \ZZ_L = P'$ since $P''$ was above $P'$, and $P'\notdivides I$ and thus $P'\not\contains I$.
:::

:::{.definition title="Splitting completely"}
Suppose $L/K$ is an extension of number fields and say $P\supseteq \ZZ_K$ is a nonzero prime ideal.
Then $P$ **splits completely** in $L$ (or $\ZZ_L$) if $e(G/P) = f(G/P) = 1$ for all $G$ above $P$ in $\ZZ_L$.
Equivalently, there is a factorization $P\ZZ_L = \prod_{i\leq n} Q_i$ with the $Q_i$ distinct and $n \da [L: K]$.
:::

:::{.proposition title="?"}
Let $M/L/K$ be a tower of number fields and $P\in \spec \ZZ_K$.
Then if $P$ splits completely in $M$, it splits completely in $L$.
:::

:::{.slogan}
Splitting completely in an extension implies splitting completely in every intermediate extension.
:::

:::{.proof title="?"}
Suppose $P$ splits completely in $M$, we then want to show that for any $P'$ in $\ZZ_L$ above $P$, then $e=f=1$.
We know $P'$ has some prime factor, so choose any $P''$ in $\ZZ_M$ above $P'$:

\begin{tikzcd}
	M && {\exists P''} \\
	\\
	L && {P'} \\
	\\
	K && P
	\arrow[no head, from=1-1, to=3-1]
	\arrow[no head, from=3-1, to=5-1]
	\arrow[no head, from=5-3, to=3-3]
	\arrow[no head, from=1-3, to=3-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMCw0LCJLIl0sWzAsMiwiTCJdLFswLDAsIk0iXSxbMiw0LCJQIl0sWzIsMiwiUCciXSxbMiwwLCJcXGV4aXN0cyBQJyciXSxbMiwxLCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzEsMCwiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFszLDQsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbNSw0LCIiLDIseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV1d)

Since $P''$ is above $P$ and $e(P''/P) = f(P''/P) =1$, we can use multiplicativity in towers:
\[
e(P'/P) \divides e(P''/P) = 1 &\implies e(P'/P) = 1 \\
f(P'/P) \divides f(P''/P) = 1 &\implies f(P'/P) = 1 \\
.\]
:::

:::{.definition title="?"}
If $L/K$ is an extension of number fields and $P\in \spec \ZZ_K$ is nonzero, then $P$ **ramifies** in $L$ (or $\ZZ_L$) if $e(Q/P)>1$ for some $Q\in \spec \ZZ_L$ above $P$.
:::

:::{.proposition title="?"}
For $M/L/K$ a tower of number fields and $P\in \spec \ZZ_K$ is nonzero and unramified in $M$, then $P$ is unramified in $L$.
:::

:::{.proof title="?"}
Same as the last proof.
:::

## Galois Theory and Prime Decomposition

:::{.remark}
This will lead into defining the Frobenius element and the fundamental theorem of algebraic number theory: Chebotarev Density.
Some setup/notation:

- $L/K$ will be a Galois extension of number of fields
- $P \in \spec \ZZ_K, Q\in \spec \ZZ_L$ will be nonzero prime ideals.
  We may or may not require $Q$ to lie above $P$.
:::

:::{.remark}
\envlist

- Any $\sigma \in G(L/K)$ is an automorphism of $L$ fixing $K$, which restricts to an automorphism of $\ZZ_L$ since it preserves algebraic integers.
  - It also preserves prime ideals.
- Suppose $Q$ lies above $P$, then $P\ZZ_L = Q^{e(Q/p)} \cdots$.
  $\sigma$ fixes $K$ pointwise, so applying it to both sides yields $P\ZZ_L = \sigma(Q)^{e(Q/P)}\cdots$ where $\sigma(Q)$ shows up with the exact same power since it preserves distinctness of ideals.
  - In particular, $\sigma(G)$ lies above $P$, so $G(L/K)$ acts on the set of primes above $P$, and it turns out (very importantly) to be transitive.
:::

:::{.theorem title="?"}
The action of $G(L/K)$ on primes above $P$ is transitive.
:::

:::{.proof title="?"}
Let $Q_1, Q_2$ be primes above $P$ and suppose toward a contradiction that $\Orb(Q_1)$ does not contain $Q_2$.
We may choose $\alpha\in \ZZ_L$ such that 

- $\alpha \equiv 0 \mod Q_2$
- $\alpha \equiv 1 \mod Q$ for all $Q$ of the form $Q = \sigma(Q_1)$, noting that none are $Q_2$

This system can be solved by the CRT, since the $Q_i$ are pairwise comaximal, since nonzero primes of a number ring are maximal.
These congruences say $\alpha\in Q_2$ but not any other $Q$, since $0\equiv 1$ only in the unit ideal.
Define $\beta = \prod_{\sigma \in G(L/K)} \sigma( \alpha) \in \ZZ_K$, and observe that $\beta\not\in Q_1$.
If it were in $Q_1$, one $\sigma(\alpha) \in Q_1$ for some $\sigma \in G(L/K)$ which would force $\alpha = \sigma \sigma\inv ( \alpha) \in \sigma\inv (Q_1)$, which contradicts the choice of $\alpha$ since \( \sigma\inv(Q_1) \) is one of the $Q$s appearing in the second congruence.
On the other hand we have $\beta\in Q_2$ since the identity is an element of $G(L/K)$ and $\alpha\in Q_2$ and $\beta$ is a multiple of $\alpha$.
We know $\beta\in \ZZ_K$, so $\beta \in Q_2 \intersect \ZZ_K = P$.
But this is fishy because $Q_1, Q_2$ both lie above $P$ and both must contain $P$:

\begin{tikzcd}
	{Q_1} && {Q_2} \\
	\\
	& P
	\arrow[no head, from=1-1, to=3-2]
	\arrow[no head, from=1-3, to=3-2]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMSwyLCJQIl0sWzAsMCwiUV8xIl0sWzIsMCwiUV8yIl0sWzEsMCwiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFsyLDAsIiIsMix7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XV0=)

Since $\beta$ is in every ideal above $P$, in particular $\beta \in Q_1$ since $Q_1 \contains P$, and this is a contradiction.
:::

:::{.proposition title="?"}
If $Q, Q'$ both lie above $P$, then $e(Q/P) = e(Q'/P)$ and $f(Q/P) = f(Q'/P)$.
:::

:::{.proof title="?"}
Pick $\sigma \in G(L/K)$ with $\sigma(Q) = Q'$.
Factor $P\ZZ_L = Q^{e(Q/P)} J$ where $J$ is a product of primes not equal to $Q$.
Applying $\sigma$ yields $P\ZZ_L = Q'^{e(Q/P)} J'$ where $J'$ is a product of primes not equal to $Q'$.
This factors $P\ZZ_L$ into primes, and by uniqueness of prime factorization this exponent has to be $e(Q'/P)$ by definition. 

For the $f$s, note that $\sigma$ induces a ring morphism between the residue fields:
\[
\bar\sigma: \ZZ_K/Q &\to \ZZ_L/Q' \\
\alpha \mod Q &\mapsto \sigma(\alpha) \mod Q'
.\]
This is well-defined since $\sigma(Q) = Q'$, and is an isomorphism since the inverse comes from \( \sigma
^{-1}\).
This will imply that the $f$s are the same: we're looking at the degree of these extensions over $\ZZ_K/P$.
An element of $\ZZ_L/Q$ also belonging to $\ZZ_K/P$ (as a subfield) has the form $\alpha \mod Q$ where $\alpha\in \ZZ_K$, and under $\sigma$ this is sent to $\sigma(\alpha)\mod Q' = \alpha \mod Q'$, which is an element of the copy of $\ZZ_K/P\injects \ZZ_L/Q'$.
So $\sigma$ identifies the copies of $\ZZ_K/P$ in either side, and
\[
[\ZZ_L/Q : \ZZ_K/P] = [\ZZ_L/Q: \ZZ_K/P]
.\]
:::

:::{.theorem title="efg theorem for Galois extensions"}
Let $L/K$ be a Galois extension of number fields and $P\in \spec \ZZ_K$ be nonzero.
Let $e,f$ be the common $e, f$ for all $Q$ above $P$, and let $g$ be the number of distinct $Q$ above $P$.
Then
\[
efg = n \da [L:K]
.\]
:::

:::{.proof title="?"}
Using the previous efg theorem, $\sum_{i=1}^g e_i f_i = n$, but by the previous proposition, all of the $e_i$ are the same and all of the $f_i$ are the same.
:::

:::{.example title="?"}
Let $L = \QQ(\zeta_5)$ over $K= \QQ$, which is Galois with $G(L/K) \cong (\ZZ/5)\units \cong \ZZ_4$.
We know $\ZZ_L = \ZZ[\zeta_5]$, which actually happens for any $n$, and $\min_{\zeta_5}(x) = x^4 + x^3 + x^2 + x + 1$.
Since $\ZZ_L$ is a monogenic, we can apply Dedekind-Kummer.

- Factoring $2\ZZ_L = P$ yields a single factor, so $g=1$.
  Since $e$ is the common exponent, $e=1$, so $f=4$.
  If you factor $\min_{\zeta_5}$ mod 2, in order to get a single prime D-K says this must be irreducible and $f_i$ are the degrees of the irreducible factors.

- $19\ZZ_L = P_1 P_2$, so $g=2, e=1$, and so $f=2$.

- $11\ZZ_L = P_1 P_2 P_3 P_4$ yields $g=4, e=1, f=1$.

- $5\ZZ_L = P_1^4$ so $g=1,e=4,f=1$.

Can the combination $(e,f,g) = (2,2,1)$ occur?
This would require a prime factoring as $P^2$, but the answer is know.
In fact $e>1$ only happens for $5$ since this corresponds (for all applicable primes $p$, which is all primes since $\ZZ_L$ is monogenic here) to the polynomial having a repeated factor mod $p$.
This would require $\discriminant(\min_{\zeta_5}(x)) \equiv 0 \mod p$, but you can show that $5$ is the only prime that divides this discriminant.
So $5$ is the only prime that could ramify, so $e=2$ never happens.

In general, for cyclotomic extensions, $p$ is always totally ramified and $e=p-1$.
:::

## Decomposition, Inertia, Frobenius

:::{.definition title="Decomposition"}
Let $L/K$ be a Galois extension of number fields and let $Q \in \spec \ZZ_L$ be nonzero.
Then $Q$ lies above a unique prime $P \in \spec \ZZ_K$, where $P = Q \intersect \ZZ_K$.
Define the **decomposition group of $Q$** is defined as 
\[
D(Q) = D(Q/P) = \ts{ \sigma \in G(L/K) \st \sigma(Q) = Q} 
.\]
:::

:::{.definition title="?"}
Better notation: define $\FF_Q \da \ZZ_L/Q$ and $\FF_P = \ZZ_K/P$, so $\FF_P \subseteq \FF_Q$.
:::

:::{.remark}
We know that the Galois group will take $Q$ to some prime above $P$, these are the ones that take $Q$ to itself.
These are the automorphisms that make sense "modulo $Q$": there is a group morphism
\[
\red_{Q/P}: D(G/P) &\to G(\FF_Q/\FF_P) \\
\sigma &\mapsto \bar\sigma 
.\]
where we reduce $\sigma$ mod $Q$, so $\bar\sigma(\alpha \mod Q) \da \sigma(\alpha) \mod Q$.
One can check

- For $\sigma \in D(Q/P)$, $\bar\sigma$ is a well-defined automorphism of $\FF_Q$ fixing $\FF_P$, since the elements in $\FF_P$ are of the form $\alpha \mod Q$ where $\alpha\in \ZZ_K$ and $\sigma$ fixes $K$.
  This crucially uses that $\sigma$ fixes $Q$, otherwise it won't be well-defined.

- $\red_{Q/P}$ defines a group morphism.

We'll see later that this is in fact a surjective morphism.
So all automorphisms in the Galois group $G(\FF_Q/\FF_P)$ are reductions mod $Q$ of automorphisms in the decomposition group, which are the only ones that make sense to reduce mod $Q$ anyway!
:::