# Frobenius (April 20)

:::{.remark}
Setup and reminders:

- $L/K$ a Galois extension of number fields,
- $P \in \spec K$ unramified in $L$
- $Q\in \spec L$ above $P$, so $e(Q/P) = 1$.
- The Frobenius was defined as $\red_{Q/P}\inv(x\mapsto x^q)$ for $q \da N(P)$.
- The characterization theorem for Frobenius: $\sigma = \Frob_{Q/P}$ iff $\sigma( \alpha) \equiv \alpha^q \mod Q$, so Frobenius acts like that $q$th power map mod $Q$.
- When $L/M/K$ with $M/K$ Galois with $Q/R/P$, $\Frob_{R/P} = \ro{\Frob_{Q/P}}{M}$
:::

:::{.proposition title="?"}
Let $L/K$ be Galois and let $Q,Q'$ be two primes lying above $P$:

\begin{tikzcd}
	L && Q && {Q'} \\
	\\
	K &&& P
	\arrow[no head, from=1-1, to=3-1]
	\arrow[no head, from=1-3, to=3-4]
	\arrow[no head, from=1-5, to=3-4]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMCwwLCJMIl0sWzAsMiwiSyJdLFszLDIsIlAiXSxbMiwwLCJRIl0sWzQsMCwiUSciXSxbMCwxLCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzMsMiwiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFs0LDIsIiIsMix7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XV0=)


Since Galois acts transitively, we can write $Q' = \sigma(Q)$ for some $\sigma \in G(L/K)$.
Then
\[
\Frob_{Q'/P} = \sigma \circ \Frob_{Q/P} \circ \sigma\inv
.\]
:::

:::{.proof title="Sketch"}
An exercise, use the characterization theorem.
Set $\tau$ to be the right-hand side,  and check that $\tau(\alpha) \equiv \alpha^p \mod Q'$ for all $\alpha \in \ZZ_L$ and $q\da N(P)$.
:::

:::{.remark}
What can we do when we are given a $P$ but not a $Q$?
There are many choices of $Q$, all related by conjugation.
:::

:::{.definition title="Frobenius Conjugacy Class"}
Define the **Frobenius conjugacy class** as
\[
\qty{L/K \over P} = \ts{ \Frob_{Q/P} \st Q \text{ lies above } P } \subseteq G(L/K)
.\]
This is a conjugacy class of $G(L/K)$.
:::

:::{.remark}
Note that this collapses to a single element when $G(L/K)\in \Ab$!
By abuse of notation, we'll identify $\qty{L/K \over P}$ with that common element (despite it being a singleton set).
:::

:::{.proposition title="Order of Frobenius"}
Every element of $\qty{L/K \over P}$ has order $f$, where $f = f(Q/P)$ for any $Q/P$.
:::

:::{.proof title="?"}
Let $Q/P$ be a prime over $P$, then $\Frob_{Q/P} \da \red_{Q/P}\inv(x\mapsto x^q)$, where the $q$th power map is the generator of $G(\FF_Q/\FF_P)$, which has size $[\FF_Q : \FF_P] = f$.
Since $\red$ was an isomorphism, we're done.
:::

## Cyclotomic Fields

:::{.remark}
Fix $m$.
Recall that the $m$th cyclotomic field $K$ is defined by
\[
K \da \QQ(\zeta_m) && \zeta_m = e^{2\pi i / m}
.\]
This is the splitting field of $x^m-1$, and in characteristic zero this implies $K/\QQ$ is Galois.
:::

:::{.proposition title="?"}
\[
[K : \QQ] = \phi(m)
.\]
:::

:::{.definition title="Cyclotomic Polynomials"}
Define the **$m$th cyclotomic polynomial** as
\[
\Phi_m(x) \da \prod_{\substack{a\mod m \\ (a, m) = 1}} (x - \zeta_m^a)
.\]
:::

:::{.remark}
Note that $\deg \Phi_m(x) = \phi(m)$, since this is precisely how many terms show up in the indexing set.
We'll show that this is the minimal polynomial of $\zeta_m$

- The coefficients are algebraic integers, since the roots are all roots of $x^m-1$, which is monic in $\ZZ[x]$.
  Since $\bar\ZZ$ is a ring, we have $\Phi_m(x) \in \bar\ZZ[x]$.

- If $\sigma \in G(K/\QQ)$, then $\sigma(\zeta_m) = \zeta_m^b$ for some $b$ coprime to $m$.
  Then
  \[
\sigma(\Phi_m(x)) = \prod_{\substack{a\mod m \\ (a, m) = 1}} (x - \zeta_m^{ab}) = \Phi_m(x)
  ,\]
  since as $a$ runs through the number coprime to $m$, so does $ab$.
  Thus $\Phi_m(x) \in \QQ[x]$, since its coefficients are fixed by every element of the Galois group.

- Combining these, the coefficients are in $\bar\ZZ[x] \intersect \QQ[x] = \ZZ[x]$.
:::

:::{.proposition title="?"}
$\Phi_m(x) = m(x) \da \min_{\zeta_m}(x)$ is the minimal polynomial of $\zeta_m$ over $\QQ$.
:::

:::{.proof title="?"}
Clearly $m(x) \divides \Phi_m(x)$ in $\QQ[x]$ since $\Phi_m(x)$ vanishes at $\zeta_m$.
So every root of $m(x)$ is a primitive $m$th root of unity, so we just need to show that every primitive $m$th root of unity is a root of $m$, i.e. we get all of them.
Observe that if $\zeta$ is *any* $m$th root of unity, then $\zeta \in \ZZ_K$: $\zeta$ is an algebraic integer, as a root of $x^m-1$, and is in $K$ since it's a power of $\zeta_m$.
Also note that $m(x) \in \ZZ[x]$ since the minimal polynomial of *any* algebraic integer has rational integer coefficients.
So $m(\zeta)\in \ZZ_K$ since $\ZZ_K$ is a ring and $\zeta\in \ZZ_K$, and $N(m(\zeta)) \in \ZZ$.


:::{.claim}
There is an $M\in \ZZ^{\geq 0}$ such that if $q$ is any prime with $q> M$ and $\zeta$ is any root of $m(x)$, then $\zeta^q$ is still a root of $m(x)$.

I.e. the roots of $m(x)$ are closed under taking $q$th powers for large enough $q$.
Moreover, it's enough to pick any $M> \max_{\zeta} \abs{N(m(\zeta))}$, taking $\zeta$ over all $m$th roots of 1.
:::

:::{.proof title="of claim"}
Let $q>M$ be chosen as above and let $\zeta$ be a root of $m(x)$.
We know $\zeta$ is a primitive $m$th root of unity and in $\ZZ_K$.
Work modulo $Q$: since $0 = m(\zeta)$ we have 
\[
0 
\equiv m(\zeta)^q \equiv m(\zeta^q)
.\]
Hence 
\[
q \divides N(q) \divides N(m(\zeta^q))
,\]
using that $N(q) = q^d$ since it's a rational integer where $d$ is the degree of the number field.
But we have $q > \abs{ N(m(\zeta^q))}$, which can only happen if the right-hand side is zero.
The only element of norm zero in a number field is zero, so $m(\zeta^q) = 0$.

:::

So if $q>M$ and $\zeta$ is a root of $m(x)$, so is $\zeta^q$.
$\zeta_m$ is a root of $m(x)$, thus so us $\zeta_m^{q_1}, \zeta_m^{q_1 q_2}, \cdots, \zeta_m^{A}$ for any $A \in \ZZ^{\geq 0}$ which can be written as a product of primes bigger than $M$.


:::{.exercise title="?"}
Show that we can choose any $a\in \ZZ$ with $\gcd(a, m) = 1$, and choose $A$ as above with $A \cong a \mod m$.
:::

With this, $\zeta_m^a = \zeta_m^A$ will be a root of $m(x)$ and we're done.
:::

:::{.remark}
Note that the exercise does follow from Dirichlet's theorem for arithmetic progressions, but there are easier proofs.
:::

## Galois Theory of $\QQ(\zeta_m) / \QQ$ and The Frobenius

:::{.remark}
Let $\sigma \in G(K/\QQ)$, then we saw that $\sigma(\zeta_m) = \zeta_m^a$ for some $a$ coprime to $m$.
So we know $\# G(K/\QQ) < \phi(m)$, since there are only that many possibilities for the right-hand side.
Since $[K:\QQ] = \phi(m)$, for each $a$ coprime to $m$ there is a $\sigma_a$ with $\sigma_a(\zeta_m) = \zeta_m^a$, so we can write
\[
G(K/\QQ) = \ts{ \sigma_a \st a\in (\ZZ/m)\units } 
.\]
Noting that $\sigma_a \sigma_b (\zeta_m) = \zeta_m^{ab} = \sigma_{ab}$, so we get an isomorphism
\[
G(K/\QQ) &\mapsvia{\sim} (\ZZ/m) \units \\
\sigma_a &\mapstofrom a\mod m
,\]
so $K/\QQ$ is an abelian extension.
:::

:::{.remark}
Consider $K \da \QQ(\zeta_m)$, then for every rational prime $P$ unramified in $K$, there is a well-defined element in $\qty{K/\QQ \over P} \in G(K/\QQ) = (\ZZ/m)\units$.
For a given $P$, which element in $(\ZZ/m)\units$ do you get?
:::

:::{.proposition title="?"}
If $p\notdivides m$, then $p$ is unramified in $\QQ(\zeta_m)$.
:::

:::{.remark}
The converse is more or less true: $\QQ(\zeta_2) = \QQ$ since $\zeta_2 = -1$, but 2 is not ramified in $\QQ$ since nothing is ramified in itself.
So if you avoid $m\equiv 2 \mod 4$, the converse becomes true.
We'll just prove the mentioned direction.
:::

:::{.lemma title="?"}
\[
\ZZ_K = \ZZ[\zeta_m]
.\]
:::

:::{.exercise title="?"}
Show that $p$ ramifies in $K$ iff $\ZZ_K/p\ZZ_K$ has a nonzero nilpotent element.
:::

:::{.proof title="of proposition, cute!"}
Suppose $p\notdivides m$, then it suffices to show that $\ZZ[\zeta_m] / p\ZZ[\zeta_m]$ has no nonzero nilpotents.
Let \( \alpha\in \ZZ[\zeta_m] \) with \( \alpha \mod p \) nilpotent in the quotient, we'll show it must be zero in the quotient.
By the lemma, we can write \( \alpha = \sum_{i=0}^{d-1} a_i \zeta^i \), using that $\ZZ_K = \ZZ[\zeta]$ where all of the $a_i$ are in $\ZZ$, we've set $\zeta\da \zeta_m$ and $d = \phi(m) = [K : \QQ]$.

Look mod $p$, then
\[
\alpha^p &\equiv \sum_{i=0}^{d-1} a_i \zeta^{ip} \mod p
\implies \alpha^{p^f} &\equiv \sum_{i=0}^{d-1} a_i \zeta^{ip^f} \mod p && \forall f
.\]
If $f$ is large enough, $\alpha^{p^f} \equiv 0 \mod p$, since some power of $\alpha$ is zero.
On the other hand, $p\notdivides m$, so there are powers of $p$ that are $1\mod m$ that show up regularly, and we can choose $f$ so that $p^f \cong 1 \mod m$, e.g. by choosing $f$ to be any multiple of the order of $p$ mod $m$.
But then the second line above reduces to $\sum a_i\zeta^{ip^f} \equiv \sum a_i \zeta^{i} = \alpha$ since $p^f \equiv 1 \mod m$, so $\alpha^{p^f} \equiv \alpha$, but we know $\alpha^{p^f} \equiv 0$.
:::

:::{.theorem title="?"}
Suppose $p\notdivides m$, which guarantees $p$ does not ramify in $K$.
Then
\[
\qty{K/\QQ \over P} = \sigma_p && \sigma_p(\zeta_m) \da \zeta_m^p
.\]
:::

:::{.proof title="?"}
Choose any $Q$ lying above $p$.
We'll show that for all $\alpha\in \ZZ_K$, $\sigma_p( \alpha) = \alpha^p \mod Q$, which by the characterization theorem will show $\sigma_p = \Frob_{Q/P}$, and in the abelian case $\qty{K/\QQ \over p}$ reduces to this single element.
We know that $Q\contains p\ZZ_K$ since $Q$ is above $p$, so it's enough to show $\sigma_p( \alpha) \equiv \alpha^p \mod p$ for all $\alpha\in \ZZ_K$ -- this is because the difference is a multiple of $p$ in $\ZZ_K$, which is a subset of $Q$.
Using $\ZZ_K = \ZZ[\zeta_m]$ to write $\alpha = \sum_{i=0}^{d-1} a_i \zeta_m^{i}$ where $d$ is the degree of the extension. 
Now reducing $\mod p$ and applying the Freshman's dream yields
\[
\sigma_p( \alpha) 
&= \sum_{i=0}^{d-1} a_i \zeta_m^{ip} \\
&\equiv \qty{ \sum_{i=0}^{d-1} a_i \zeta_m^{i}}^p \mod p \\
&\equiv \alpha^p
.\]
:::

:::{.remark}
This can be used to determine how all of the primes $p\notdivides m$ factor in $\QQ(\zeta_m)$, i.e. easily determining the relevant values of $f$ and $g$.
:::