# Chebotarev Density (April 26)

## Setup

:::{.remark}
 Today: the Frobenius element, cyclotomic extensions of $\QQ$, Chebotarev density theorem, and some applications.
 We'll then look at applying Frobenius to understanding squares modulo a prime and get a proof of quadratic reciprocity.

 Setup: 
 let $L/K$ be a Galois extension of number fields, $P \in \spec K$ unramified in $L$, and $Q\in \spec L$ lying above $p$.
 
 - There is a unique element $\Frob_{Q / P} \in \Gal(L/K)$ satisfying 
 \[
\Frob_{Q/P}(\alpha) = \alpha^p \mod q
&&
p \da \# \ZZ_K/p
 .\]

 - As $Q$ ranges over the primes of $L$ over $P$, then $\Frob_{Q/P}$ ranges over an entire conjugacy class, which we denoted $\symb{L/K}{P}$.
  For abelian Galois groups, we'll identify this class with its single representative.

:::
 
:::{.remark}
How does this theory play out for cyclotomic extensions?
We've seen that for $K\da \QQ(\zeta_n)$, the extension $K/\QQ$ is Galois of degree $\phi(m) = \# (\ZZ/m)\units$, and moreover these are isomorphic via
\[
(\ZZ/m)\units &\to \Gal(K/\QQ)\\
a\mod m &\mapsto \sigma_a: \zeta_a \mapsto \zeta_m^a
.\]
:::
 
:::{.proposition title="?"}
 If $p$ is prime and $p\notdivides m$, then $p$ is unramified in $K$.
 We proved this last time, and it follows that $\symb{K/Q}{P} \in \Gal(K/\QQ) \cong (\ZZ/m)\units$.
Moreover, we saw that 
\[
\symb{K/Q}{P} \equiv p \mod m
.\]
:::

:::{.remark}
Thus for any two primes not dividing $m$ which become equal modulo $m$, then they split in the same way in the cyclotomic field. 
So the way a rational prime splits in $\QQ(\zeta_m)$ is entirely determined by its residue class $\mod m$.
:::

:::{.corollary title="?"}
 If $p$ is a prime not dividing $m$, then $e,f,g$ depend only on $p\mod m$.
:::
 
:::{.proof title="?"}
 If $p$ doesn't divide $m$, $p$ is unramified and $e=1$.
 Recall that $f$ is the order of any Frobenius element above $p$ in the Galois group, $\symb{K/Q}{P} = p \mod m$ in $(\ZZ/m)\units$. 
 Note that $efg = \phi(m)$ is the degree of the field, and solving yields $g = \phi(m)/f$.
:::

:::{.remark}
 In class field theory, there are analogs of this, foreshadowing something that happens in general for abelian extensions of number fields.
 In fact, every such extension is a subfield of the cyclotomic field.
:::
 
## Chebotarev's Theorem 

:::{.question}
A classical question: let $a\mod m$ be a coprime residue class, are there infinitely many (positive) rational primes $p$ with $p\equiv a\mod m$?
For instance, are there infinitely many primes $p$ satisfying $p \equiv 5 \mod 11$?
:::

:::{.remark}
Note that only coprime residue classes are interesting here.
Considering, say, $p\equiv 5 \mod 10$ would only yield multiples of 5 and thus $p=5$ is the only solution.
In general, for $a\mod m$ with $a, m$ not coprime, there is at most one solution.
:::

:::{.answer}
For $a, m$ coprime, it is a theorem of Dirichlet that there are infinitely many.
This proof uses $L\dash$functions and their analytic behavior.
:::

:::{.theorem title="Chebotarev's Density Theorem: The Fundamental Theorem of Algebroanalytic Number Theory"}
Let $L/K$ be a Galois extension of number fields and $C$ a conjugacy class of $\Gal(L/K)$.
Note that for every prime $P$ of $K$ that is unramified in $L$, we can associate the Frobenius conjugacy class.
There are only finitely many primes of $K$ that ramify in $L$, so all but finitely many primes yield a conjugacy class -- is it this fixed conjugacy class $C$?

The theorem is that there are infinitely many primes $P$ of $K$ for which $P$ is unramified in $L$ and the Frobenius conjugacy class satisfies $\symb{L/K}{P} = C$.
:::

:::{.remark}
This is qualitatively stated, but one can make this quantitative -- there is a well-defined proportion of primes for which this is true. 
Letting $P$ range over the primes of $K$, the *proportion* with $\qty{L/K \over P} = C$ will be $\# C / [L:K]$.
Here *proportion* is defined in the following way:
\[
\lim_{x\to\infty} 
\qty{ \# \ts{ P \in \spec K \st N(P) \leq x, \, \qty{L/K \over P} = C} \hspace{1em} \over \# \ts{ P\in \spec K \st N(P) \leq x} }
,\]
which is sometimes called the *natural density* of these primes.
What Chebotarev proved is slightly weaker and involved the *Dirichlet density* instead.
:::

:::{.corollary title="?"}
We recover Dirichlet's theorem on primes $p\equiv a \mod m$ by taking $L \da \QQ(\zeta_m)$, $K=\QQ$, and $C \da \ts{a\mod m}$.
The density theorem yields infinitely many primes $P$ with Frobenius equal to $C$, but then $\qty{L/K\over p} = \ts{p\mod m}$, yielding $p\equiv a \mod m$.
Moreover the proportion of such primes is $1/\phi(m)$, the degree of $[L:K]$, so the coprime residue classes are essentially distributed uniformly.
:::

:::{.remark}
This is useful when one needs infinitely many primes of a certain form, for which one can apply the density theorem to a well-chosen extension of number fields.
:::

:::{.question}
Consider primes $p$ and consider the multiplicative order of $2\mod p$, which is a divisor of $p-1$.
How often is this order even or odd?
One might expect this to happen half of the time, but this is not true.
Instead:
:::

:::{.corollary title="of a strong Chebotarev theorem, due to Hasse"}
The proportion of $p$ for which $2$ has even order $\mod p$ is $17/24$.
:::

:::{.remark}
Note that 2 is special, and this proportion changes for $3\mod p$ to something around $2/3$.
Interesting algebraic number theory governs why this is *not* $17/24$, and involves how different number fields intersect.
:::

## Residues $\mod p$ and Quadratic Reciprocity

:::{.remark}
If $G$ is a cyclic group of even order, there is a unique morphism 
\[
\phi_G:G &\to \ts{\pm 1} \\
g &\mapsto 
\begin{cases}
1 &  g \in G^2
\\
-1 & \text{else}.
\end{cases}
\]

:::

:::{.definition title="Legendre Symbol"}
Let $p$ be an odd prime, and let $G \da (\ZZ/p)\units$ which is cyclic of order $p-1$.
For each $a\in \ZZ$ coprime to $p$, define the **Legendre symbol**
\[
\symb{a}{p} \da \phi_G(a\mod p) 
=
\begin{cases}
1 &  a = a_0^2\mod p \text{ for some } a_0\in G
\\
-1 & \text{else]}.
\end{cases}
\]
Note that this is a group morphism, and thus multiplicative and we get
\[
\symb{ab}{p} = \symb{a}{p} \symb{b}{p} && a,b \in \ZZ \text {coprime to} p
.\]
One can also extend the symbol by defining $\symb{n}{p} = 0$ when $n$ is not coprime to $p$.
:::

:::{.question}
Given $a$, can we characterize the primes $p$ (not dividing $a$) for which $\symb{a}{p} =1$.
:::

:::{.remark}
Let $K = \QQ(\sqrt d)$ where $d$ is squarefree, then $K/\QQ$ is abelian and the Galois group can be thought of as $G \da \Gal(K/\QQ) = \ts{\pm 1}$.
Each rational prime $P$ unramified in $K$ will yields a well-defined Frobenius conjugacy class in $K$, and since $G$ is a abelian this is a single element -- which element is it?

If $p$ is an odd prime not dividing $d$, we know from our study of quadratic fields that $p$ is unramified in $K$.
:::

:::{.proposition title="?"}
Let $p$ be an odd prime not dividing $d$, then
\[
\symb{K/\QQ}{p} = \symb{d}{p}
,\]
where we've identified the Galois group as $\ts{\pm 1}$.
Thus is $d$ is a square $\mod p$, the Frobenius is trivial, and conversely if $d$ is not a square then the Frobenius is nontrivial.
:::

:::{.proof title="?"}
Since $p$ is odd not dividing $d$, $p$ doesn't ramify and thus $e=1$.
Since we're in a quadratic field, the only other things that can happen are being split (completely) or inert.
So when does $p$ split?
This is precisely when $g=2$, so there are two primes above $p$, and we know $efg=2$ is the degree of the extension, so $f=1$.
But $f$ is the order of any Frobenius element, so the order of $\symb{K/\QQ}{p} = 1$, making it the identity.
So $p$ splits $\iff \symb{K/\QQ}{p} = 1$.

On the other hand, when we worked out how primes factor in quadratic fields, we used Dedekind-Kummer, and if $p$ is odd not dividing $d$ then $\symb{d}{p} = 1$.
Equivalently, $p$ splitting involves looking at how $x^2-d$ factors in $K$, and that's what yields this condition.
So $p$ splits $\iff \symb{d}{p} = 1$.

> Note that looking at how arbitrary primes factor involves looking at the minimal polynomial of $\tau = \sqrt{d}, {1 + \sqrt d \over 2}$, but we can throw out the second case here if we're willing to throw out primes dividing the index $[\ZZ_K : \ZZ[\sqrt d] ] \in \ts{1, 2}$.
  So for odd primes, the way $p$ factors is exactly the way $x^2-d$ factors.

Thus $p$ splits iff both of these quantities are $+1$ and doesn't split iff they're both $-1$, so they must be equal.
:::

:::{.proposition title="?"}
If $p$ is an odd prime,
\[
\symb{-1}{p} = 
\begin{cases}
1 & p\equiv1\mod 4 
\\
-1 & \text{else}.
\end{cases}
\]
:::

:::{.proof title="?"}
Consider $K \da \QQ(\sqrt{-1})$.
If $p$ is an odd prime, we can identify $\Gal(K/\QQ) = \ts{\pm 1}$ to obtain 
\[
\symb{K/\QQ}{p} = \symb{-1}{p}
.\]
Note that $K$ is also the cyclotomic extension $K = \QQ(\zeta_4)$, and for cyclotomic extensions we can compute the Frobenius elements as $\symb{K/\QQ}{p} \equiv p \mod 4$, viewing $\Gal(K/\QQ) = (\ZZ/4)\units$.

From the first perspective, we have $\symb{K/\QQ}{p} = \id \iff \symb{-1}{p} = 1$, and from the second perspective this happens $\iff p \equiv 1 \mod 4$.
:::

:::{.remark}
This is perhaps overkill, but it's nice that it follows easily from the general theory!
What about $2$ instead of $-1$?
:::

:::{.proposition title="?"}
If $p$ is an odd prime,
\[
\symb{2}{p} = 
\begin{cases}
1 & p\equiv \pm 1\mod 8
\\
-1 & \text{else}.
\end{cases}
\]
:::

:::{.remark}
The usual elementary proof uses Gauss' lemma, and there are other proofs that use clever counting.
:::

:::{.proof title="?"}
We have a tower of extensions $K/L/\QQ \da \QQ(\zeta_8) / \QQ(\sqrt 2) / \QQ$, and consider corresponding primes $Q/R/P$.
Since the overall extension is Galois, by the Galois correspondence we can write $\QQ(\sqrt 2) = \QQ(\zeta_8)^H$ as the fixed field of some $H\leq G\da \Gal(\QQ(\zeta_8)/\QQ)$ of index 2.

We know that $G = (\ZZ/8)\units$ which has size 4, so $H$ has 2 elements: the identity and an order 2 element.
All of the $g\in G$ are order 2, so which is it?
We look for an automorphism of $K/\QQ$ fixing $L$, and taking $\sigma$ to be complex conjugation works since $\sigma(\zeta_8) = \zeta_8\inv \neq \zeta_8$, so $\sigma \neq \id$ is nontrivial.
So $\sigma = -1\mod 8$ and we can write $H = \ts{\pm 1\mod 8}$.

Now note that 
\[
\symb{2}{p} = 1 \iff \symb{\QQ(\sqrt 2)/\QQ}{p} = \id
,\]
since the Legendre and Frobenius symbols are equal here.
This happens $\iff \Frob_{R/P} = \id \iff \ro{\Frob_{Q/P}}{\QQ(\sqrt 2)} = \id$, using what we've previously proved about how Frobenius behaves in towers of extensions.
This happens $\iff \symb{\QQ(\zeta_8)/\QQ}{P} \in H \iff p\mod 8 \in \ts{\pm 1 \mod 8}$.
:::

:::{.remark}
For a general $a\in \ZZ_{\neq 0}$, we can factor $a = (\pm 1)(2^k) \prod p_i$ where the $p_i$ are odd primes.
Using multiplicativity of the Legendre symbol, since we understand the first two terms, it only remains to understand $\symb{q}{p}$ for $q$ an odd prime -- this is precisely what quadratic reciprocity gives us.
:::