# Thursday, December 01

## Chebotarev Density 

:::{.remark}
Main theorems to catch up on: Dedekind's different theorem, Chebotarev density, Dirichlet unit theorem.
To discuss: Frobenius elements.
Take a standard NT1 square $B/A, L/K$ with Galois group $G$, $p\in \spec A$, and $\kappa(p) = A/p \cong \FF_q, \ell(P) = B/P \cong \FF_{q^f}$.
Assume $p$ is unramified in $B$, so $I(P/p) = \ts{1}$, let $\eta: D(P/p)\iso \Aut(\FF_{q^f} / \FF_q)$ which has a canonical generator $F_q: x\mapsto x^q$.
Its pullback is denoted $\tau(P/p) \da \eta\inv(F_q)$, the **Frobenius element**.
Note that $\size \tau(P/p) = f$ since $D(P/p)$ is cyclic of order $f$, and

- $p$ splits completely in $B$ iff $\tau(P/p) = 1$, where it suffices to check one conjugate of $P$.
- $p$ is inert iff $pB$ is prime iff $f = N \da \size G$, forcing $G$ to be cyclic.
  So if $G$ is not cyclic, there are no inert primes!

A simple consequence: let $K = \QQ(\sqrt{d_1},\cdots, \sqrt{d_n})$ with $[K: \QQ] = 2^n$, then $G = C_2^n$ and if $n \geq 2$ then no prime remains inert in $K$.
:::

:::{.remark}
Let $L/K/\QQ$ be finite degree extensions of number fields and let $p\in K$.
Write $\norm{p} \da \Norm_{K/\QQ} = \size \ZZ_K/p$, the size of the residue field.
We know that $\size \ts{p\in \spec \ZZ_K \st \norm{p} \leq N} < \infty$ for any $N$, so let $\Sigma \subset \mspec \ZZ_K$ and say $\Sigma$ has **density** $\delta$ iff
\[
{\size \ts{p\in \Sigma \st \norm p \leq x} \over \size\ts{p\in\mspec \ZZ_K \st\norm p \leq x }} \converges{x\to\infty}\too \delta
.\]

Another definition: if $G$ is a group, a **normal subset** $S \subseteq G$ is any subset closed under conjugation, so $gSg\inv = S$ for all $g\in G$.
:::

:::{.theorem title="Chebotarev density, 1922"}
Take $L/K/\QQ$ with $G \da \Gal(L/K)$ and let $X \subseteq G$ be a normal subset.
Let $C(p)$ be the conjugacy class of Frobenius $\tau(P/p)$, and define
\[
\Sigma_X \da \ts{p\in \mspec \ZZ_K \st p \text{ is unramified in } L,\, C(p) \subseteq X}
.\]
Then $\Sigma_X$ has density in $\mspec \ZZ_K$ equal to $\size X/\size G$.
Thus every $g\in G$ is of the form $\tau(P/p)$ for infinitely many $P$, and the density of $p$ that split completely is $1/\size G$, setting $X = \ts{1}$.
:::

:::{.remark}
Easy to remember in the commutative case: then $C(p) = \tau(P/p)$ is a single element and $X=x$ is a single element, so the probability that $\tau(P/p) = x$ is $1/\size {G}$.
:::

:::{.example title="?"}
Let $N\geq 2$ and $\zeta_N \da e^{2\pi i /N}$ and let $K = \QQ(\zeta_N)$.
Fact: $\ZZ_K = \ZZ[\zeta_N]$, and $\Gal(K/\QQ) = C_N\units$, where every $\sigma\in \Aut(K/\QQ)$ is of the form $\zeta_N \mapsto \zeta_N^{\sigma a}$ for $a\in (\ZZ/N\ZZ)\units$.
By Dedekind-Kummer, the odd primes that ramify in $\ZZ_K$ are those $p$ such that $p\divides N$, and $2$ ramifies iff $4\divides N$.

Let $p > 2$ with $p\not\divides N$, so $p$ is unramified in $K$.
The Frobenius element at $p$ is $\tau(P/p) = p\in (\ZZ/N\ZZ)\units$ for any $P$ above $p$, corresponding to $\zeta_N \mapsto \zeta_N^P$ which reduces mod $N$ to $x\mapsto x^p$.
By Chebotarev density, for any fixed $a\in (\ZZ/N\ZZ)\units$, the proportion of primes $p$ such that $\tau_p = a$ is $1/\phi(N)$.
Since this implies $p\equiv a\mod N$, this says there are infinitely many primes satisfying this congruence and they are equidistributed -- this is Dirichlet's theorem on primes in arithmetic progressions.
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## Dedekind's Different Theorem

:::{.theorem title="Dedekind's Different Theorem"}
Consider $L/K$ with $B/A$, $P\in \spec B, p\in \spec A$,
let $\Delta = \Delta_{B/A}$ be the different ideal and let $e \da e(P/p)$.

a. If $e\not\in P$ and $\ell(P)/\kappa(p)$ is separable (tame ramification, ramification index is not divisible by the characteristic) then $v_P(\Delta) = e-1$.
b. Otherwise $v_P(\Delta) \geq e$.
:::

:::{.exercise title="?"}
Let $K$ be a field, $A\in \kalg^\fd$ which is a local Artinian principal ring (so Dedekind domains modulo prime powers), and define $e$ to be the nilpotency index of $A$, so the smallest power $e$ such that $P^e = 0$.
Note that in our case $A = B/P^e$.
Then the trace map $T: A\to k$ is identically zero (so $A$ is an étale algebra) iff $A/p$ is inseparable over $k$ or $\characteristic(k) \divides e$ (so $e\equiv 0$ in $k$).
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:::{.proof title="of theorem"}
By localizing we can assume $A$ is a DVR with $P = \gens{p}$.
ETS

1. $P^{e-1}\divides \Delta$,
2. $P^e\divides \Delta \iff$ the ramification is not tame, i.e. $e\in p$ or $\ell(P)/\kappa(p)$ is inseparable
3. Apply a lemma: every integral ideal $I \in \mathrm{Int}(B)$ satisfies 
\[
I\divides \Delta \iff \Delta\subseteq I \iff I\inv \subseteq \Delta\inv = B\dual \\
\iff \Trace_{L/K}(I\inv) = \Trace_{L/K}(I\inv B) \subseteq A
.\]

**Step 1**:
To show (1), NTS $\Trace)P^{-1-e} \subseteq A$, so $\gens{p} = pB = P^{e-1} a$ for $a\in \mathrm{Int}(B)$ with $P\divides a$.
Note $\Trace(P^{-1-e}) = \Trace\qty{{1\over P} a}$,
so we want $P^{-1-e} = {1\over p} a$, i.e. $\Trace(a) \subseteq pA$.
Let $\alpha\in a$, then $\Trace_{L/K}(\alpha) = \Trace_{B/A}(\alpha)$ and $\Trace_{B/A}(\alpha)\mod p = \Trace_{(B/pB)/(A/pA)}$.
Since $\mfp B = pB$ and $a$ are both divisible by the same primes of $B$, we have $\sqrt{\mfp B} = \sqrt{a}$ and thus there exists an $N$ such that $\alpha^N \in \mfp B$.
Then $\bar\alpha \da \alpha \mod \mfp B$ is nilpotent and thus $\Trace(\bar\alpha)=0$.

**Step 2**:
Let $\mfp B = P^\ell b$ with $P\not\divides b$.
As above, we have $P^e\divides \Delta \iff \Trace(b) \subseteq pA \iff \forall \beta\in b$ we have $\Trace_{(B/pB) / (A/pA) } (\bar\beta) = 0$.
Now $p^e$ and $b$ are coprime, so by the CRT we have $B/pB = B/p^e \times B/b$, so write $x\in B/pB$ as $x=(x_1,x_2)$ to obtain $\Trace(x) = \Trace_{B/p^e}(x_1) + \Trace_{B/b}(x_2)$.
If $x\in b$ then $x_2 = 0$ and the second trace is zero, so $x\in b\implies \Trace(x) = \Trace_{B/p^e}(x_1)$.
For all $x\in B$ there exists $x\in \bar B$ such that $x\equiv y \mod p^e$ and $x\equiv 0\mod b$.
Then $\Trace_{B/p^e}(\bar y) = \Trace_{B/p^e}(\bar x) = \Trace_{B/pB}(\bar x)$, and 
\[
\Trace_B(b) \subseteq p \iff \Trace_{B/pB}(b) = 0 \iff \Trace_{B/p^e}(B/p^e) = 0
.\]
Now $B/p^e$ is an algebra over $\kappa(p) = A/p$, and by the exercise this holds iff $B/p$ over $A/p$ is inseparable and the characteristic of $\kappa(p)$ is divisible by $e$ iff $e\in p$.
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:::{.remark}
On Hensel's bound: assume $\ell(P)$ is separable over $\kappa(p)$ (e.g. if $\kappa(p)$ is finite or characteristic zero), then
\[
v_P(\Delta) \leq e-1 + v_P(e)
.\]
Idea: reduce to the case of monogenic extensions $B/A$, so $B = A[\alpha]$ (generated as an $A\dash$algebra).
In this case, if $f\in A[t]$ is the minimal polynomial for $\alpha$, then $\Delta = \gens{\dd{f}{t}(\alpha) }$ is generated by the derivative.

Note that even if $A\in \DVR$, $B/A$ need not be monogenic. 
However, there is the following:
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:::{.theorem title="?"}
If $L/K$ with $B/A$ with $A,B\in \DVR$, then $B = A[\alpha]$ is monogenic.
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:::{.remark}
This situation may or may not occur.
NTII solves this issue: instead of localizing at $p$, complete at $p$.
Then if $A$ is a *complete* DVR, then (by a version of Hensel's lemma), $B$ is a complete DVR and $B/A$ is monogenic.
By a variant of local-to-global principles for lattices, $\Delta$ can be computed after a completion.
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