# Ch. XYZ: Galois Theory and Prime Decomposition (April 13)

:::{.remark}
Two recommended resources:

- Samuel, Algebraic Theory of Numbers (Dover)
- Matt Baker's [online notes on Algebraic number theory](https://people.math.gatech.edu/~mbaker/pdf/ANTBook.pdf).

:::

:::{.remark}
Setup:

- $L/K$ a Galois extension of number fields.
- $P \in \mspec K$
- $Q\in \mspec L$

We saw that $G(L/K)$ acts on the prime ideals above $P$, making all of the $e,f$ values the same, and thus the decomposition into prime ideals in this type of extension is simpler than in a general extension.
We'll look at decomposition and inertia groups today.

Recall that if $Q$ lies above $P$, then the **decomposition group** $D(Q/P)$ is the set of $\sigma\in G(L/K)$ such that $\sigma(Q) = Q$.
Note that $P$ is redundant in this notation, since $Q \intersect \ZZ_K = P$.
Also recall that $\FF_Q \da \ZZ_L/Q$ is the residue field associated to $Q$ and $\ZZ_P \da \ZZ_L/P$ and we view $\FF_P \subseteq \FF_Q$.
Such an extension of finite fields is always Galois, and the Galois groups turn out to be related to the decomposition groups.
For $\sigma \in D(Q/P)$ we defined $\bar\sigma: \FF_Q \to \FF_Q$ where $\alpha \mod Q \mapsto \sigma(\alpha) \mod Q$.
It's easy to check this is 

- Well-defined, precisely by the definition of $D(Q/P)$, 

- An automorphism of $\FF_Q$: it must be surjective since everything in $\ZZ_L$ is in the image of $\sigma$ since $\sigma$ was an automorphism of $\ZZ_L$ to begin with.
  But then it's a surjective morphism from a finite field to itself, hence an automorphism.
- $\sigma$ fixed $\FF_P$ pointwise
  Now consider applying $\sigma$ to an element of $\FF_P$, which are of the form $\alpha \mod Q$ where $\alpha$ comes from $\ZZ_K$.
  But $\sigma$ fixed $\ZZ_K$ pointwise, so $\sigma$ fixed $\FF_P$ pointwise.

So each $\sigma\in D(Q/P)$ yields a $\bar\sigma \in G(\FF_Q/\FF_P)$ and we get a group morphism
\[
\red_{Q/P}: D(Q/P) &&\to G(\FF_Q/\FF_P) \\
\sigma &\mapsto \bar\sigma
.\]
The following is the deeper and more important fact about this morphism, which requires a technical proof:
:::

:::{.theorem title="?"}
$\red_{Q/P}$ is surjective.
:::

:::{.proof title="?"}
We can assume $\FF_P$ is a proper subfield of $\FF_Q$, since the result follows immediately otherwise.
By the primitive element theorem, since $|FF_Q/\FF_P$ is separable we can write $\FF_Q = \FF_P(\bar\alpha)$ where $\alpha\in \ZZ_L$ and $\bar{\wait}$ denotes reducing mod $Q$.
Note $\alpha\not\in Q$, since this would mean $\bar\alpha = 0$ and thus $\FF_Q = \FF_P$.
On the other hand, we can assume $\alpha\in Q'$ for all other $Q'\neq Q$ above $P$.
Why?
Sketch: the equation $\FF_Q = \FF_P(\bar\alpha)$ only depends on $\alpha\mod Q$, so if $\alpha$ is not in some $Q'$, just adjust $\alpha$ modulo $Q'$ without affecting its class mod $Q$ to get a new $\alpha$ in $Q'$.
So if not, replace $\alpha$ with $\alpha'$ satisfying
\[
\alpha' \cong \alpha \mod Q \\
\alpha'\cong 0 \mod Q' && \forall Q'\neq Q \text{ above }P
.\]

Look at the minimal polynomial $\bar m(x)$ of $\bar\alpha$ over $\FF_P$, some monic polynomial in $\FF_P[x]$, where $\FF_P$ is the reduction mod $P$ of elements in $\ZZ_K$.
So we can think of $\bar m(x)$ as some $m(x) \in \ZZ_K[x]$ whose coefficients have been reduced mod $Q$.
We'll show that for each root $r$ of $\bar m(x)$ in $\FF_Q$, there is some $\sigma\in D(Q/P)$ such that $\bar\sigma(\bar\alpha) = r$
This is enough, since any automorphism of $\FF_Q$ fixing $\FF_P$ is determined by the image of $\bar\alpha$, which has to go to some other root of $\bar m (x)$.
If we can show this statement, this means that $\bar\sigma$ has to hit every automorphism in $G(\FF_Q/\FF_P)$.

Define $g(x) = \prod_{\sigma\in G(L/K)} (x- \sigma( \alpha))$.
Where does $g$ live?
$\alpha$ came from $L$, so $\sigma( \alpha) \in L$, and multiplying over all $\sigma$ puts the coefficients in $K$.
Even better, since $\alpha\in \ZZ_L$ is algebraic, this will have $\ZZ_K$ coefficients, so $g\in \ZZ_K[x]$.
Now reduce mod $Q$ to get $\bar g(x) \in \FF_P[X]$, and moreover $\bar g(\bar \alpha) = \bar{g(\alpha)} = 0$ in $\FF_Q$, since $g(\alpha) = 0$ by definition since one $\sigma$ in the product is the identity.
Thus we know $\bar m(x) \divides \bar g(x)$ in $\FF_P[x]$.
Notice that in $\FF_Q[x]$, if we take $g$ and reduce mod $Q$ we get
\[
\bar g(x) = \prod_{\sigma \in G(L/K)} (x - \bar{\sigma(\alpha)} )
.\]
Since $\bar m(x)$ divides $\bar g(x)$, every root of $\bar m$ has the form $\bar{\sigma( \alpha)}$ for some $\sigma\in G(L/K)$.
We want this to be $\bar\sigma(\bar \alpha)$ instead to conclude the proof, so take a root of $\bar m$ in $\FF_Q$ and write it as $\bar{\sigma(\alpha)}$ with $\sigma\in G(L/K)$

:::{.claim}
$\sigma \in D(G/P)$ has to be in the decomposition group.
:::
If this is true, we're done since $\bar{\sigma( \alpha)} = \bar \sigma (\bar \alpha)$.

Suppose toward a contradiction that $\sigma\not\in D(G/P)$, so neither $\sigma$ nor $\sigma\inv$ fixes $Q$, so $\sigma\inv(Q) \neq Q$.
By choice of $\alpha$, we have $\alpha \in Q' \da \sigma\inv(Q)$, so $\sigma(\alpha)\in Q$.
Then $\bar{\sigma(\alpha)} = 0$, but $0$ is not a root of $\bar m(x)$ since it is irreducible where $\bar m(x) \neq x$ since $[\FF_Q: \FF_P] \geq 2$ and thus $\deg \bar m(x) \geq 2$.
:::

:::{.remark}
By the first isomorphism theorem, we have 
\[
G(\FF_Q/\FF_P) \cong D(Q/P) / \ker \red_{Q/P}
.\], where we call the kernel the **inertia group**:
\[
\ker \red_{Q/P} = I(Q/P) \da \ts{ \sigma \in D(Q/P) \st \bar \sigma = \id_{\FF_Q} }
.\]
These are the elements $\sigma$ in the decomposition group such that $\sigma(\alpha) \equiv \alpha \mod Q$ for every $\alpha$.

Since we have a group action of the Galois group on primes above $P$, we can apply Orbit-Stabilizer: we have $\Stab(Q) = D(Q/P)$ and the orbit is all primes above $P$, so
\[
G(L/K) / D(Q/P) \cong \ts{\text{Primes above } P}
.\]
Taking cardinalities, 
\[
[L:K] / \# D(Q/P) = g \implies \# D(Q/P) = {[L:K] \over g} = {efg \over g} = ef
.\]
We also have
\[
{ef \over \# I(Q/P)} = \#G(\FF_Q/\FF_P) = [\FF_Q: \FF_P] = f
.\]
In summary, 

- $\# D(Q/P) = f$,
- $\# I(Q/P) = e$,
- If $P$ is unramified, $e=1$ and $I(Q/P)$ is trivial and $\red_{Q/P}$ is an isomorphism.

This map is usually an isomorphism, since there are only finitely many $P$ ion $\ZZ_K$ that ramify in $\ZZ_L$.
:::

## Inertia and Decomposition Fields

:::{.remark}
It will turn out that the fixed fields appearing here have number-theoretic interpretations.
:::

:::{.definition title="?"}
If $Q$ lies above $P$, then define the **inertia field** corresponding to $Q/P$, written $L^{I(Q/P)}$, and the **decomposition field** $L^{D(Q/P)}$.

:::

:::{.remark}
Suppose $L/K$ is Galois, and consider an intermediate field $L/M/K$.
We can find intermediate primes:

\begin{tikzcd}
	L && Q \\
	\\
	M && {R = Q \intersect \ZZ_M} \\
	\\
	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=3-3, to=1-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMCwwLCJMIl0sWzAsMiwiTSJdLFswLDQsIksiXSxbMiw0LCJQIl0sWzIsMiwiUiA9IFEgXFxpbnRlcnNlY3QgXFxaWl9NIl0sWzIsMCwiUSJdLFswLDEsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbMSwyLCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzMsNCwiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFs0LDUsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XV0=)

We can consider $e, f$ of $R/P$ in this picture. 
Going to bigger extensions then $M$ makes $R$ bigger, so $e$ increases as $M$ gets bigger.
:::

:::{.theorem title="?"}
The inertia field $L^{I(Q/P)}$ is the largest field extension $M$ for which $e(R/P) = 1$.
More precisely, 
\[
e(R/P) = 1 \iff M \subseteq L^{I(Q/P)}
.\]
:::

:::{.remark}
From Galois theory, $L/M$ is also Galois, so the key to proving this theorem involves understanding $D, I$ of $Q/R$.
Noting that $G(L/M) \subseteq G(L/K)$, 
We have
\[
D(Q/R) 
&\da \ts{\sigma\in G(L/M) \st \sigma(Q) = Q} \\
&= \ts{\sigma\in G(L/K) \st \sigma(Q) = Q} \intersect G(L/M) \\
&= D(Q/P) \intersect G(L/M)
,\]
so the decomposition groups are related by restriction.
Suppose $\sigma \in D(Q/R)$, then $\red_{Q/R}(\alpha)$ is an automorphism of $\FF_Q$.
We can also get an automorphism of $\FF_Q$ by taking $\red_{Q/P}(\sigma)$ -- it turns out these are the same automorphism.
Why?
Both map $\alpha \mod Q$ to $\sigma(\alpha) \mod Q$, which doesn't involve $R$ or $P$.
We can thus write
\[
I(Q/R) 
&= \ts{ \sigma \in D(Q/R) \st \bar{\sigma} = \id_{\FF_Q} } \\
&= \ts{ \sigma \in D(Q/P) \intersect G(L/M) \st \bar{\sigma} = \id_{\FF_Q} } \\
&= \ts{ \sigma \in D(Q/P) \st \bar{\sigma} = \id_{\FF_Q} }\intersect G(L/M) \\
&= I(G/P) \intersect G(L/M)
.\]
:::

:::{.remark title="Upshot"}
\[
D(Q/R) &= D)Q/P) \intersect G(L/M) \\
I(Q/R) &= I)Q/P) \intersect G(L/M) \\
.\]
:::

:::{.proof title="of theorem"}
Let $L/M/K$ with $Q/R/P$, we want to show $e(R/P) = 1 \iff M \subset L^{I(Q/P)}$:

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

> [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMCwwLCJMIl0sWzAsMiwiTSJdLFswLDQsIksiXSxbMiwwLCJRIl0sWzIsMiwiUiJdLFsyLDQsIlAiXSxbMiwxLCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzAsMSwiIiwyLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFs1LDQsIiIsMix7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbNCwzLCIiLDIseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV1d)

We'll use multiplicativity of $e$ in towers.
Recall that $e(R/P) = e(Q/R) e(R/P)$, so $e(R/P) = 1$ iff $e(Q/P) = e(Q/R)$.
Interpreting this as the size of inertia, this happens iff $\# I(Q/P) = \# I(Q/R)$, iff $\#I(Q/P) = \#( I(Q/P) \intersect G(L/M) )$.
This happens iff $I(Q/P) \subseteq G(L/M)$, iff $M \subseteq L^{I(Q/P)}$.
:::

:::{.theorem title="?"}
The decomposition field $L^{D(Q/P)}$ is the largest $M$ for which *both* $e(R/P) = f(R/P) =1$.
:::

:::{.proof title="?"}
Replace $e$ in the previous proof with the product $ef$ and $I$ replaced by $D$, using $\#D =ef$.
:::

:::{.remark}
Next time: why these theorems are interesting!
How ramification and splitting completely behaves in fields vs their Galois closures.

:::