# Ramification index and also inertia level

Let $L,K$ be number areas and also $L|K$ a galois expansion. Allow $(0)\neq Q$ a prime perfect in $\mathcal O_L$ (= ring of integers in $L$) and also $P=Q \cap \mathcal O_K$.

$Z_Q $ represents the disintegration area of $Q$ and also $T_Q$ represents the inertia area of $Q$.

Currently placed $Q' :=Q\cap Z_Q$ and also $Q'' :=Q\cap T_Q$.

Just how does one confirm, that $e(Q|Q'')=e(Q|P)$ and also $f(Q''|Q')=f(Q|P)$, if $e$ represents the implication index and also $f$ the inertia level?

This is basically the argument in Daniel Marcus is *Number Fields * (which I really extremely advise, specifically for its workouts), Theorem 28 on web page 100.

Allow $G=\mathrm{Gal}(L/K)$ be the Galois team. The disintegration team of $Q$ is $$D = \{ \sigma\in G\mid \sigma Q = Q\}$$ and also the inertia team of $Q$ is $$E = \{\sigma\in G\mid \sigma(a)\equiv a\pmod{Q}\text{ for all $a\in \mathcal{O}_L$}\}.$$ Then $Z_Q$ is specified to be the set area of $D$, and also $T_Q$ is the set area of $E$.

Customarily, allow $[L:K]=n = efr$, where $e$ is the implication level and also $f$ is the inertia level.

First, I assert that $[Z_Q:K]=r$. Without a doubt, by the Fundamental theory of Galois Theory, $[Z_Q:K] = [G:D]$. If $\tau\in G$, after that every component of the coset $\tau D$ maps $Q$ to $\tau Q$ ; in addition, if $\tau Q=\rho Q$, after that $\rho^{-1}\tau \in T$. So we have a one - to - one document in between the cosets of $D$ in $G$, and also the tops over $P$ of the kind $\tau Q$ with $\tau\in G$. Given that $L$ is Galois over $K$, the activity of $G$ is transitive on the tops existing over $P$, and also there are $r$ of them. So the index $[G:D]$ amounts to $r$, therefore $[Z_Q:K]=r$, as asserted.

Next, we show that $e(Q'|Q)=f(Q'|Q)=1$. Notification that $Q$ is the only prime of $\mathcal{O}_L$ that exists over $Q'$ : due to the fact that $\mathrm{Gal}(L/Z_Q)=D$ by the Fundamental Theorem of Galois Theory, and also $D$ acts transitively on the tops of $\mathcal{O}_L$ existing over $Q'$ ; yet every component of $D$ maps $Q$ to itself, so $Q$ is the only prime existing over $Q'$. Given that $[L:Z_Q]=e(Q|Q')f(Q|Q')r(Q|Q') = e(Q|Q')f(Q|Q')$, and also given that $erf=[L:K]=[L:Z_Q][Z_Q:K]=[L:Z_Q]r$, after that $[L:Z_Q]=ef$. And also given that $e(Q|Q')\leq e$ and also $f(Q|Q')\leq f$, it adheres to that we have to have $e(Q|Q')=e$, $f(Q|Q')=f$. And also given that $e=e(Q|P) = e(Q|Q')e(Q'|P)$, and also $f=f(Q|P) = f(Q|Q')f(Q'|P)$, after that $e(Q'|P)=f(Q'|P)=1$.

Specifically, given that $1 = e(Q'|P) = e(Q'|Q'')e(Q''|P)$, after that $e(Q'|Q'')=1$. Hence, $e(Q|Q'') = e(Q|Q')e(Q'|Q'') = e = e(Q|P)$, which confirms the first equal rights.

Ultimately, we show that $f(Q|Q'') = 1$, or equivalently that $\mathcal{O}_{T_Q}/Q''$ amounts to $\mathcal{O}_L/Q$. For this it is adequate to show the equivalent Galois team is unimportant. If we can develop this, after that we will certainly have $f = f(Q|Q') = f(Q|Q'')f(Q''|Q') = f(Q''|Q')$, which will certainly offer the 2nd equal rights you desire.

To show that $\mathcal{O}_L/Q$ is the unimportant expansion of $\mathcal{O}_{T_Q}/Q''$, it suffices to show that for every single $\overline{a}\in\mathcal{O}_L/Q$, the polynomial $(x-\overline{a})^m$ hinges on $\mathcal{O}_{T_Q}/Q''[x]$ for some favorable $m$. If this holds true, after that every component of the Galois team have to send $\overline{a}$ to itself. Select any kind of preimage $a\in \mathcal{O}_L$ of $\overline{a}$. After that the polynomial $$\prod_{\sigma\in E}(x - \sigma(a))$$ has coefficients in the set area of $E$, that is, in $T_Q$ ; lower modulo $Q''$ to get a polynomial with coefficients in $\mathcal{O}_{T_Q}/Q''$ ; given that $\sigma(a)\equiv a \pmod{Q}$ for all $\sigma\in E$, the lowered polynomial is of the kind $(x-\overline{a})^m$, with $m=|E|$. This confirms that every component of $\mathcal{O}_L/Q$ is dealt with by every component of the Galois team, so the expansion is unimportant, therefore the inertia level is $1$. This confirms the 2nd equal rights, as laid out over.