I'm studying number theory on Marcus book and at a certain point I'm required to prove the following facts about the Frobenius automorphism.
We start with a lemma and then are required to specialize it
Lemma: Let K be a number field, and L,M two finite extensions of it, let now P, Q, U, V be primes in K, M, L ,ML such that Q, U lie over P and V lies over Q and U. Then if P is unramified in M $\phi(V|Q)$ restricted to M is equal to $\phi(U|P)^{f(Q|P)}$
Let now $$\begin{matrix} K&\subset& L&\subset& M&\\ \cup & & \cup && \cup\\ P&\subset& Q&\subset& U\end{matrix} $$, here P is unramified in M so that we can use the lemma.
I want to prove the following two things:
$\phi(U|Q)=\phi(U|P)^{f(Q|P)}$
If L is normal of K then $\phi(Q|P)$ is the restriction of $\phi(U|P)$ to L.
I don't really get what is meant by restriciton to [] and so I'm finding hard to find an answer to my questions.
Can you help me?
The first statement you have to prove is related to the property of the norm of the ideal $P$, we have that, since $Q$ lies over $P$, $||Q||=||P||^{f}$ (This happens because P is unramified, so $PO_L=Q_1\cdots Q_r$ then $||PO_L||=||P||^{rf}=||Q||^{r}$). For the second point, you have $\phi (U|P)$ and $\phi (Q|P)$ the first morphism acts (comes from a morphism of the Galois group) on the field $O_M/U$ sending an element $a$ to $a^{||P||}$. The second morphism does the same thing but acts on the field $O_L/Q$. What you have to prove is, when you know that a morphism of the first group is associated to a certain morphism in the Galois group, you take this last one and restrict to $L$. Now you check whether or not this morphism in the new galois group correspond to the automorphism of the finite field you were looking for.