Uniqueness of the Frobenius automorphism Let $L/K$ be a Galois extension of algebraic number fields with rings of integers $\mathcal{O}_L$ and $\mathcal{O}_K$ respectively.
Let $\mathfrak{p} \subset \mathcal{O}_K$ be a prime ideal and $\mathfrak{P} \subset \mathcal{O}_L$ a prime above $\mathfrak{p}$.
It can be shown that there exists an automorphism $\varphi_{\mathfrak{P}} \in \textrm{Gal}(L/K)$ satisfying
$$
\varphi_{\mathfrak{P}}(x) \equiv x^q\ (\textrm{mod}\ \mathfrak{P}) \qquad \forall x \in \mathcal{O}_L \tag{1},
$$
where $q:=\lvert \mathcal{O}_K/\mathfrak{p} \rvert$.
This is the Frobenius automorphism of $L/K$ corresponding to the prime $\mathfrak{P}$.
I want to show that the element in $\textrm{Gal}(L/K)$ satisfying $(1)$ is unique.
My idea was to assume that $\sigma, \tau \in \textrm{Gal}(L/K)$ satisfy $(1)$, then define $\upsilon:=\sigma^{-1}\circ \tau$, so that
$$
\upsilon(x) \equiv x\ (\textrm{mod}\ \mathfrak{P})\qquad \forall x \in \mathcal{O}_L. \tag{2}
$$
Or in other words:
$$
\mathfrak{P} \mid \upsilon(x)-x \qquad \forall x \in \mathcal{O}_L. \tag{3}
$$
Then somehow deduce from (3) that
$$
\mathfrak{P} \mid \upsilon(x)-x \qquad \forall x \in L, \tag{4}
$$
from which one would conclude that
$$
\upsilon(x)-x = 0 \qquad \forall x \in L, \tag{5}
$$
and thus find that $\upsilon = \textrm{Id} \in \textrm{Gal}(L/K)$, and hence $\sigma = \tau$.
But as I am sure you will agree, the above is more reminiscent of wishful thinking than actual mathematics.
Does anyone know if there is a standard "book proof" of the statement, or if one might prove it along lines similar to the above?
Many thanks.
 A: Your proof is almost correct. The problem is that the fact that you are trying to prove is false: the Frobenius element need not be unique! For example, in the extension $\mathbb Q(\sqrt{2})/\mathbb Q$, if $\mathfrak p = (2)$ and $\mathfrak P = (\sqrt 2)$, then both elements of $\mathrm{Gal}(\mathbb Q(\sqrt{2})/\mathbb Q)$ are Frobenius elements. However, if $\mathfrak P/\mathfrak p$ is unramified, then the Frobenius element will be unique and your argument is a part of the proof.
Let $D=D_{\mathfrak {P/p}} = \{\sigma\in \mathrm{Gal}(L/K):\sigma(\mathfrak P)= \mathfrak P\}$ be the decomposition group. If $\sigma \in  D$, then $\sigma$ acts on $\mathcal O_L$ and fixes $\mathfrak P$, so descends to an automorphism of $\mathcal O_L/\mathfrak P$. In this way, we get a homomorphism
$$D_{\mathfrak{P/p}}\to \mathrm{Gal}(\mathbb F_\mathfrak P/\mathbb F_{\mathfrak p}).$$
This map is surjective, and any element in the preimage of the Frobenius element of $\mathrm{Gal}(\mathbb F_\mathfrak P/\mathbb F_{\mathfrak p})$ is a Frobenius element in $\mathrm{Gal}(L/K)$.
The kernel of this isomorphism, called the inertia group is
$$I=I_{\mathfrak{P/p}} = \{\sigma \in \mathrm{Gal}(L/K) : \sigma(x) = x\pmod {\mathfrak P}\; \forall x\in\mathcal O_L\}.$$
Note that this group being trivial is exactly what you need to be able to deduce $(5)$ from $(4)$ in your sketch proof.
However, the inertia group need not be trivial: its order is exactly $e(\mathfrak {P/p})$, the ramification index of $\mathfrak{P/p}$. Hence, the Frobenius element is unique if and only if $\mathfrak{P/p}$ is unramified.
