Frobenius element

Remarks

Existence and uniqueness: such a $\sigma$ fixes $\mathfrak{P}$, so it lies in the decomposition group $D_{\mathfrak{P}}$. The map from $D_{\mathfrak{P}}$ to the Galois group of the residue extension is onto, with kernel the inertia group, which is trivial because $\mathfrak{p}$ is unramified. So exactly one element of $D_{\mathfrak{P}}$ maps to the Frobenius automorphism $y \mapsto y^q$ of the residue field. $\mathrm{Frob}_{\mathfrak{P}}$ generates $D_{\mathfrak{P}}$, and its order is the residue degree $f$.

For $\tau \in G$, $\mathrm{Frob}_{\tau(\mathfrak{P})} = \tau \, \mathrm{Frob}_{\mathfrak{P}} \, \tau^{-1}$. So the Frobenius elements of the primes over $\mathfrak{p}$ form one conjugacy class $\mathrm{Frob}_{\mathfrak{p}}$ of $G$; if $G$ is abelian, it is a single element, the Artin symbol of $\mathfrak{p}$.

Example: $K = \Q$, $L = \Q(\zeta_n)$ with $\zeta_n$ a primitive $n$-th root of unity, and $p$ a prime not dividing $n$. The automorphism $\sigma_p$ with $\sigma_p(\zeta_n) = \zeta_n^p$ satisfies $\sigma_p(x) \equiv x^p$ modulo every prime over $p$, since $(a + b)^p \equiv a^p + b^p$ modulo $p$; so $\mathrm{Frob}_p = \sigma_p$.

Non-example: in $\Q(i)/\Q$, the prime $2$ is ramified. Both elements $\sigma$ of $G$ satisfy $\sigma(x) \equiv x^2 \pmod{(1 + i)}$, because the residue field is $\F_2$ and $i \equiv -i \equiv 1 \pmod{(1+i)}$; so there is no unique Frobenius element.