Frobenius automorphism of a quadratic extension I'd like to calculate the Frobenius automorphism of the quadratic field $K=\mathbb{Q}(\sqrt{d})$ where $d\in\mathbb{Z}$ is squarefree.
As the prime $p$ should be umramified, we have $p\mathcal{O}_K=R_1R_2$ or $p\mathcal{O}_K=R$ with $f(R|p)=2$.
The first case implies $\big(\frac{d}{p}\big)=1$, i.e. $\sqrt{d}\in\mathbb{F}_p$ and $\sqrt{d}^p\equiv \sqrt{d}\mod p$. Thus the identity is the Frobenius automorphism.
What do we get in the second case?
 A: In the second case $\mathcal O_K/p\mathcal O_K$ is a finite field of order $p^2$, it follows that the reduction map $\operatorname{Gal}(K) \to \operatorname{Gal}(\mathcal O_K/p\mathcal O_K)$ is an isomorphism and the Frobenius element is the only non-trivial element in $\operatorname{Gal}(K)$, namely the conjugation $\sqrt{n} \to -\sqrt{n}$.
Ur to put it differently, because $O_K/p\mathcal O_K$ has order $p^2$, the Frobenius $O_K/p\mathcal O_K \to O_K/p\mathcal O_K, \overline{x} \mapsto \overline{x}^p$ is not the identity map, as in a field with $p^2$ elements, there is an element $a$ with $a^p \neq a$, thus the frobenius automorphism of $K$ can't be the identity map, but since $\operatorname{Gal}(K)$ has only two elements, this leaves only one possibility.
A: I wanted to answer to your comment in the previous answer, but I realized that it was too long to be a comment. So I will write it as an answer. But this is just an addition to the previous answer, and therefore I would ask you not to verify this as the valid answer! The answer of the previous user looks perfect to me.
As you said in the comment, the Frobenius automorphism $\sigma \in G := \text{Gal}(K \mid \mathbb{Q})$ is the unique element in the decomposition group $D:=D(R\mid p) \subseteq G$ (in this case $D=G$) with that property. But I think that the nicer conceptual definition is the following:
In any field $F$ of characteristic $p>0$, the Frobenius endomorphism is given by $x\mapsto x^{p}$. In our context, the residue field of $\mathcal{O}_{K}$ at $R$ is a finite separable field extension of the field $\mathbb{F}_{p}$, hence is itself a field of characteristic $p>0$. The Frobenius endomorphism of $\mathcal{O}_{K}/R$ is then a generator of the corresponding Galois group $G':=G(\mathcal{O}_{K}/R \mid \mathbb{F}_{p})$.
On the other hand, the fundamental exact sequence of Galois theory tells us that $D\cong G'$, because $p$ being unramified implies that the inertia subgroup is trivial. Under this isomorphism, the Frobenius endomorphism on $\mathcal{O}_{K}/R$ corresponds to a generator of the decomposition group $D$. And this is the Frobenius element of $R$ over $p$.
In our particular case, since we are looking at $R=p\mathcal{O}_{K}$, we have $D=G$, so the Frobenius element must be a generator of $G$. And there is only one generator, so we are done.
A: In the particular case  of an abelian field $K$, given a prime $P$ of $O_K$ over a prime number $p$ unramified in $K$, the Frobenius automorphism $\phi_{p,K}$ of $K$ (it depends only on $p$) is the unique automorphism of $K$ s.t. $\phi_{p,K} (x) \equiv x^p$ mod $P$ for any $x \in O_K$. In other words, $\phi_{p,K}$ is a lift to $O_K$ of the usual Frobenius automorphism relative to the extension of finite fields $(O_K/P)/\mathbf F_p$. Let us now consider a quadratic field $K=\mathbf Q(\sqrt d)$. The prime $p$ is unramified iff $p $ does not divide the discriminant of $K$, which is $d$ or $4d$. If $p$ splits in $K$, obviously $(O_K/P)=\mathbf F_p$ and the two Frobenius are the identity. If $p$ is inert in $K$, obviously  $(O_K/P)=\mathbf F_{p^2}=\mathbf F_p (\sqrt {\bar d})$, where $\bar d$ is the class of $d$, and the two Frobenius send resp. $\sqrt d$ and $\sqrt {\bar d}$ to their opposites.
This is so plain that I guess that your question is actually about some precise relationship between the Frobenius automorphism $\phi_{p,K}$ and the Legendre symbol $(\frac dp)$. This is a natural question : $(\frac dp)$ can obviously be viewed as a generator of $Gal(\mathbf F_p (\sqrt {\bar d})/\mathbf F_p)\cong (\pm 1)$, $\phi_{p,K}$ is generator of $Gal(\mathbf Q(\sqrt d)/\mathbf Q)\cong (\pm 1)$, and what we saw above amounts to $\phi_{p,K}=(\frac dp)$. But that's not all. Take two distinct odd primes $p,q$ , and consider the cyclotomic field $F=\mathbf Q_{\zeta_q}$ (in which $p$ is necessarily unramified). It is classically known that $F/\mathbf Q$ is a cyclic extension of degree $(q-1)$, hence it contains a unique quadratic subfield $K$, and the usual calculation of the discriminant of $F$ implies that $K=\mathbf Q(\sqrt {q^*})$, where $q^*=(-1)^{\frac {q-1}2}$. The same argument as before shows that $\phi_{p,K}=(\frac {q^*}p)$. Moreover, the restriction to $K$ of $\phi_{p,F}\in Gal(F/\mathbf Q)\cong \mathbf F^*_{q}$  coincides naturally with $\phi_{p,K}\in Gal(K/\mathbf Q)\cong (\pm 1)$, so that $\phi_{p,K}=1$ iff $\phi_{p,F}\in Gal(F/K)\cong {\mathbf F^*_{q}}^2$ (= the unique subgroup of index $2$ of $\mathbf F^*_{q}$. This means  that $\phi_{p,K}=(\frac pq)$. Transforming the equality $(\frac pq)=(\frac {q^*}p)$ using Euler's criterion, one gets   $(\frac pq)(\frac qp)=(-1)^\frac{(p-1)(q-1)}4$. Among the numerous proof of the quadratic reciprocity law, this one stands out as the very first embryo of the so called reciprocity for "power residue symbols" which appeared later in class field theory.
