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  "body": "\n## Definition\n\nLet $L/K$ be a finite [[Galois extension|0QMZAQ6QQ4]] of [[algebraic number fields|MWBDFS2KJ5]] with [[Galois group|0XTY86BZKY]] $G$, and let $\\mathcal{O}_K \\subseteq \\mathcal{O}_L$ be the rings of integers.\nLet $\\mathfrak{p}$ be a nonzero prime ideal of $\\mathcal{O}_K$ that is [[unramified|H5RC67SD24]] in $L$, and let $\\mathfrak{P}$ be a prime ideal of $\\mathcal{O}_L$ over $\\mathfrak{p}$.\nLet $q = N(\\mathfrak{p}) = \\abs{\\mathcal{O}_K/\\mathfrak{p}}$ be the [[norm|M54ADM0CVF]] of $\\mathfrak{p}$.\nThe *Frobenius element* of $\\mathfrak{P}$ is the unique $\\sigma \\in G$ such that\n$$\\sigma(x) \\equiv x^{q} \\pmod{\\mathfrak{P}} \\quad \\text{for all } x \\in \\mathcal{O}_L.$$\nIt is written $\\mathrm{Frob}_{\\mathfrak{P}}$ or $\\left(\\frac{L/K}{\\mathfrak{P}}\\right)$.\n\n## Remarks\n\nExistence and uniqueness: such a $\\sigma$ fixes $\\mathfrak{P}$, so it lies in the [[decomposition group|7X92G8CQ33]] $D_{\\mathfrak{P}}$.\nThe 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.\nSo exactly one element of $D_{\\mathfrak{P}}$ maps to the [[Frobenius automorphism|0CGCPX8Z70]] $y \\mapsto y^q$ of the residue field.\n$\\mathrm{Frob}_{\\mathfrak{P}}$ generates $D_{\\mathfrak{P}}$, and its order is the residue degree $f$.\n\nFor $\\tau \\in G$, $\\mathrm{Frob}_{\\tau(\\mathfrak{P})} = \\tau \\, \\mathrm{Frob}_{\\mathfrak{P}} \\, \\tau^{-1}$.\nSo the Frobenius elements of the primes over $\\mathfrak{p}$ form one [[conjugacy class|0DEBXAJ5SA]] $\\mathrm{Frob}_{\\mathfrak{p}}$ of $G$; if $G$ is abelian, it is a single element, the *Artin symbol* of $\\mathfrak{p}$.\n\nExample: $K = \\Q$, $L = \\Q(\\zeta_n)$ with $\\zeta_n$ a primitive $n$-th root of unity, and $p$ a prime not dividing $n$.\nThe 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 $\\mathrm{Frob}_p = \\sigma_p$.\n\nNon-example: in $\\Q(i)/\\Q$, the prime $2$ is ramified.\nBoth 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.\n",
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