= Solution
Use arithmetic <Frobenius automorphisms>: residues are raised to the size of the base <residue field>. Let $S$ contain the finite primes ramified in the finite <abelian extension> $L/K$. The <Artin map> is the <group homomorphism>
$$
\operatorname{Art}_{L/K}:I_K^S\longrightarrow\operatorname{Gal}(L/K),\qquad
\prod_{\mathfrak p\notin S}\mathfrak p^{n_{\mathfrak p}}
\longmapsto\prod_{\mathfrak p\notin S}\operatorname{Frob}_{\mathfrak p}^{n_{\mathfrak p}},
$$
where $I_K^S$ is the <group> of <fractional ideals> supported outside $S$. The <Artin symbol> $\operatorname{Frob}_{\mathfrak p}$ is independent of the prime above $\mathfrak p$, since the extension is abelian. <Commutativity> makes the displayed extension from <prime ideals> well defined. Defining this map does not yet assert the <Artin reciprocity law>, which describes its congruence <group kernel>.
For an intermediate <field> $K\subseteq E\subseteq L$, restriction gives
$$
\boxed{\operatorname{Art}_{L/K}(\mathfrak a)|_E=\operatorname{Art}_{E/K}(\mathfrak a).}
$$
To prove it, start with an unramified prime $\mathfrak p$ and a prime of $L$ above it. Its <Frobenius automorphism> acts on the residue of every <integral element> of $E$ by raising it to $N\mathfrak p$. Its restriction is therefore the <Frobenius automorphism> for $E/K$ at the restricted prime. The identity for all allowed <fractional ideals> follows by the <group homomorphism> property. On <Galois groups>, restriction is the quotient by $\operatorname{Gal}(L/E)$, so passing to a subextension quotients the <Artin map> accordingly.
Now let $E/K$ be any finite extension and put $M=LE$. This is the translated extension: $M/E$ is abelian, and restriction identifies $\operatorname{Gal}(M/E)$ with $\operatorname{Gal}(L/(L\cap E))$. A prime $\mathfrak q$ of $E$ above a prime $\mathfrak p\notin S$ is unramified in $M/E$, because forming a <compositum> with $E$ preserves an <unramified extension>. Put $r=f(\mathfrak q/\mathfrak p)$. The residue-field sizes satisfy $N\mathfrak q=(N\mathfrak p)^r$. Consequently the restriction to $L$ of Frobenius at $\mathfrak q$ acts on its integral residues by raising them to $(N\mathfrak p)^r$, giving
$$
\operatorname{Frob}_{\mathfrak q}(M/E)|_L
=\operatorname{Frob}_{\mathfrak p}(L/K)^r.
$$
But the <ideal norm> is $N_{E/K}\mathfrak q=\mathfrak p^r$. Multiplying this prime-by-prime identity proves
$$
\boxed{\operatorname{Art}_{M/E}(\mathfrak a)|_L
=\operatorname{Art}_{L/K}(N_{E/K}\mathfrak a).}
$$
This is <restriction and norm compatibility of the Artin map>. The <fractional ideal> $\mathfrak a$ must be supported over primes unramified in $L/K$. No assumption that $E/K$ itself is unramified or Galois is needed. With geometric rather than <arithmetic Frobenius>, both maps are inverted and the compatibility identities remain unchanged.
Back to article page
