Existence of Conductor for Cyclotomic Extension (pg 200, Serge Lang A.N.T.) Let $\zeta$ be a primitive $m$th root of unity, $m \not\equiv 2 \pmod 4$, and $K = \mathbb{Q}(\zeta)$.  For $p$ prime and unramified i.e. $(m,p) = 1$, I know that the Artin symbol $(p, K/\mathbb{Q})$ is exactly the map $\zeta \mapsto \zeta^p$.  Therefore for $a \in \mathbb{Q}$, relatively prime to $m$, I believe that $$(a, K/\mathbb{Q}) = (\zeta \mapsto \zeta^{\prod\limits_p p^{\nu_p(a)}})$$ where by $p^{-1}$ in the exponent I mean any integer $x$ which is an inverse of $p$ modulo $m$.  Serge Lang claims that whenever $d > 0$ is a natural number for which $a \equiv d \pmod{^{\ast}m}$ (meaning $\nu_p(a-d) \geq \nu_p(m)$ for all $p$ dividing $m$), it follows that $$(a, K/\mathbb{Q}) = (\zeta \mapsto \zeta^d)$$ 

Why is this?

Also, from this claim it follows that if $a \equiv 1 \mod{^{\ast} m}$, then $(a, K/\mathbb{Q})$ is in the kernel of the Artin map.  Serge Lang claims that the converse is also true.  Why is this?
 A: An answer I think mostly answers my question.  The Artin map for a cyclotomic extension is easier to understand if we write it as a composition $$P(m) \rightarrow (\mathbb{Z}/m\mathbb{Z})^{\ast} \xrightarrow{\cong} Gal(K/\mathbb{Q})$$ where $P(m)$ is the group of principal (fractional) ideals relatively prime to $m$.  Writing any $(a) \in P(m)$ as $(\frac{b}{c})$ for relatively prime integers $b, c$, this is just the map $(\frac{b}{c}) \mapsto bc^{-1} \mapsto (\zeta \mapsto \zeta^{bc^{-1}})$, where $c^{-1}$ is an integer whose product with $c$ is $\equiv 1 \pmod m$.
There is another way of thinking about notion of $\mod^{\ast}$ congruence, namely that for any prime $p$ dividing $m$, $\frac{b}{c}$ lies in $\mathbb{Z}_p$, and $a \equiv d \pmod{^{\ast} m}$ (for a fixed integer $d$) means that $$a \equiv d \pmod{p^{\nu_p(m)}\mathbb{Z}_p}$$ for all $p$ dividing $m$.  Hence $\frac{b - cd}{p^{\nu_p(m)}} \in \mathbb{Z}_p \cap \mathbb{Q} = \mathbb{Z}$, so $b \equiv cd \pmod{p^{\nu_p(m)}}$.  By CRT, this gives us $b \equiv cd \pmod m$, or $bc^{-1} \equiv d \pmod m$.  Conversely $bc^{-1} \equiv d \pmod m$ gives you $\frac{b}{c} \equiv d \pmod{^{\ast} m}$, and this is just as easy to see.  
