Where have I gone wrong in calculating norm residue symbol. It is well known that the reciprocity map in local class field theory gives reciprocity map in Global class field theory. Namely, if $(\cdot,L_\beta/K_P):K_P^\times \rightarrow G(L_\beta/K_P)$ is the local reciprocity map for place $P$ in global field $K$, then global Artin map $[\cdot,L/K]$:$I_K\rightarrow G(L/K)$ is given by $\prod_P (\cdot,L_\beta/K_P)$. The kernel of this contains $K^\times$ and I understand this. But I tried to verify something explicitly and getting negative result so I would like to know where I have gone wrong. 
So now let $K$ be a rational function field over one variable over finite field $\mathbb{F}_q$ where $q=p^n$ for a prime $p$ and $L$ be its constant extension of degree $m$.
Let $Q$ be a place in $K$ which does not split in $L$ and $f_Q$ be corresponding irreducible polynomial.
Then $[f_Q,L/K]=\prod(f_Q,L_\beta/K_P)$ where $\beta|P$. All factors in the right hand side is 1 except for the case $P=Q$ and $P=\infty$ as all extension $L_\beta/K_P$ is constant and hence unramifed and image of $f_Q$ in those field is a unit. In the infinity case I get $\phi^{-deg(Q)}$ and when $P=Q$ we get $\phi$ where $\phi$ is used to denote the generator of $G(L/K)$. This is because $G(L/K)\cong G(L_{\beta'}/K_Q)\cong G(L_\infty/K_\infty)$ as those places do not split. (Also because $(a,\bar{\mathbb{F}_q}((T))/\mathbb{F}_q((T)))=\phi^{v(a)}$ where in this case I mean $\phi$ as a generator of $G(\bar{\mathbb{F}_q}((T))/\mathbb{F}_q((T)))$ 
But there is no reason $-\deg(Q)+1$ must be divisible by $n$. Obviously I am wrong somewhere but I can't seem to figure out at the moment.
 A: You are beginning with a global field $K$ of char. $p$, containing $\mathbb F_q$ as its maximal finite subfield, and forming the (everywhere unramified) extension $L = \mathbb F_{q^m} \otimes_{\mathbb F_q} K.$
If $\mathfrak p$ is a prime of $K$, with residue field $\kappa(\mathfrak p)$,
then the primes $\mathfrak q$ of $L$ over $\mathfrak p$ correspond to the
various field factors of the tensor product $\mathbb F_{q^{m}} \otimes_{\mathbb F_q} \kappa(\mathfrak p).$
Note that Gal$(L/K)$ is canonically isomorphic to Gal$(\mathbb F_{q^m}/\mathbb F_q)$. If $\mathfrak q$ lies over $\mathfrak p$, then the extension $\kappa(\mathfrak q)/\kappa(\mathfrak p)$ is a direct factor of $\mathbb F_{q^m} \otimes_{\mathbb F_q} \kappa(\mathfrak p).$  This tensor product has an obvious action of Gal$(\mathbb F_{q^m}/\mathbb F_q)$ (via its action on the left factor in the tensor product), 
and the decomposition group $D_{\mathfrak q}$ is the subgroup which preserves
$\kappa(\mathfrak q)$.
In particular, if $\kappa(\mathfrak p)  = \mathbb F_q$, then the tensor product is just equal to $\mathbb F_q$, and so there is a single prime $\mathfrak q$ over $\mathfrak p$, of degree $n$.   Furthermore, in this case $D_{\mathfrak q} = $ Gal$(L/K)$.

As in your post, let's now take $K = \mathbb F_q(x)$.
If $\mathfrak p$ corresponds to the point $x = 0$, then the residue field is $\mathbb F_q,$ and so there is a single prime lying over infinity, of degree $m$.
If $\mathfrak p$ is the point at infinity, then $\kappa(\mathfrak p) = \mathbb F_q,$
and thus there is also a single prime lying over infinity, of degree $m$.
In each case the decomposition group is equal to all of Gal$(L/K) =$ Gal$(\mathbb
F_{q^m}/\mathbb F_q)$.
In particular, in each case the Frobenius element at $\mathfrak p$ is equal
to the Frobenius element Frob${}_q$ in Gal$(\mathbb F_{q^m}/\mathbb F_q)$.
Let's take $f = x$.  Then the norm residue symbol of $x$ at $\mathfrak p$ is trivial except for $\mathfrak p$ corresponding to $0$ or $\infty$.
At $0$, we see that $x$ has a zero of order $1$, while at $\infty$, it has a pole of order $1$.  Thus the Artin map sends $x$ to Frob${}_q \cdot $Frob${}_q^{-1}$, i.e. to $1$, which is what you wanted to check.
You should try some other examples yourself.  E.g. if $f$ is irreducible of degree two, and $n$ is odd, then there will again be a unique prime $\mathfrak q$ over $\mathfrak p$, and $\kappa(\mathfrak q) = \mathbb F_{q^{2n}}.$   The Frobenius element Frob${}_{\mathfrak p}$ is then equal to Frob${}_{q^2} = $Frob${}_{q}^2$.
The order of the pole  of $f$ at $\infty$ is $2$, so in this case the global Artin map takes $f$ to Frob${}_{q^2}\cdot $ Frob${}_q^{-2} = $ Frob${}_q^2 \cdot$
Frob${}_q^{-2} = 1$ (as it must!).
A: I am not 100 percent sure but I think I figured it out.
This happens because image of $\phi$ in $G(L_{\beta'}/K_Q)$ isn't necessarily cannonical Frobenius generator. As those two groups are isomorphic it will be a generator but not necessarily the one we want to make use of local theory.
