I am reading about Quadratic residues codes and i dont understand this part..
Theorem:
The polynomials $g_Q(x)$ and $g_N(x)$ belong to $\mathbb{F}_l[x].$
Proof: It's sufficient to show that each cofficient of $g_Q(x)$ and $g_N(x)$ belong to $\mathbb{F}_l$
Let $g_Q(x)=a_0+a_1x+...+a_kx^k,$ where $a_i\in \mathbb{F}_{l^m}$ and $k=(p-1)/2$. Raising each coefficient to its $l$th power, we obtain
$$a_0^l+a_1^lx+...+a_k^lx^k= \prod_{r\in Q_p}(x-\alpha^{lr} )= \prod_{j\in lQ_p}(x-\alpha^{j} )= \prod_{j\in Q_p}(x-\alpha^{j} )=g_Q(x)$$
Note that we use the fact that $lQ_p=Q_p$
Why it is sufficient to show that each cofficient of $g_Q(x)$ and $g_N(x)$ belong to $\mathbb{F}_l$
and why this equal $$a_0^l+a_1^lx+...+a_k^lx^k=\prod_{r\in Q_p}(x-\alpha^{lr} )= \prod_{j\in lQ_p}(x-\alpha^{j} )$$
In the finite field $\mathbb{F}_{l^n}$ we have an automorphism $\rho : c \mapsto c^l$ fixing the subfield $\mathbb{F}_{l}$ (it is called the Frobenius).
You can check that if $\sigma$ is an automorphism and if we have a monic polynomial $\sum_{m=0}^k a_m x^m= \prod_{i=1}^k (x-\alpha_i) $ then $\sum_{m=0}^k \sigma(a_m) x^m=\prod_{i=1}^k (x-\sigma(\alpha_i))$ (think to the $a_m$ as polynomial functions of the roots $\alpha_i$).
Thus, if $\sigma$ acts by permuting the $\alpha_i$ then $$\sum_{m=0}^k a_m x^m= \prod_{i=1}^k (x-\alpha_i) =\prod_{i=1}^k (x-\sigma(\alpha_i))=\sum_{m=0}^k \sigma(a_m) x^m$$ ie. $a_n = \sigma(a_n)$ which means $a_n $ belongs to the fixed field of $\sigma$.
In the case of $\sigma = \rho$, its fixed field is $\mathbb{F}_l$, whence $$\sum_{m=0}^k a_m x^m = \sum_{m=0}^k \rho(a_m) x^m \qquad \implies \qquad \sum_{m=0}^k a_m x^m \in \mathbb{F}_l[x]$$