Let $q$ be a power of a prime $p$, $m$ be an integer greater than $0$. Does there exist polynomials $P_0,\cdots,P_m$ of $\mathbb F_q[T]$ not all in $\mathbb F_q$ such that there exist polynomials $Q_0,\cdots, Q_m$ of $\mathbb F_q[T]$ $\mathbb F_q$-linearly independent satisfying the following condition: for all integer $k$ large enough $$\deg\left(P_0Q^{q^k}_0+\cdots+P_mQ^{q^k}_m\right)<q^k\max_{0\le i\le m}\deg(Q_i).$$
Thanks in advance
Yes. Let $P_0 = T$, $P_1 = -T$, $Q_0 = T^2+1$, $Q_1=T^2$. The degree of $(T^2+1)^{q^k}-(T^2)^{q^k}$ is at most $2q^k-2$ (the terms of degree $2q^k$ cancel), so the degree of $T(T^2+1)^{q^k}-T(T^2)^{q^k}$ is at most $2q^k-1 < 2q^k$.
If the $P_i$ are not $\mathbb{F}_q$-linearly dependent, then no. For sufficiently large $k$, all of the high-degree terms in $Q_i^{q^k}$ (except the $q^k$th power of the highest-degree term in $Q_i$) will have a factor of $q$ in their coefficients (by the multinomial theorem) and therefore be $0$. So to ensure all of the highest-degree terms sum to $0$, the sum of $c_i^{q^k}P_i$ (where $c_i$ is the coefficient of the highest-degree term of $Q_i$) must be $0$, which implies the $P_i$ are linearly dependent over $\mathbb{F}_q$.
