Let $P$ be an irreducible polynomial of $\mathbb F\_q[T]$ of degree $2$. Does there exist two polyomials $\alpha,\beta\in\mathbb F\_q[T]$ (not both zeroes) such that the sequence $(\beta T^{q^{2n}}-\beta T+\alpha)\_{n\in\mathbb N}$ tends to $0$ for the topolgy induced by the $P$-adic valuation?



Converging sequence of polynomials
The answer is no:


The difference of the $(n+1)$-st and the $n$-th element in the sequence is
$\beta T^{q^{2(n+1)}}-\beta T^{q^{2n}}=\beta(T^{q^2}-T)^{q^{2n}}$ which is divisible by $P^{q^{2n}}$ as $P$ is irreducible of degree $2$.


For the sequence converging to zero P-adically, one therefore needs
$\alpha+\beta ( T^{q^{2n}}-T)\equiv 0 \mod P^{q^{2n}}$ for every $n$.


But given $\alpha$ and $\beta$ with degrees $a$ and $b$, respectively, one can choose $n$ such that $q^{2n}>\max\{a,b\}$. Then $\deg(\alpha)=a<\deg(\beta ( T^{q^{2n}}-T))=b+q^{2n}<2\cdot q^{2n}=\deg(P^{q^{2n}})$.
For $\alpha+\beta ( T^{q^{2n}}-T)$ being congruent to zero modulo $P^{q^{2n}}$, it therefore has to be identically zero, i.e. $\alpha=\beta=0$.


