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?
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- 2$\begingroup$ Could you please give some context that explains where this question is coming from? The formulation does not sound like a random question asked purely out of curiousity. $\endgroup$KConrad– KConrad2019-05-23 19:23:33 +00:00Commented May 23, 2019 at 19:23
- 1$\begingroup$ $\alpha=\beta=0$? $\endgroup$Fedor Petrov– Fedor Petrov2019-05-23 19:29:22 +00:00Commented May 23, 2019 at 19:29
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1 Answer
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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. $\alpha=\beta=0$.
