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Is there a method to factorise a polynomial, for $k \leq m$ and $a_i \in \mathbb{F}_p$, $$ 1 + t^k(1 + a_1 t + a_2 t + \ldots + a_m t^m)^k $$ as a product $$ (1 + t^k)^{x_1} \cdots (1 + t^l)^{x_l} \cdots (1 + t^m)^{x_m} \pmod{t^{m+1}}$$ where $p \nmid l$. In other words, is there a method to find explicit formulae for the exponents $x_i$ in the above expression?

Here $\mathbb{F}_p$ denotes the prime field of characteristic $p > 0$.

Thanks.

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  • $\begingroup$ A local field is a locally compact topological field with respect to a non-discrete topology. Finite fields don't qualify, do they? $\endgroup$
    – Gerry Myerson
    Aug 24, 2018 at 9:58
  • $\begingroup$ This question arose out of understanding the factorisation of an element of $1 + t^k$ under the substitution in $t$ of an element in the maximal ideal of $\mathbb{F}_p((t))$, the laurent series in one variable $t$ over $\mathbb{F}_p$. $\endgroup$
    – Vanya
    Aug 24, 2018 at 10:04

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You can do it by induction. Just note that $(1+t^r)^a=1+at^r+O(t^{r+1})$, so in your case $x_1 = 1$ and you compute $(1+t^k +ka_1t^{k+1}+\cdots)/(1+t^k) = 1+ka_1t^{k+1}+\cdots$, so you divide by $(1+t^{k+1})^{ka_1}$ and continue. I.e. $x_2=ka_1$ and $l=k+1$ if $a_1\ne0$. If $a_1=0$ then you need to figure out the first non-zero coefficient and continue.

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  • $\begingroup$ Thank you for the answer. I am actually looking for a closed formula for the exponents. $\endgroup$
    – Vanya
    Aug 24, 2018 at 4:48
  • $\begingroup$ @user49908 Do you have any reason to expect a closed form? $\endgroup$
    – Felipe Voloch
    Aug 24, 2018 at 7:43

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