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I am going through a paper, and theres one point in the proof I've been having problems understanding.

We have $\alpha \in G$ with minimal polynomial $h$, where $G$ is the algebraic closure of a field $F$, a finite field of prime order $q$. Also take $[F(\alpha):F]=d$, and take $P_q(n)$ as the set of all monic polynomials over the field of order $q$ of order $n$. We have $1\leq k

I'm having problems showing that: $$[ \prod_{i=1}^n(x-x^{q^i})]^k \cdot \sum_{f\in P_q(n), h \nmid f}\frac{1}{f^k} \equiv 0 \pmod {(x-\alpha)^{k\lfloor \frac{n}{d} \rfloor}}$$

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    The algebraic closure of a field is always infinite. Since $F$ is finite, we will have $[G:F]=\infty$2012-10-23
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    What is $k$? An arbitrary integer?2012-10-23
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    Any what is $n$, another arbitrary integer? By the way the $n$ in $P_q(n)$ is probably meant to do something; is it the degree of the monic polynomials?2012-10-23
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    What paper is it? This might help us2012-10-23
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    Fixed some things that should answer some questions. Also, the title of the paper is "Sums of REciprocals of Polynomials over Finite Fields" by Kenneth Hicks, Xiang-dong Hou, and Gary L. Mullen.2012-10-23

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