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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 $1\leq k<q$.

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 $[G:F]=\infty$ –  Dennis Gulko Oct 23 '12 at 15:55
    
What is $k$? An arbitrary integer? –  Thomas Andrews Oct 23 '12 at 16:11
    
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? –  Marc van Leeuwen Oct 23 '12 at 16:19
    
What paper is it? This might help us –  Alexander Gruber Oct 23 '12 at 16:39
    
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. –  Frank White Oct 23 '12 at 16:59

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