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I'm looking for a way to prove the following proposition, which is an exercise from a book about Algebraic Geometric Codes that I'm trying to read:

If $\mathcal P$ is a set of $n$ points in $\Bbb F_q$ (a field with cardinality $q$) and $g_i(x)=\prod_{P_j\in\mathcal P,P_j\neq P_i} (x-P_j)^{-1}$, then for any polynomial $f$ with $\deg f\leq n-2$ we have that $$\sum_{P_i\in\mathcal P}f(P_i)g_i(P_i)=0 $$

Note that this is somewhat similar to the residue formula over $\Bbb C$ in the sense that the function $g$ is holomorphic everywhere except for a finite set of points (maybe including the point at infinity in the case of $\Bbb C$), and because we may think of the $i$'th summand as $\text{Res}_{P_i}fg$ where $g=\prod_{P_i\in\mathcal P}(x-P_i)^{-1}$

I tried to mimic the residue formula proof for $\Bbb C$ but I couldn't make sense of how to cope with the integral along a circle (what is the radius, and how to even translate these points to being inside/outside of a circle).

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  • $\begingroup$ "except for a finite set of points" ... There are only finitely many points in $\Bbb{F}_q$. So even if $\mathcal{P} = \Bbb{F}_q$, $g$ is "holomorphic" (whatever that means when there are no complex numbers in sight) "everywhere except for a finite set of points". Of course, every $\Bbb{F}_1 \smallsetminus \mathcal{P}$ is a finite set, so this statement tells us very little. $\endgroup$
    – Eric Towers
    Mar 16, 2021 at 9:41
  • $\begingroup$ Of course. That was meant more for the analogous $\Bbb C$ version, in my case it is kind of moot. $\endgroup$
    – NL1992
    Mar 16, 2021 at 9:53
  • $\begingroup$ There appears to a typo somewhere in what you've written. If $\ q=5\ $, $\ P_1=0, P_2=1\ $, and $\ f(x)\ $ is the degree zero polynomial $\ f(x)=2\ $, then $\ g(x)=\frac{1}{x(x-1)}\ $, and $\ \sum_\limits{i=1}^2f(x)g(x)(x-P_i)=\frac{2}{x-1}+\frac{2}{x}=\frac{2(2x-1)}{x(x-1)}\ $. This expression is zero only for $\ x=3\ $. It is undefined at $\ x=0\ $ and $\ x=1 $, and has values $\ 3\ $ and $\ 2\ $ at $\ x=2\ $ and $\ x=4\ $ respectively, so not even when it's well-defined is it always zero. $\endgroup$
    – lonza leggiera
    Mar 16, 2021 at 11:31
  • $\begingroup$ @lonzaleggiera Thank you, I did make a mistake, I was trying to make this look as similar to the complex version as possible and didn't plug in $P_i$ into $g$. Edited. $\endgroup$
    – NL1992
    Mar 16, 2021 at 13:36
  • $\begingroup$ Rather than comparing with Cauchy's integral formula, you may get a lot more mileage comparing with the Lagrange interpolation formula (which observation was only possible once you "plug[ged] $P_i$ into $g$"). However, this is probably not an effective way to make progress. $\endgroup$
    – Eric Towers
    Mar 16, 2021 at 18:51

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