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Consider the field $\mathbb{F}_q$ where $q=p^k$ for some prime $p$. I have some identities related to binomial coefficients over such a field, which I wish to prove. So, can someone tell me a source where I could read up on these?

An example of the identities that I am looking out for is the following. For some $a$, such that $0 \leq a < q$, $${q(q-1)\choose(q-a)(q-1)} + {(q-1)(q-1)\choose(q-a)(q-1)} + {(q-2)(q-1)\choose(q-a)(q-1)} + . . . + {(q-a)(q-1)\choose(q-a)(q-1)} = 1$$

in $\mathbb{F}_q$. I have some more similar identities that I wish to prove and I would like it if someone could give me a hint / a strategy / a reference for the same. (Please comment if you wish to see more context)

MathManiac
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  • What's the role of the field $\Bbb{F}_q$ here? It seems to me that $a$ is just an integer, so you are just looking at binomial coefficients modulo $p$. That is, in the field $\Bbb{F}_p$. In that case you may benefit from studying Lucas Correspondence. For an on-site proof of the identity see for example here (Sorry about linking to my own answer. It is just the one I could find fast). – Jyrki Lahtonen Jun 03 '16 at 05:18
  • Yeah, there's no role of $\mathbb{F}_q$ doesn't have any role here. It plays a role in the original problem that I am working on (not mentioned here) – MathManiac Jun 03 '16 at 05:35
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    Makes sense :-) And I'm having second thoughts about Lucas... :-( – Jyrki Lahtonen Jun 03 '16 at 05:40

1 Answers1

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I found the solution to the above problem. A general term of the above sum looks like ${(q-t)(q-1) \choose (q-a)(q-1)}$ which is the coefficient of $x^{(q-a)(q-1)}$ in $(1+x)^{(q-t)(q-1)}$. Hence, the sum is the coefficient of $x^{(q-a)(q-1)}$ in $(1+x)^{q(q-1)}+(1+x)^{(q-1)^2}+...+(1+x)^{(q-a)(q-1)}$. We can continue this sum up till 1 as none of those terms will alter any coefficient of $x^{(q-a)(q-1)}$. So, we are looking for the coefficient of $x^{(q-a)(q-1)}$ in:

$$(1+x)^{q(q-1)}+(1+x)^{(q-1)^2}+...+(1+x)^{(q-a)(q-1)} + (1+x)^{(q-a-1)(q-1)} + . . . + 1$$ $$=\frac{(1+x)^{(q-1)(q+1)}-1}{(1+x)^{q-1}-1}$$ $$=\frac{\frac{(1+x)^{q^2}}{1+x}-1}{\frac{(1+x)^q}{1+x}-1}$$ $$=\frac{\frac{1+x^{q^2}}{1+x}-1}{\frac{1+x^q}{1+x}-1}$$ $$=\frac{x^{q^2}-x}{x^q-x}$$ $$=\frac{x^{q^2-1}-1}{x^{q-1}-1}$$ $$=1+x^{q-1}+x^{2(q-1)}+...+x^{q(q-1)}$$

And hence, the required coefficient is $1$.

MathManiac
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