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I have this exercise :

$q = p^r$ (it is not clearly precised in the exercise that $r = n$). and $\mathbb{F}_q$ is the finite field of cardinal $q$. Let's $P \in \mathbb{F}_q[X_1, ..., X_n]$ of degree $d < n$. Show that $p$ divides the number of solutions in $\mathbb{F}_q^n$ of $P(x_1, ..., x_n) = 0$.

Hint : We can consider the polynomial $P^{q-1}$ and use the fact that if $r < q-1$, then $\sum_{x\in\mathbb{F_q}} x^r = 0$.

But I don't see how to do...

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For all $x_1,x_2,\ldots,x_n\in \mathbb{F}_q$ we either have $P(x_1,x_2,\ldots,x_n)=0$ or $P(x_1,\ldots,x_n)^{q-1}=1$. This is because all non-zero elements of $\mathbb{F}_q$ are roots of the equation $x^{q-1}=1$.

Consequently the number of zeros of $P$ in $\mathbb{F}_q^n$, when viewed as an element of $\mathbb{F}_p$, i.e. reduced modulo $p$, is $$ \sum_{x_1,x_2,\ldots,x_n\in\mathbb{F}_q}P(x_1,x_2,\ldots,x_n)^{q-1}. $$

You are expected to find the value of this sum. I recommend doing it for each term in the multinomial expansion of $P(x_1,\ldots,x_n)^{q-1}$ separately.

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It is $- \sum_{x_1,x_2,\ldots,x_n\in\mathbb{F}_q}P(x_1,x_2,\ldots,x_n)^{q-1}$, isn't it ? But I don't manage to show that it equals to $0$ mod $p$ : the monomials are of degree $\leqslant (q-1)d$, which can be a multiple of $q-1$. – Arnaud Apr 20 '13 at 14:00
@Arnaud: You are on the right track. Next use the assumption $d<n$, and do a partial sum over single variable. – Jyrki Lahtonen Apr 20 '13 at 16:57
I think that I've the solution : if we expand $P(x_1,x_2,\ldots,x_n)^{q-1}$, each term looks like $x_1^{i_1}...x_n^{i_n}$, with $i_1 + ... + i_n \leqslant (q-1)d < (q-1)n$. So there is $k$ such as $i_k < q-1$. So the partial sum over $i_k$ is zero mod $p$, and we have the conclusion. Is it right ? – Arnaud Apr 20 '13 at 17:09
That's it! Well done, Arnaud! – Jyrki Lahtonen Apr 20 '13 at 17:11
I believe this trick is due to either Warning or Chevalley. At least I learned about it in the context of Chevalley-Warning's theorem. – Jyrki Lahtonen Apr 20 '13 at 17:14

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