Number of solutions of $P(x_1, ..., x_n) = 0$ in $\mathbb{F}_q^n$

Question

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...

Answer 1

Hints:

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.