Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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...

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer
    
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
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.