Chevalley's theorem proof I'm trying to prove Chevalley's theorem stating that
$$ \text{If } f \in \mathbb{Z}[x_1, \dots, x_n] \text{ is a form of degree } r < n \text{,}$$
$$ \text{then there exists a nonzero solution of }  f = 0 \pmod{p}$$
To do that, it is sufficient to prove that in this situation the equation can't have exactly one solution (since $ f $ is a form, it has one zero solution). 
I know that for every polynomial (considered $ \pmod{p} $) there exists exactly one polynomial $ \overline{f} $ satisfying:
(a) $ f = \overline{f} $ on all points in $ \mathbb{Z}_p^n $
(b) in all monomials in $ \overline{f} $ the variables $ x_i $ are in powers smaller than $ p $
(c) $\deg \overline{f} \leq \deg f$
Knowing that, I can continue the proof by contradiction:
Suppose we have exactly one solution $ a = (a_1, \dots, a_n) $ of $ f = 0$. Then the polynomial
$$ g = 1 - f^{p-1} $$
Has the property that admits the value $ 1 $ in $ a $ and $ 0 $ everywhere else in $ \mathbb{Z_p} $. I can write down another polynomial with this property:
$$ h = \prod\limits_{i=1}^n \left(1 - (x_i - a_i)^{p-1}\right) $$
It also satisfies condition (b) from the previous lemma. If I knew for certain that $ h = \overline{g} $, then the proof would be complete (since $ n(p-1) \leq r (p-1) $ and it's a contradiction).
But I'm confused here - is it true that if a polynomial satisfies (a) and (b) then it must satisfy (c)? Or is it so just in that case? Or maybe I'm approaching it all wrong.
I would appreciate a hint.
 A: Let's denote $g$ the image in $ \mathbb{F}_p[x_1, \dots, x_n]$, which is still a form of degree $r' < n$, with $r'\leq r$. Now let $V$ the set of zero of $g$ in $(\mathbb{F}_p)^n$. As $(0,\ldots,0)$, $Card(V)\geq 1$. Set $P = 1-g^{p-1}$ and let $x\in (\mathbb{F}_p)^n$. If $x\in V$ you have $P(x)=1$, and if $x\not\in V$, $g(x)$ is not zero, so that $g(x)^{p-1} = 1$ by small Fermat theorem, and thereof $P(x)=0$. For any $h\in \mathbb{F}_p[x_1, \dots, x_n]$ set $S(h) = \sum_{x\in (\mathbb{F}_p)^n} h(x)$. What preceeds shows that $Card(V)\equiv S(P) \textrm{ mod } p$. So that if we show that $S(P) = 0$, we win. But $r'<n$ so that $deg(P)<n(p-1)$, so that $P$ is linear combination (over $\mathbb{F}_p$) of the monomials $X_1^{a_1} \ldots X_n^{a_i}$ with $\sum_{i=1}^{n} a_i < n(p-1)$. If suffices therefore to show that $S(X_1^{a_1} \ldots X_n^{a_i}) = 0$ when $\sum_{i=1}^{n} a_i < n(p-1)$, by additivity of $S$. $S(X_1^{a_1} \ldots X_n^{a_i}) = \prod_{i=1}^{n} S(X_i^{a_i}) $. As $\sum_{i=1}^{n} a_i < n(p-1)$, one of the $a_i$ is $<p-1$, and to conclude, it is sufficient to show that $S(X_i^{a_i}) = 0$ to conclude. As $\mathbb{F}_p^{\times}$ is cyclic of order $p-1$, let $y\in\mathbb{F}_p^{\times}$ such that $y^{a_i} \not=1$. (Possible because $a_i < p$.) As $x\mapsto yx$ is a bijection, you have $$S(X_i^{a_i}) = \sum_{x\in \mathbb{F}_p} x^{a_i} = \sum_{x\in \mathbb{F}_p} (yx)^{a_i} = \sum_{x\in \mathbb{F}_p} y^{a_i} x^{a_i}= y^{a_i}  \sum_{x\in \mathbb{F}_p} y^{a_i} x^{a_i} = y^{a_i}  S(X_i^{a_i})$$ which implies that $(1 - y^{a_i}) S(X_i^{a_i}) = 0$, and as $y^{a_i} \not=1$, you have $S(X_i^{a_i}) = 0$. So we finally proved that $Card(V)$ is zero mod $p$. But we saw that as $(0,\ldots,0)$, $Card(V)\geq 1$. Therefore $Card(V) \geq p$, implying the existence of a non zero solution in $\mathbb{F}_p^n$, which in turn implies the existence of a non zero solution of your equation.
Remark. This extends trivialy to the case of a finite number of polynomials such that the sum of their degrees is $<n$, and shows that they must have a non-zero common zero.
