TITLE: Intuition regarding Chevalley-Warning Theorem
QUESTION [10 upvotes]: Three versions of the theorem are stated on pages 1-2 in these notes by Pete L. Clark:
http://math.uga.edu/~pete/4400ChevalleyWarning.pdf
Could anyone offer some intuitive way to think about this theorem?
Thanks!

REPLY [12 votes]: Near the end of his notes, Pete mentions some connections to $p$-adic cohomology, which is a very sophisticated alebraic geometry topic. I'm going to sketch how that connection works, and then show that you can actually understand these connection using "only" the level of algebraic geometry you'd learn in Hartshorne. Of course, motivation that goes up to the level of Hartshorne may or may not be what you are looking for!
For simplicity, take the case of a single polynomial $P$ which is homogenous of degree $d$. (Challenging exercise: Show that proving CW in this case implies the full result in Pete's notes.)
Let $Z_{aff}$ be the number of $P$ in $\mathbb{F}_q^{n}$. The center of every proof I know is showing that $Z_{aff} \equiv 0 \mod p$.   Let's count zeroes in the projective space $\mathbb{P}_{\mathbb{F}_q}^{n-1}$;  call this count $Z_{proj}$. So we have $Z_{aff} = (q-1) Z_{proj} +1$ and $Z_{aff} \equiv 0 \mod p$ if and only if $Z_{proj} \equiv 1 \mod p$. 
Let $X$ be the hypersurface $P=0$ in $\mathbb{P}^{n-1}$. In the $p$-adic approach Pete sketches, one would define cohomology groups $H^i(X)$, for $0 \leq i \leq 2 \dim X = 2(n-2)$. These would come with an action of the $q$-power Frobenius $F$, and we would have
$$\# X(\mathbb{F}_q) = \sum_{i=0}^{2 \dim X} (-1)^i \mathrm{Tr} \left(F^* : H^i(X) \to H^i(X) \right).$$
One would then show that the first term is $1$, and the other terms are divisible by $p$, completing the proof. Additionally, there would be complications resulting from the fact that $X$ may not be smooth. 
However, you don't have to learn these complicated cohomology theories in order to give a cohomological proof of the CW theorem. Instead, you can use Fulton's fixed point formula! This states:
$$\# X(\mathbb{F}_q)  \equiv \sum_{i=0}^{\dim X} (-1)^i \mathrm{Tr} \left( F^*: H^i(X, \mathcal{O}) \to H^i(X, \mathcal{O}) \right) \mod p$$
Here the cohomology groups are sheaf cohomology, as covered in Hartshorne. Note that the right hand side only makes sense modulo $p$. One advantage to this formula, besides that it uses a more elementary cohomology theory, is that there are no smoothness hypotheses on $X$. 
Now, $H^0(X, \mathcal{O})$ is $1$ dimensional and the frobenius action is trivial, so that contributes $1$. By the Lefschetz hyperplane theorem, $H^j(X, \mathcal{O})$ vanishes for $0 < j < \dim X$. So we have
$$\# X(\mathbb{F}_q) \equiv 1 + (-1)^{\dim X} \mathrm{Tr} \left( F^*: H^{\dim X}(X, \mathcal{O}) \to H^{\dim X}(X, \mathcal{O}) \right) \mod p$$
So far, we have not used that $d<n$. Using this, we show that $H^{\dim X}(X, \mathcal{O})$ vanishes so $\# X(\mathbb{F}_q) \equiv 1 \mod p$ and we win.
A fun challenge is to work out the case that $d=n$. The elementary computation in Pete's notes shows that $\#X(\mathbb{F}_q)$ is something like $1$ plus the coefficient of $x_1^{q-1} \cdots x_n^{q-1}$ in $P^{q-1}$. (I'm probably off by a sign somewhere.) Challenge: Show that, when $d=n$, $H^{\dim X}(X, \mathcal{O})$ is one dimensional and the scalar by which Frobenius acts is the coefficient of $x_1^{q-1} \cdots x_n^{q-1}$ in $P^{q-1}$.