Intuition regarding Chevalley-Warning Theorem Three versions of the theorem are stated on pages 1-2 in these notes by Pete L. Clark:
http://alpha.math.uga.edu/~pete/4400ChevalleyWarning.pdf
Could anyone offer some intuitive way to think about this theorem?
Thanks!
 A: I don't know if this is relevant, but the Chevalley-Warning theorem does show that finite fields have a rather important property that comes up in geometric applications (in etale cohomology, for instance).
A field $k$ is called C1 or quasi-algebraically closed if every polynomial in $n$ variables with coefficients in $k$ and homogeneous of degree $n>d$ has a nontrivial root. Being C1 means that the field's Brauer group is trivial, and consequently controls the Galois group quite a bit: if the field is C1, then the Galois cohomology of a torsion module vanishes in degrees $>1$. So it's quite useful to know that a field is C1.
Clearly algebraically closed fields are C1, but there are three other important examples:


*

*Finite fields (this is the Chevalley-Warning theorem)

*A field of transcendence degree one over an algebraically closed field (i.e. the rational function field of an algebraic curve). This is "Tsen's theorem."

*The maximal unramified extension of a local field (this is in Serge Lang's thesis).


Note by contrast that a field like $\mathbb{R}$ is not C1, and in fact its absolute Galois group $\mathbb{Z}/2\mathbb{Z}$ has torsion cohomology in arbitrarily large (i.e., all even) degrees.
A: I am now interpreting the question as "How might one come to believe that the Chevalley-Warning Theorem is true?"  (More specifically, I will try to put myself in Emil Artin's shoes when he conjectured this result in the 1930's.)
Let us say that a field $K$ satisfies property $C_1(d)$ if any homogeneous polynomial equation $f(x_1,\ldots,x_n) = 0$ in $n$ variables of degree $d$ has a solution $(x_1,\ldots,x_n) \neq (0,\ldots,0)$ if $n > d$.  Then a field is $C_1$ if it is $C_1(d)$ for all positive integers $d$.
Here are some observations about this property:
1) Any field is $C_1(1)$: a homogeneous linear equation in at least two variables has a nontrivial solution.
2) Suppose that a field $K$ is such that for all positive integers $d$, every homogeneous polynomial in $n = d$ variables has a nontrivial zero.  Then $K$ is algebraically closed.  
Indeed, if $K$ is not algebraically closed it admits for some $d > 1$ a degree $d$ field extension $L/K$.  The norm form associated to $L/K$ is a homogeneous polynomial of degree $d$ in $d$ variables which has only the trivial solution.
This explains the condition that $n > d$: this is necessary to go beyond algebraically closed fields.
3) Now let's look at the property $C_1(2)$ of a field $K$: this says that any quadratic form in at least three variables over $K$ is isotropic: has a nontrivial solution.  When the characteristic is not $2$, this is equivalent to the following: for any $a,b,c \in K^{\times}$, the equation $ax^2 + by^2 = c$ has a solution $(x,y) \in K^2$.  (In other words, any plane conic has a $K$-rational point.)  This type of equation was studied by number theorists at least since Fermat in the 1600's.  Legendre for instance well knew that the field $\mathbb{Z}/p\mathbb{Z}$ has this $C_1(2)$ property.  It is possible to give a very elementary proof by counting arguments: you rewrite the equation as $ax^2 -c = by^2$ and observe that the left hand side and right hand side each run through $\frac{p-1}{2} + 1 = \frac{p+1}{2}$ distinct elements of $\mathbb{Z}/p\mathbb{Z}$, hence they must take a common value.  This proves Chevalley-Warning when $d = 2$ (and the characteristic is odd; you can certainly prove it in even characteristic by brute force if you want).
4) The next case of Chevalley-Warning is $d = 3$: that is, any cubic equation in at least four variables over a finite field has a nontrivial zero.  The general study of cubic surfaces is already very hard, and I'm not sure what classical work there was on this.  However, I believe that when Artin conjectured that C-W would hold he knew about Hasse's result that a smooth cubic equation in $3$-variables over a finite field has a nonzero rational point.  Here the number of equations and variables are both equal to $3$, and so by 2) above this cannot possibly hold for all cubic equations in three variables: one can take the norm form of a cubic extension to get a counterexample.  
But from a geometric perspective this counterexample is not very worrisome: over the algebraic closure, what one has is three planes meeting at the origin, but with the Galois group acting so that there is no component which is rationally defined over the ground field.  Therefore the only possible $K$-rational points are the intersection points of the components which, if you're thinking about it affinely, are only the origin (and if you're thinking about it projectively, there are no intersection points).  This is very different from a geometrically irreducible cubic equation, to which Hasse's Theorem applies. 
Moreover, if you try to geometrically build the same example one dimension higher, it doesn't work: if you have three hyperplanes in $K^4$ all passing through the origin, their common intersection is a line, not just a point.  
So I think it's at least plausible that Hasse's Theorem implies that a finite field is $C_1(3)$.  Maybe you could actually prove this along the lines given above: I haven't tried, but it would be interesting.
5) Another thing you can do of course is pick your favorite values of $q,n,d$ and do the finite amount of computations necessary to see whether indeed every form of degree $d$ in $n$ variables over $\mathbb{F}_q$ must have a nontrivial zero.  Presumably Artin did some of this.  In this regard I cannot help mentioning Exercise 10.16 in Ireland & Rosen's classic number theory text: "Show by explicit calculation that every cubic form in $2$ variables over $\mathbb{F}_2$ has a nontrivial zero."  As many students have discovered over the years, explicit calculation rather shows that this is false: e.g. $x_1^3 + x_1^2 x_2 + x_2^3$ has no nontrivial zero.  See this MO answer for a bit more on this: it is in fact easy to show that for every finite field $\mathbb{F}_q$ and every pair $(n,d)$ with $n \leq d$, there is a form over $\mathbb{F}_q$ of degree $d$ and in $n$ variables with only the trivial zero.
6) It is easy to show that any $C_1$ field has vanishing Brauer group: for any central division algebra $D$ over $K$, the reduced norm is a homogeneous polynomial in $d^2$ variables of degree $d$ with only the trivial zero.  The vanishing of the Brauer group of every finite field is equivalent to the statement that every finite division ring is a field, a celebrated 1907 theorem of Wedderburn that Artin was certainly very familiar with.  This property really is weaker than $C_1$ in the sense that we now know of fields which have vanishing Brauer group of every finite extension but are not $C_1$ (I believe the first examples were constructed by Ax), but for "familiar" fields the two properties tend to be equivalent.  Thus this gives at least a lot of nontrivial examples of Chevalley-Warning.
Beyond that, it seems to me that Artin made a fairly gutsy conjecture here, unless he knew more than he let on.  It is perhaps worth noting that he also conjectured that every $p$-adic field has property $C_2$ (a homogeneous form in $n$ variables of degree $d$ has a nontrivial zero if $n > d^2$) and that the maximal abelian extension of $\mathbb{Q}$ has property $C_1$.  The former conjecture was eventually proven false by Terjanian (although there is a lot of truth in it...), and the latter conjecture remains wide open to this day, and I don't know any really compelling reasons for (or against) it.
A: 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 $P^{q-1}$.
