Why study algebraic varieties over an algebraic closure of the finite field? Let $\mathbb{F}_q$ be a finite field and $I = \langle f_1,\ldots,f_r \rangle \subseteq \mathbb{F}_q[x_1,\ldots,x_n]$ an ideal. 
Let me write $V(I)$ for the set $\{x \in \mathbb{F}_q^n : f(x) = 0 \text{ for all } f \in I\}$.
I am interested in computing the number of solutions to the system of equations $f_1=\cdots=f_r=0$, which is given by $|V(I)|$. Put $I_q = \langle x_1^q-x_1, \ldots, x_n^q-x_n\rangle$. Since $|V(I)| < \infty$, I know that $|V(I)| = \dim_{\mathbb{F}_q} \mathbb{F}_q[x_1,\ldots,x_n]/(I + I_q)$. 
In many of the texts that I have found online, the object $V(I)$ is actually defined over an algebraic closure of $\mathbb{F}_q$, even if the ultimate goal is to study the finite number of solutions in $\mathbb{F}_q^n$. I have noticed that something like the Krull dimension of $V(I)$, which is zero in the setting above, can all of a sudden be non-zero over the algebraic closure and that the dimension of $V(I)$ over an algebraic closure equals its dimension as a vector space if $f_1,\ldots, f_r$ are linear forms, which does not seem to be the case in the setting above. My question is: why move to an infinite field in order to study something that is finite? Is this simply because the theory is "nicer" in this case?
 A: Maybe the principal reason is because it is nicer to work with the algebraic closure for algebro-geometric reasons. But there are more reasons as well.
First, the algebraic closure of $\mathbb{F}_q$ is not a bizarre object actually, it is just $$\overline{\mathbb{F}_q}=\bigcup_{k\geq 1} \mathbb{F}_{q^k}.$$ So, if you denote by $V_K(I)$ the solutions in $K^n$ to the system of equations $f_1=\dots=f_r=0$, you have that 
$$ V_{\overline{\mathbb{F}_{q}}}(I)=\bigcup_{k\geq 1} V_{\mathbb{F}_{q^k}}(I)$$
Now, if you take a point $x\in V_{\overline{\mathbb{F}_{q}}}(I)$, the smallest $k$ such that $ x\in (\mathbb{F}_{q^k})^n$ is $k=[\mathbb{F}_q(a_1,\dots,a_n):\mathbb{F}_q].$ Moreover, the field extension $\mathbb{F}_q(a_1,\dots,a_n)$ has an algebro-geometric interpretation, it is the residue field of the point $x$ inside the variety $V_{\overline{\mathbb{F}_{q}}}(I)$ and is called the degree of the point $x$. So in this language, you are interested in the number of points of $V_{\overline{\mathbb{F}_{q}}}(I)$ of degree 1. But a better question is to count the number of points of degree $k$ for all $k\geq 1$, because there is an analogy between counting this points and the Riemann Zeta function.
This analogy is easier to understand in terms of the Dedekind Zeta function: If $K$ is a finite extension of $\mathbb{Q}$ and $\mathcal{O}_K$ is its ring of integers, then the Dedekind zeta function is the sum
$$\zeta_K(s)=\sum_{I\subseteq \mathcal{O}_K}\frac{1}{\#(\mathcal{O}_K/I)^s}$$
where the sum goes over the ideals of $\mathcal{O}_K$ and $s$ is a complex number (let's not focus on the convergence issues here). Using the unique factorization of ideals as products of prime ideals it is possible to show that 
$$\zeta_K(s)=\prod_{P\subseteq \mathcal{O}_K}\frac{1}{1-\#(\mathcal{O}_K/P)^{-s}} \tag{$\star$}$$
where the product goes over the maximal ideals of $\mathcal{O}_K$. Notice that $\#(\mathcal{O}_K/P)=p^{f(P)}$ where $p\in \mathbb{Z}$ is the prime below $P$ and $f(P)$ is the degree of inertia of $P$.
Now, there is an analogy from algebraic geometry between maximal ideals of a ring and points in a variety. So we can try to squeeze this analogy in order to construct something like $(\star)$. For this you can replace "maximal ideal of $\mathcal{O}_K$" by "point of $V_{\overline{\mathbb{F}_{q}}}(I)$" and the number $p^{f(P)}$ above by the number $q^{\deg(x)}$ (both are the numbers of elements on the respective residue fields). Then we get
$$\zeta_{V_{\overline{\mathbb{F}_{q}}}(I)}(s)=\prod_{x\in V_{\overline{\mathbb{F}_{q}}}(I)}\frac{1}{1-(q^{\deg(x)})^{-s}}$$
and by doing $T=q^{-s}$ and some formal computations we obtain
$$\zeta_{V_{\overline{\mathbb{F}_{q}}}(I)}(s)=\exp\left ( \sum_{k=1}^\infty \frac{\#(V_{\mathbb{F}_{q^k}}(I))T^k}{k}\right ).$$
So, the problem to obtain the zeta function of the variety is equivalent to the problem of counting $V_{\mathbb{F}_{q^k}}(I)$ that is, to count the number of solutions of your system of equation over all the finite field extensions and not only on $\mathbb{F}_{q}$. 
As with all Zeta functions, there is a series of problems such as to find its functional equation or to prove some sort of Riemann hypothesis with it. In this particular case all this theory goes over the name of Weil Conjectures and they were one of the principal motors in the development of algebraic geometry in the last part of the past century (the conjectures are already solved by the way).
