Affine varieties, as the common zeroes of polynomials over some field, are irreducible algebraic sets and they're relatively easy to understand when polynomials are over $\Bbb C$. However, how can I determine that an algebraic set is a variety (i.e. irreducible) over a finite field $\Bbb F_q$, where q is a power of a prime. And more importantly, how to find its dimension?
For instance, in order to make everything easy, let $f=x^2+3$ and our finite field be $\Bbb F_7$. The algebraic set defined by $f$ over $\Bbb F_7$ is the set $$V= \{y\in acl(\Bbb F_7) : f(y)=0 \}$$
Therefore $2,5,9,12 \in V$, as $2^2+3 \equiv5^2+3 \equiv 9^2+3 \equiv 0 $ (mod $7$), and there are infinitely such elements in $V$. Equivalently, $V$ consists of the numbers $2$ mod $7$ and $5$ mod $7$. So, I think that $V$ is defined by union of smaller algebraic sets $x-2$ and $x-5$. So, I think that V is not a variety. But, are these smaller components of $V$ varieties? If they're so, what are their dimension? More generally, how can I find dimension of a variety defined over a finite field. For example, what is dimension of the variety defined by $x-2$ over $\Bbb F_7$ ?
Your $V$ consists of the union of two zero-dimensional varieties, namely the points $y = 2$ and $y = -2$. There are many ways to define the dimension of a variety, such as the Krull dimension (in the affine case) or the transcendence degree of the function field.
You can determine whether an affine variety is irreducible by checking whether its ring of functions is an integral domain.