The adjective "affine" is roughly synonymous with linear. So why don't they just use linear? I don't know. I think they want a clear distinction between an affine variety and a projective variety. (I'll talk about projective varieties after we plow through affine varieties.) Unless otherwise stated, or indicated by context, a variety (with no adjective) is an affine variety.
The desirable properties of a variety are topological, and that means we need to understand the topological space that they live in.
Let K be a field and let C be its algebraic closure. A variety is an algebraic set, and an algebraic set is a region R in Cn, such that a set of functions vanishes on R. The functions are polynomials in n variables, with coefficients in K. These functions form a ring, and for some ideal S in this ring, the functions of S vanish on R. In other words, S and R correspond.
By strong nullstellensatz, S is a radical ideal. The closed regions in Cn are the algebraic sets R, and the closed ideals in K[x1,x2,x3,…xn] are the radical ideals in the polynomial ring. We already proved this is a valid topology.
Irreducible sets can exist in any topological space. To summarize, a set R is reducible if it can be covered by two proper closed sets. The two closed spaces combine to form R. A space R is irreducible if it is not the finite union of proper closed sets.
Consider the 0 ideal. Since polynomials form an integral domain, this is a radical ideal. And it is satisfied by every point in Cn. Thus Cn is a closed algebraic set. Is it reducible? Suppose two proper closed sets combine to form the entire space. The union of these spaces corresponds to the product of the two ideals, and in an integral domain, this product is nonzero. The union cannot cover the entire space. We have found our first algebraic variety. The 0 ideal, and/or the space Cn, forms a variety.
For all other algebraic sets, the region is irreducible iff the ideal is prime. We have already proved this for zariski rings in general, but just for grins let's prove it again here; it only takes a few lines. Assume A and B are radical ideals. Let P be prime, hence another radical ideal, and suppose its region is the union of two closed sets determined by A and B. Thus P = A∩B, P contains A*B, P contains B, and B vanishes on all of R. Conversely if P is not prime, let A*B lie in P. Replace A with A+P and B with B+P; the product is still in P. Push A and B up to radical ideals; the product is still in P. Product equals intersection when ideals are radical, so the intersection lies in P. Since A and B contain P, the intersection equals P. The union of A′ and B′ is P′, and each of A′ and B′ is properly contained in P′. Therefore R is irreducible iff R′ is a prime ideal. Varieties and prime ideals coincide.
Some books require K to be algebraically closed, i.e. K = C, for R to be a variety. This is necessary for some theorems, but not for others. I prefer to leave K unspecified - provided C is the algebraic closure of K.