Algebraic Varieties, An Introduction


An affine variety is an algebraic set with some rather nice properties. But before we dive into the math, let's have some words about words.

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.


A variety is a closed irreducible set. The first part is implied. If a variety is an algebraic set it is closed. But we also want it to be irreducible - and what does that mean?

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.


The field K is noetherian, and the ring of polynomials over K is noetherian. Ascending chains of radical ideals are finite, thus descending chains of closed sets in Cn are finite. This makes Cn a noetherian space. It is therefore compact, and every closed set is uniquely a union of irreducible components. In other words, every algebraic set is uniquely a union of varieties. All this we get for free, simply by invoking the relevant theorems from topology.


If Cn is hausdorff then its varieties, i.e. its irreducible regions, are all individual points. If a prime ideal vanishes on more than one point the space cannot be hausdorff. Zero vanishes everywhere, hence the space is not hausdorff.


An open subset of a variety is a quasi variety. Like any other subspace of a noetherian space, this is compact. And since a variety is irreducible, a quasi variety is irreducible.

Prime Spectrum

Zariski has placed a topology on Cn, and on the prime ideals of any ring, including the ring of polynomials over K. Turning to the topology of the ring, one can equate prime ideals with varieties in Cn. Closed sets in the prime spectrum come from radical ideals, which define algebraic sets in Cn. In other words, the closed regions are the same, whether we are looking at points in spec r, or varieties in Cn.