Algebraic sets, rings and Hilbert's Nullstellensats

Affine varieties

All rings, unless stated otherwise are commutative.

Recall that an integral domain or domain is a ring where ab = 0 implies a = 0 or b = 0. Moreover, if R is a domain and a ≠ 0, then if ab = ac then b = c, i.e. the law of cancellation holds in a domain. This is an equivalent definition. We say that the domain is a principal ideal domain or PID if every ideal is generated by one element.

As an example ℤ is a PID. Indeed, if I is a non-zero ideal, then we can choose elements a and b in I such that b > a > 1. Then we can write b = qa + c for some integers q and c satisfying q ≥ 0 and 0 ≤ c < a. As c is non-negative and a is the smallest positive integer in I, we see that c = 0, i.e. b = qa. This ring is also a unique factorization domain, which is a ring in which each non-zero element decomposes as a unique product of primes.

We call a non-zero element a of R a unit if there exists some b such that ab = 1. Ideals generated by units are the entire ring. A non-unit which is not the product of any two non-units is called an irreducible element. A unique factorization domain or UFD is a ring in which every non-zero, non-unit element is a unique product of irreducible elements. Hence, ℤ is a UFD. We also have the following well-known fact, which is easy to prove.

We say R is Noetherian if every ideal is finitely generated. We have the following important theorem.

We use the notation R[X] instead of R[X₁, …, X_n] when it is convenient and it does not cause confusion.

There are two proofs. We give a sketch of both. The first way is to take an ideal I in R[X] and take the set of leading coefficients in I which is a finitely generated ideal in R. We call this ideal J. The idea of the proof is to decompose J into a sum which has a nice pre-image in R[x].

The better way of approaching this is to use the concept of a Grobner basis. Unfortunately, I could not find a precise definition, but a Grobner basis is basically a finite generating set of I with "nice algorithmic properties". Thankfully, the exact definition does not matter for us. Basically, if F = ∑a_iX₁^i_i⋯X_k^i_k, and G₁, …, G_m are a set of polynomials in R[X], then we can write F = ∑A_iG_i + C, where deg C < deg F. This can be done by a sort of Euclidean algorithm using lexicographic ordering, called Bucherberg’s Algorithm.

Given a subset X of Rⁿ, we define its ideal I(X) to be the set of polynomials f ∈ R[X] are zero on X. Similarly, for an ideal I of R, we define its locus to be the set of points p ∈ Rⁿ such that fp = 0 for all f ∈ I. In general, these maps are not inverses of each other, but we will get to their relationship quite soon. However, we can see some interesting properties. First, we say an algebraic set is irreducible if it is not a union of two non-empty algebraic sets, where the two sets in the union are unequal. It is easy to show that V is irreducible if and only if I(V) is prime. Moreover, by Hilbert’s basis theorem, we have

We continue to get more interesting properties of algebraic sets. Suppose I ⊂ J are ideals. Then V(J) ⊂ V(I). The situation is simple when k is algebraically closed. If I is a proper ideal of k[X], then there is some maximal ideal J containing I. But we know J is generated by ⟨x₁ − a₁, …, x_n − a_n⟩. So a = (a₁, …, a_n) ∈ V(I). Hence

Next we define the radical of an ideal I as rad(I) = {f ∈ R[X] : f^r ∈ I for some r > 0}.

When k is an algebraically closed field, I claim that I(V(I)) = rad(I). If we had this, then this would allow us to reduce a generating ideal of an algebraic set to a minimal one. Showing that rad(I) ⊂ I(V(I)) is straightforward. For the other direction, suppose f ∈ I(V(I)). Let f₁, …, f_k be the generators of I. We wish to find some r such that f^r = ∑a_if_i where a_i are members of k[X₁, …, X_n]. Then the ideal J generated by f₁, …, f_k and 1 − x_n + 1f is the entire ring, since if it were not, then V(J) would have at least one element, which it does not, because f vanishes whenever each f_i does. Thus one can write 1 = g₁f₁ + ⋯ + g_kf_k + h(1 − x_n + 1f) for some g_i and h in k[x₁, …, x_n + 1]. Substituting 1/f(x₁, …, x_n) for x_n + 1 results in the equation $$1 = \sum g_i(x_1,\ldots,x_n,\frac{1}{f(x_1,\ldots,x_n)}) f_i(x_1,\ldots,x_n).$$ We can multiply both sides by a high enough power of f to remove f from the denominator of the rational expression on the right hand side, yielding f^r = ∑c_if_i for some c_i in I. Thus we have the following theorem

This important theorem characterizes gives us a one-to-one correspondence between the radical ideals of 𝔯 and the algebraic sets in 𝔸ⁿ. We can go even further. We say 𝔯 is reduced if it contains no non-zero nilpotent elements.

Thus we have a one-to-one correspondence between the following sets radical ideals of 𝔯 ↔︎  reduced quotients of 𝔯 ↔︎  alg. sets The reader may have noticed, or already knows that the "reduced quotient" is just a ring k[x₁, …, x_n]/I(X) where X is an algebraic set. We call this the coordinate ring, and delay a full study to the next section.

Let us sketch out some more properties of algebraic sets before we finish this section. We will finish with giving 𝔸ⁿ a topology, and we will introduce the notion of dimension of algebraic sets.

As an example, note that the Zariski-closed sets of 𝔸¹ are the finite subsets of 𝔸¹. The closed sets in 𝔸² consist of finite sets, lines and curves.

We define the Krull dimension of a ring to be the supremum of lengths of proper chains of prime ideals, and we denote this by dim 𝔯. For a prime ideal 𝔭 ⊂ 𝔯, we define its height to be the supremum of its proper chains of prime ideals.

For example, the dimension of 𝔸ⁿ is n.

We leave the proof as an exercise to the reader.