Proof: The proof will proceed by induction on the number of variables. Since we know that k[x₁] is a Euclidean domain, hence a principal ideal domain, we know the result with only one variable. (In this case, each ideal is generated by just one element.) The induction step is made more managable by introducing the following definition. (Notice that the definition contains within itself the statement of a proposition, that its two parts are equivalent. This proposition is left as an exercise.)

Proof: We will actually prove the contrapositive. So, assume that R[x] is not noetherian. Let I be an ideal in R that is not finitely generated. Choose an element f₁∈ I of minimal positive degree. Inductively, we can choose elements f_k+1 ∈ I\ (f₁,…, f_k) that have minimal positive degree. Let n_k denote the degree of f_k, and let a_k denote its leading coefficient. Observe first that

n₁ ≤ n₂ ≤ n₃ ≤ ….

Now consider the chain of ideals

(a₁)⊂ (a₁, a₂)⊂(a₁,a₂,a₃) ⊂…

contained in R. This chain of ideals cannot be stationary. For, suppose that (a₁, …, a_k) = (a₁, …, a_k, a_k+1). Then we can write

a_k+1 = ∑_i=1^k b_i a_i

for some b_i∈ R. Now consider the polynomial

g = f_k+1 - ∑_i=1^k b_i x^n_k+1-n_i f_i.

By the construction of the sequence of f_i's, we know that g ∈ I \ (f₁, …, f_k). However, we have constructed g in a way that kills off the leading coefficient, leaving a polynomial of degree smaller than the degree of f_k+1. This is a contradiction. So, we have shown that if R[x] is not noetherian, then R is not noetherian. The theorem follows.

Proof: Obvious.

Proof: By the previous lemma, it suffices to prove this when I is a maximal ideal. Let L=k[x₁, …, x_n]/I be the quotient field. If L=k, then we're done. To see this, write a_i∈ k for the image a_i ≡ x_i mod I. Thus, each polynomial x_i-a_i∈ I and

(x₁-a₁, …, x_n-a_n) ⊂ I.

Since the former ideal is maximal, this containment is an equality, and it is easy to see that Z(I)={(a₁, …, a_n)}. It now suffices to prove the following proposition.

Proposition Let k be a field, and let L be an extension field of k that is finitely generated as a k-algebra. Then L is an algebraic extension of k.

Proof: We can write L=k[a₁,…,a_n]. If n=1, then either a₁ is an algebraic element or L≈ k[a₁] is not a field. So, by induction, we may assume that L=k(a₁)[a₂, …, a_n] is an algebraic extension of the field k(a₁). We may also assume that a₁ is transcendental over k (otherwise, we are done). For all 2 ≤ i ≤ n, we have an equation

a_i^n_i + b_i,1 a_i^n_i-1 +…+b_{i, n_i} = 0,

with coefficients b_i,j∈ k(a₁). We can choose an element b∈ k(a₁) large enough to clear all the denominators of the b_i,j, so that

(ba_i)^n_i +bb_i,1 (ba_i)^n_i-1 +…+b^n_ib_{i,n_i} = 0.

In particular, L is an integral extension of k(a₁). (In other words, it is generated by the integral elements ba₂,…,ba_n.) In particular, given any element z∈ L, there exists a non-negative integer N such that b^Nz is integral over k[a₁]. To see this, one needs a series of lemmas.

Lemma If s is an integral element over a ring R, then R[s] is finitely generated as an R-module.

Proof: Obvious.

Lemma Let R⊂ S be an extension of commutative rings with 1. If S is finitely generated as an R-module, then every element s∈ S is integral over R.

Proof: Let t₁,…,t_m be module generators. Write each product

st_i = ∑_j=1^m r_ijt_j,

with coefficients r_ij∈ R. Rewriting this with Kronecker deltas yields

∑_j=1^m (δ_ijs-r_ij)w_j = 0

for all i=1, …,m. Looking at this equation in an appropriate vector space over the field of fractions of S, one gets a nontrivial solution to a certain matrix equation. Therefore, one has

0 = det(δ_ijs-r_ij) = s^m + (terms of lower degree),

a monic equation satisfied by s with coefficients in R.

Lemma The R⊂ S be an inclusion of commutative rings with 1. Then the set of elements of S that are integral over R forms a subring of S.

Proof: Let a, b∈ S be integral elements. Since b is certainly integral over R[a], we have that R[a]/R and R[a,b]/R[a] are finitely generated modules. Thus, R[a,b] is a finitely generated R-module. Thus, the elements a+b and ab∈ R[a,b] are integral over R, and the result follows.

Returning to the proof of the proposition, we see that any z∈ L can be written as a polynomial in a₂,…,a_n; multiplying by a large power of b writes it as a combination of things that are known to be integral over the polynomial ring k[a_i], and it follows that b^Nz is also integral. Now suppose f∈ k[a₁] is any polynomial that is relatively prime to b. Then 1/f∈ L, but none of the elements b^N/f can possibly be integral over k[a₁]. This is a contradiction; therefore, a₁ could not have been transcendental, and the proposition (and the Weak Nullstellensatz) follows.

The previous theorem shows that every proper ideal cuts out a nonempty algebraic set. The next theorem will characterize the ideals associated to algebraic sets.

Definition Let I be an ideal in a ring R. The radical of I is defined as

Rad(I) = {f∈ R : ∃ N>0, f^N∈ I}.

Remark The radical Rad(I) is always an ideal. Any ideal that satisfies I=Rad(I) is called a radical ideal. If X is any subset of Aⁿ, then I(X) is a radical ideal.

Theorem (Hilbert's Nullstellensatz) Let k be an algebraically closed field. Let J⊂ k[x₁, …, x_n] be any ideal. Then

I(Z(J)) = Rad(J).

Proof: Since I(X) is always radical and J⊂ I(Z(J)), the inclusion Rad(J)⊂ I(Z(J)) is trivial. So, let g∈ I(Z(J)) Now use Hilbert's Basis Theorem to choose generators f₁,…,f_s of J. Consider the auxiliary ideal

J' = (f₁,…,f_s,gx_n+1-1) ⊂ k[x₁,…,x_n+1].

Since g=0 whenever all f_i=0, we see that Z(J')=∅. By the Weak Nullstellensatz, J'=k[x₁,…,x_n+1]. Thus, there is a relation

1 = (∑₁^s a_if_i) + b(gx_n+1-1)

for some b, a_i ∈ k[x₁,…,x_n+1]. Substitute y=1/x_n+1 in this equation and clear denominators to get

y^m = ∑ a'_i f_i + b'(g-y)

with coefficients now living in k[x₁,…,x_n,y]. Now specialize by setting y=g to obtain

g^m = ∑ a"_i f_i ∈ J ⊂ k[x₁,…,x_n].

Thus, g∈ Rad(J) and the theorem is proved.

Corollary Two algebraic sets X₁, X₂ ∈ Aⁿ are equal if and only if I(X₁)=I(X₂).

Corollary There is a natural one-to-one inclusion-reversing correspondence between algebraic sets in Aⁿ and radical ideals in k[x₁,…,x_n]. Under this correspondence, points are associated with maximal ideals.

Definition Let X be an algebraic set in Aⁿ. Define the ring of regular functions on X to be

A(X) = k[x₁, …, x_n]/I(X).

Remark The elements of A(X) can be interpreted as functions on X with values in k. To, see this, note that A(Aⁿ) = k[x₁, …, x_n] is the full polynomial ring. Clearly, we can view a polynomial in n variables as a well defined function on the set of n-tuples from k. When X is an algebraic set, we can restrict any polynomial function on affine n-space to a function on X. Two polynomials restrict to the same function on X if and only if their difference lives in the ideal I(X).

Notice that the assignment X|→ A(X) shows that X determines A(X) uniquely. However, unlike the ideal I(X), the ring A(X) does not directly determine X as a closed subset of Aⁿ. The problem, of course, is that there is no canonical way to recover the generators of the polynomial ring from the quotient ring. If you choose two different sets of generators for A(X), you get two distinct algebraic sets, possibly embedded in different affine spaces, with the same coordinate ring. We can already see, however, that A(X) has the potential to recover X in a more abstract form. For example, the points of X are in one-to-one correspondence with the maximal ideals of k[x₁, …, x_n] that contain I(X); by the usual yoga of the isomorphism theorems, these are, in turn, in one-to-one correspondence with the maximal ideals of A(X).

Comments on this web site should be addressed to the author:

Kevin R. Coombes
Department of Biomedical Informatics
The Ohio State University
Columbus, Ohio 43210

Hilbert's Theorems