## Hilbert's Theorems

The previous section contained a defintion of an algebraic set as the locus defined by an ideal. The first result of this section is a crucial finiteness result.

Theorem (Hilbert's Basis Theorem) If k is a field, then every ideal in the polynomial ring k[x1, , xn] is finitely generated.

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.)

Definition A commutative ring R with 1 is called noetherian if one of the following equivalent conditions holds.
1. Every ideal in R is finitely generated.
2. Every ascending chain of ideals in R terminates.

Now the Hilbert Basis Theorem will follow from the following more general result.

Theorem If R is a noetherian ring, then so is the polynomial ring R[x].

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 fk+1 I\ (f1,, fk) that have minimal positive degree. Let nk denote the degree of fk, and let ak 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, , ak) = (a1, , ak, ak+1). Then we can write

ak+1 = i=1k bi ai

for some bi R. Now consider the polynomial

g = fk+1 - i=1k bi xnk+1-ni fi.

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

Definition Let X be any subset of An. Define its associated ideal in the polynomial ring to be

I(X) = {f k[x1, , xn] : f(x) = 0 x X}.

Lemma
1. For ideals I J, we have Z(I) Z(J).
2. For algebraic sets X Y, we have I(X) I(Y).
3. For any subset X of An, we have Z(I(X)) = X-, the closure in the Zariski topology.

Proof: Obvious.

Theorem (Weak Nullstellensatz) Let k be an algebraically closed field. Let I k[x1, , xn] be a proper ideal. Then Z(I) is a nonempty algebraic set in An.

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

(x1-a1, , xn-an) I.

Since the former ideal is maximal, this containment is an equality, and it is easy to see that Z(I)={(a1, , an)}. 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,,an]. 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, , an] 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

aini + bi,1 aini-1 ++bi, ni = 0,

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

(bai)ni +bbi,1 (bai)ni-1 ++bnibi,ni = 0.

In particular, L is an integral extension of k(a1). (In other words, it is generated by the integral elements ba2,,ban.) In particular, given any element z L, there exists a non-negative integer N such that bNz 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,,tm be module generators. Write each product

sti = j=1m rijtj,

with coefficients rij R. Rewriting this with Kronecker deltas yields

j=1m (δijs-rij)wj = 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(δijs-rij) = sm + (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,,an; 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[ai], and it follows that bNz 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 bN/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, fN 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, , xn] be any ideal. Then

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,,fs of J. Consider the auxiliary ideal

J' = (f1,,fs,gxn+1-1) k[x1,,xn+1].

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

1 = (1s aifi) + b(gxn+1-1)

for some b, ai k[x1,,xn+1]. Substitute y=1/xn+1 in this equation and clear denominators to get

ym = a'i fi + b'(g-y)

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

gm = a"i fi J k[x1,,xn].

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,,xn]. 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, , xn]/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, , xn] 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, , xn] 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).