1 Nullstellensatz

1.1 Finite generation

Let R be a commutative ring with 1.

An R-module is an abelian group M that admits left multiplication by elements from R, where this multiplication is associative and distributive over addition.

The module M is called module-generated over R by a subset N if each element m ∈ M has a representation m = ∑ r_{i}n_{i} with r_{i} ∈ R and n_{i} ∈ N.

In other words, M is module-generated over R by a subset N if M is the smallest R-submodule of M containing N.

The module M is called module-finite over R if M is module-generated over R by a finite subset N.

Now let S be a commutative ring containing the ring R. The ring S is called ring-generated over R by a subset T if each element s ∈ S has a representation s = ∑ r_{i}z_{i} with r_{i} ∈ R and each z_{i} a monomial over T, that is, of the form t_{1}^{e_{1}}...t_{m}^{e_{m}} for certain t_{1}, ..., t_{m} ∈ T and nonnegative integers e_{1}, ..., e_{m}.

In other words, S is ring-generated over R by a subset T if S is the smallest subring of S containing R and T.

The ring S is called ring-finite over R if S is ring-generated over R by a finite subset T.

For example, a finite-dimensional vector space V over a field k is modulefinite over k. And a polynomial ring k[x_{1}, ..., x_{n}] in finitely many variables is ring-finite over k.

Lemma 1.1 Let R ⊆ S ⊆ T be commutative rings. If T is module-finite over S, and S is module-finite over R, then T is module-finite over R.

Proof: Exercise. □

1.2 Integrality

Let S be a commutative ring containing the commutative ring R. An element s ∈ S is called integral over R if it satisfies a monic equation s^{n} + a_{1}s^{n−1} + ... + a_{n} = 0 with coefficients a_{i} in R.

Proposition 1.2 Let S be a domain with subring R. Let T be the set of elements in S that are integral over R. Then T is a subring of S containing R.

That T contains R is clear: r ∈ R satisfies the equation x − r = 0. That T is a subring will follow from the following lemma.

Lemma 1.3 Let S be a domain with subring R. Let s ∈ S. Equivalent are: (i) s is integral over R, (ii) R[s] is module-finite over R, (iii) there is a subring R′ of S containing R[s] that is module-finite over R.

Proof: (i) implies (ii): an equation s^{n} + a_{1}s^{n−1} + ... + a_{n} = 0 expresses s^{n} in terms of 1, s, ..., s^{n−1}, so R[s] is module-generated by these finitely many elements.

(ii) implies (iii): take R′ = R[s].

(iii) implies (i): Suppose R′ is module-generated over R by t_{1}, ..., t_{n}. Write the products st_{i} using these generators: st_{i} = ∑ c_{ij}t_{j} with c_{ij} ∈ R. Let A = sI − C where I is the identity matrix and C = (c_{ij}). Then At = 0, where t is the column vector (t_{j}). In the quotient field of S we conclude that det A = 0, and that is the required monic equation for s. □

Proof of the proposition: if a, b ∈ S are integral over R, then R[a] is modulefinite over R, and R[a, b] is module-finite over R[a], so R[a, b] is module-finite over R. Now let s be one of the elements a + b, a − b, ab. Take R′ = R[a, b] in the above lemma to find that s is integral over R. That means we proved that T is closed under addition, subtraction and multiplication. □

1.3 Weak Nullstellensatz

Theorem 1.4 Let k be an algebraically closed field. Let I be an ideal of the polynomial ring k[x_{1}, ..., x_{n}]. If 1 ∉ I, then V (I) ≠ ∅.

In the proof of this theorem, we’ll need the following proposition.

Proposition 1.5 Let L be a field containing a field k. If L is ring-finite over k, then L is module-finite over k.

Proof of the theorem: If I is made larger, V (I) gets smaller. So, it suffices to show this for a maximal ideal I. If the ideal I is maximal, the quotient L = k[x_{1}, ..., x_{n}]/I is a field. We have a natural embedding of k into L. If this embedding is an isomorphism, then there are elements a_{i} ∈ k that map to the residue classes x_{i} + I, i.e., have the property that x_{i} − a_{i} ∈ I. Now I = (x_{1} − a_{1}, ..., x_{n} − a_{n}) since the right hand side is a maximal ideal contained in I. Consequently, V (I) = {(a_{1}, ..., a_{n})} is a single point, and therefore nonempty.

Remains to show that the embedding of k into L really is an isomorphism. This follows from Proposition 1.5. Indeed, let us identify k with its image in L. In our situation L = k[x_{1}, ..., x_{n}]/I is ring-generated by the residue classes x_{i} + I, so by this proposition L is module-finite over k, and by Lemma 1.3 every element of L is integral (and in partiticular algebraic) over k. But k is algebraically closed, so L = k. □

Proof of the proposition: We have L = k[s_{1}, ..., s_{n}] for certain elements s_{1}, ..., s_{n} ∈ L, and want to find a finite set module-generating L over k. Use induction on n.

If n > 1 then put K = k(s_{1}). Then L = K[s_{2}, ..., s_{n}], and by induction on n we see that L is module-finite over K. If K is also module-finite over k, then L is module-finite over k (by Lemma 1.1) and we are done.

So, suppose K = k(s_{1}) is not module-finite over k, so that s_{1} is transcendent over k, and K ≃ k(x). Let us write x instead of s_{1}.

Since L is module-finite over K, every element of L is integral over K.

Now look at the subring T of L consisting of the elements of L that are integral over k[x]. If s ∈ L has equation s^{n}+a_{1}s^{n−1}+...+a_{n} = 0 with coefficients a_{i} in K = k(x), then these a_{i} are rational expressions a_{i} = f_{i}(x)/g_{i}(x). Let h = h(x) be a common multiple of all denominators g_{i}(x). Then hs satisfies (hs)^{n} + (ha_{1})(hs)^{n−1} + ... + h^{n}a_{n} = 0, a monic equation with all coefficients in k[x], so that hs ∈ T.

If we do this for s = s_{2}, ..., s_{n}, and take for h the least common multiple (or just the product) of all denominators encountered for all s_{j}, then we find a single h such that hs_{j} ∈ T for all j.

Let z be an arbitrary element of L. Since L = k[x, s_{2}, ..., s_{n}] this element can be written as a sum of terms, each a monomial in the s_{j}. It follows that h^{N}z ∈ T for sufficiently high exponent N.

Now that this holds for all elements z ∈ L, take a particular one, for example z = 1/f where f ∈ k[x] is an irreducible polynomial not dividing h. Now if h^{N}/f ∈ T, then we have an equation (h^{N}/f)^{m} + c_{1}(h^{N}/f)^{m−1} + ... = 0 with coefficients c_{i} ∈ k[x]. Multiplying by f^{m−1} we find that h^{Nm}/f ∈ k[x]. But that is false since f is irreducible and does not divide h.

So, the assumption that K = k(s_{1}) is not module-finite over k leads to a contradiction, and this finishes the n > 1 part of the proof by induction.

Remains to look at n = 1. Given is L = k[s]. To show that L is module-finite over k.

Consider the map from k[x] onto k[s] sending x to s. Its kernel is some ideal I, and k[x]/I ≃ k[s] = L.

If I ≠ (0), then I = (g(x)) for some polynomial g(x) since k[x] is a PID. W.l.o.g. g(x) is monic, and we see that s is integral over k because g(s) = 0, and hence L is module-finite over k as desired.

And if I = (0), then L ≃ k[x], but k[x] is not a field since it does not contain 1/x.

This proves everything. □

1.4 Nullstellensatz

Theorem 1.6 Let k be algebraically closed. Let I be an ideal in k[x_{1}, ..., x_{n}]. Then I(V (I)) = Rad(I). That is, if g ∈ k[x_{1}, ..., x_{n}] and g vanishes on V (f_{1}, ..., f_{m}), then there is an N such that g^{N} = ∑ c_{i}f_{i} for certain c_{i} ∈ k[x_{1}, ..., x_{n}].

Proof: Apply the Weak Nullstellensatz in n+1 dimensions: Look at the ideal J = (f_{1}, ..., f_{m}, x_{n+1}g − 1) in k[x_{1}, ..., x_{n}, x_{n+1}]. Since V (J) = ∅, it follows that 1 ∈ J, i.e.,

1 = ∑ a_{i}f_{i} + a.(x_{n+1}g − 1)

for certain a, a_{i} ∈ k[x_{1}, ..., x_{n}, x_{n+1}]. Put x_{n+1} = 1/y and multiply by a power of y to remove denominators:

y^{N} = ∑ b_{i}f_{i} + b.(g − y)

for certain b, b_{i} ∈ k[x_{1}, ..., x_{n}, y]. Now put y = g. □

One sees that for algebraically closed fields there is a 1-1 correspondence between closed sets (sets of the form V (f_{1}, ..., f_{m})) and radical ideals of k[x_{1}, ..., x_{n}] (ideals I that equal their radical).