Hilbert’s Nullstellensatz is a basic but foundational theorem in commutative algebra that has been discussed on the blogosphere repeatedly, but thematically now is the appropriate time to say something about it.
The idea of the weak Nullstellensatz is quite simple: the polynomial ring has evaluation homomorphisms sending for some point , so we can think of it as a ring of functions on . The ideal of functions vanishing at is maximal, so a natural question given our discussion yesterday is whether these exhaust the set of maximal ideals of . It turns out that the answer is “yes,” and there are a lot of ways to prove it. Below I’ll describe the proof presented in Artin, which has the virtue of being quite short but the disadvantage of not generalizing. Then we’ll discuss how the Nullstellensatz allows us to describe the maximal spectra of finitely-generated -algebras.
See also the relevant post at Rigorous Trivialities.
Theorem (weak Nullstellensatz for ): The map from to is a bijection.
Proof. First we observe that the result is immediate for , since all ideals are principal, so any proper ideal is generated by a polynomial which vanishes somewhere, hence is contained in some ideal .
It turns out that the general case follows almost immediately from this one. Let be a maximal ideal and consider the quotient homomorphism . If for each the kernel contains for some constant , then is of the form , so suppose there is such that this doesn’t occur.
Now, prime ideals can be characterized among ideals by the property that their complements are closed under multiplication, and this will become very important later. It follows that if no element is in , neither is any of their products, so is not in the kernel and embeds in . Since is a field, must in fact extend to an injection . (In other words, the field of fractions of an integral domain is characterized by the universal property that any homomorphism to a field factors through it.)
Now we make the following observations:
- The elements of are linearly independent, so has uncountable dimension as a vector space over .
- is the image of a countable vector space over , hence is a countable-dimensional vector space over .
- Every linearly independent subset of a countable-dimensional vector space is at most countable.
These are all fairly easy to verify, so we are done. (Note, however, that in this form the Nullstellensatz actually holds for replaced by any algebraically closed field, but the above argument fails for the algebraic closure of a finite field.)
Probably the most important corollary of the weak Nullstellensatz, at least philosophically, is the following.
Corollary: Let be an ideal of generated by a finite set of polynomials . The maximal ideals of are precisely the maximal ideals where . If is not the unit ideal, there is at least one such .
Proof. if and only if contains , so the first claim is immediate from the correspondence theorem. The second claim follows from the fact that is contained in some maximal ideal, which is in turn a corollary of Zorn’s lemma (at least, for now).
We have not proven, but it will turn out, that every ideal of is of this form by Hilbert’s basis theorem, which we’ll discuss later. Thus the study of finitely-generated -algebras is equivalent to the study of quotients of polynomial rings over , which in turn is, if not equivalent, then at least very closely related to, subsets of defined by polynomial equations, or affine varieties. These are precisely the closed sets in the Zariski topology on , and the affine variety corresponding to an ideal will be denoted .
This is a vindication of our discussion of compact Hausdorff spaces: for finitely-generated -algebras the abstract space can in fact be associated to a concrete geometric object. Note, however, that the Zariski topology is very different from the usual topology coming from ; for one thing, it is almost never Hausdorff.
With that in mind, a natural question is whether the ring of functions on a variety is enough to determine it. Let denote the ring of polynomial functions on generated by the coordinate functions , and let denote the ideal of all polynomial functions vanishing on ; then clearly . If for some ideal then clearly contains . Since, as we discussed earlier, functions on a space cannot be nilpotent, it follows that contains the radical , since any such function satisfies identically on , hence identically on . In purely algebraic terms this means that , the intersection of all maximal ideals containing , contains . As a corollary of the strong Nullstellensatz (which, once again, we will discuss later) we will find that in fact the two are equal, so and the radical ideal corresponding to a variety does in fact determine it uniquely.
Examples of varieties
- The “circle” is an affine variety defined by a single polynomial, which is irreducible since is not a square in . It follows that is an integral domain. In fact, it is isomorphic to the ring of Laurent polynomials in . The “circle” is a Riemann surface, and the above change of variables shows that it is biholomorphic to the punctured plane . It turns out more generally that any compact Riemann surface is a (not necessarily affine) variety over , although we haven’t yet described what a non-affine variety is.
- The general linear group is the complement of the variety defined by , which is a polynomial function on . It follows that is a Zariski-open subset of . The “correct” way to turn into an affine variety is to introduce an extra variable and define the subset of given by .
- The union of the -axis and the -axis is the affine variety defined by , and unlike some of the other examples it is not a complex manifold because neighborhoods of the points that are not are homeomorphic to balls in but neighborhoods of are not. This is the simplest example of a singularity called a crunode. Since is not an integral domain, one of our goals will be to figure out how to detect singularities so we can avoid such behavior.
- The variety defined by has a cusp singularity at the origin, so it is also not a complex manifold. Since is irreducible, is an integral domain, but since can both be shown to be irreducible this ring does not have unique factorization. It turns out that the reason this ring doesn’t have unique factorization is essentially the same reason that doesn’t have unique factorization, and it can be repaired by tossing in an element of the field of fractions, which in this case is . Note that and , so the resulting ring is just . This choice of extra element, in turn, defines a homomorphism defined by , which corresponds to a parameterization of by a complex parameter . In this parameterization the point is no longer singular in the sense that the parameterization moves smoothly through it.
In the next few posts we’ll talk more about the basis theorem and the strong Nullstellensatz.