Hilbert’s Nullstellensatz plays a central role in algebraic geometry. It can be seen as the fundamental link between the modern theory of schemes and the classical theory of algebraic varieties over fields. Since this is one of the first results a novice in algebraic geometry learns and is often proved very algebraically, one often does not gain a good understanding of the proof till much later.
I would like to fix my own understanding the result and it’s geometric nature in this post. I will go through a few proofs of the theorem and point out the geometric ideas behind it. The proof of Hilbert’s lemma is usually broken up into the following two steps: 1) Prove the weak Nullstellensatz and 2) Derive the strong Nullstellensatz using the Rabinowitsch or other means. I will be focusing solely on the first step in this post. Nothing in this is new to me except perhaps the presentation and mistakes.
The weak Nullstellensatz is a statement about solving polynomial equations in multiple variables over a field. The one variable version of the problem is well understood (think Galois theory) and says that any polynomial over a field will have all it’s solutions in some finite extension of . The Nullstellensatz says that this result propagates to multiple variables. That is:
Theorem 1 [Weak Nullstellensatz]: Let be a set polynomials in . Then exactly one of the following is true:
- There exist polynomials such that .
- There exist values such that for all .
Clearly, both conditions cannot be simultaneously true. Therefore, it suffices to assume that 1) is false and prove 2). This leads to the following reformulation.
Theorem 1.2 : Let be the ideal spanned by the polynomials in . If is a proper ideal, then there is a finite field extension and a homomorphism .
This reformulation is clearly equivalent to the first and furthermore, by replacing by a maximal ideal containing it, we can assume that is a finitely generated field over .
One can think of this statement as saying that any maximal ideal of is in fact of the form and therefore, the closed point of a finite type scheme over a field are precisely the points of a corresponding algebraic variety. This is the connection between scheme theory and classical algebraic geometry that I alluded to above.
We can also compare this statement to the one-dimensional case and interpret the weak Nullstellensatz as saying that any solutions to a system of polynomial equations over a field in fact lie in an algebraic extension of the field.
We can make one final simplification before moving on to proofs. It is easily seen that it is sufficient to prove the theorem for algebraically closed fields since we can then recover the above formulation by passing to the algebraic closure of . Therefore, from now on, I will assume that is algebraically closed. What will be important for us is that is infinite and perfect.
Of the above formulations, the most important to us will be the idea that we want to locate rational points (in the algebraically closed case) on finite type schemes over a field. There is a common strategy to all our proofs that runs so:
Let be the given finite type scheme on which we want find a point with values in an algebraic closure. The idea then is to find a that receives a finite map with a rational point in the image. Usually, will be some well understood variety like affine space or some open subset of it.
Given such a map, we simply the rational point on the base and base change to it. This will be finite over and hence a finite field extension which will also simultaneously be a point of . This point on is what we were looking for all along!
One of the standard ways to prove the theorem is to use something called Noether Normalization, whose proof is also often couched in commutative algebra. I will give a geometric interpretation of the proof here and use it to quickly prove the Nullstellensatz.
Theorem 2 [Noether Normalization]: Let be finitely generated ring over the field and let be the corresponding scheme. Then, there is a finite, surjective map . Note that this forces .
Proof: We can embed in some (essentially, pick generators for ). Embed further into in the standard way and let be the projective closure of .
We can assume that is not all of since otherwise take and to be the embedding map. Therefore, is not all of and we can find a rational point at infinity that is not on . (Recall that we assumed to be an infinite field). Similarly, we can find a rational hyperplane (ie, defined over ) in .
Project to through . The map is proper and hence has closed image. Since was chosen to be at infinity, this restricts to a map . If the map is surjective, take to be , otherwise inductively continue the process till it is surjective.
The map is finite since it is quasi-finite finite and proper and therefore, we have established that there is some finite map .
Remark 0: The entire content of the theorem is in ensuring that the finite map is surjective. Otherwise, one could simply take a closed embedding into some affine space!
Remark 1: If one writes out the maps in the above proof explicitly, you recover the usual algebraic proof with some minor changes (and linear substitutions throughout). There is of course a fair bit of choice involved with the point and the hyperplane but one also has to choose (essentially) in the standard proof. Taking to be as simple as possible makes the computation easy.
Remark 2: Our life above was simplified by assuming to be infinite. If is infinite, the only change to be made is to embed take a Veronese embedding of first before choosing and . This corresponds to taking a hypersurface of high degree instead of a hyperplane for .
Deriving the Nullstellensatz: Proving the weak Nullstellensatz is a very short road from here and can be done in a couple of slightly different ways. Recall that
First: Suppose is a field that is finitely generated as a ring over . Then, by Noether normalization, we can find a finite map where . However, the dimension of any field is and so, and is finite over as required.
Second: More in line with the outlined strategy, one can proceed so: Let be a finitely generated ring over . By Noether normalization, we can find a finite map .
Since our finite map is surjective, take any rational point on the base and let be the fiber of over this point. Since finiteness is maintained under pull backs, is in fact finite over the field and is therefore a finite field extension.
That is, there is a solution of the polynomials defining in a finite field extension of , to be precise, in the field corresponding to
Unwinding the proof, we are essentially doing the following: Since the are finite over , they each satisfy some monic polynomial with coefficients in . We can substitute in values of for the variables and end up with a system of monic equations for the over .
Crucially, these equations will necessarily have a common solution in since the normalization map was surjective. However, since the fibers are finite, the solutions will all lie in some finite extension as required!
Remark: This last proof is really constructive in the sense that one can follow the procedure to find solutions for the system of polynomial equations we started with. It depends of course on Noether normalization being constructive but our proof is easily seen to provide an algorithm.
Transcendence basis and Noether Normalization
We do not in fact need the entire power of Noether normalization to prove the Nullstellensatz. One can make do with a suitable weaker version that only applies to field extensions and is in fact equivalent to the existence of a transcendence basis for a finitely generated field extension.
Recall that for a field , all finitely generated field extensions can be broken up in the form where the are algebraically independent over and is algebraic. Elements in the above decomposition are called a transcendence basis.
Further, if is perfect, then by the primitive element theorem, we can in fact write where is a monic polynomial over . Recall our standing assumption that is perfect.
This can be seen as a (very) weak version of Noether normalization in the following way: We can find an affine algebraic variety such that it’s function field . Then, the existence of a transcendence basis amounts to saying that the generic point of maps to the generic point of such that the resulting field extension is finite.
In fact, this isn’t that for off from Noether normalization for the following reason. We can spread out the map on generic points to a map over some open subset of the affine plane in the following way:
Consider the coefficients of (as defined a couple of paragraphs above). They are a finite set of rational polynomials in the variables . We can assume they have a common denominator . Then, let and . It is easy to see that is a model for over in the sense that . It is equally clear that continues to be finite over .
Since corresponds to an open subset of and to a variety with function field , we might as well take . Thus, we have found a finite surjective map an open subscheme of as promised.
We can use these ideas to prove the Nullstellensatz.
The Nullstellensatz once again
Let be a finitely generated ring over that also happens to be a field. By our above result on transcendence basis, there is an isomorphism .
Now, is the union (=direct limit) of rings of the form for appropriate and a monic polynomial over . This follows directly from our discussion on spreading out in the last section since any function field is the union of the rings corresponding to distinguished open sets.
Since is finitely generated over , we can find some such that all the map to . However, since is infinite (recall our standing assumption), we can find such that . Thus, we can find a map where we evaluate at the values .
However, this tells us that there is a map which forces to itself be a finite field extension of (since ) is (and was assumed to be a field).
Remark: This proof has unmistakable similarities to the our second proof two sections ago. The key difference is that we have substituted Noether Normalization with the existence of a transcendence basis.
Both the proofs proceed by realizing our finitely generated scheme as something finite over something well understood. In the previous section, we could take the base to be all of and further demand that the map be surjective. This allowed us to pick any rational point on the base to find a point on .
However, in the proof in this section, we can only take our base to be an open subset of (in our proof, this corresponds to and accordingly, we need to take a point that lies in this open set. Since surjectivity is basically obtained from our construction of spreading out, this is not too hard to do if we insist our field be infinite. After this, the two proofs proceed in exactly the same way.
Doing away with Transcendental basis
By modifying our last proof a little, we can in fact even do away with any reliance on transcendental basis. Instead, we can proceed by induction on the number of variables in our algebra.
Let be a field as usual. We want to show that is a finite field extension of . Equivalently, we want to show that the are algebraic over (thinking of the as elements in ). We will do this by induction on . Note that for a fixed $latex $
The case of is tautological. So let us suppose that . Denote by the image of in . I will henceforth write . For contradiction, we can suppose that is transcendental over .
By our inductive hypothesis, this implies that the are algebraic over . By a similar spreading out argument as before, we can in fact suppose that the image of is of the form where is a polynomial over $\latex k$.
Therefore, we have shown that there is a finite map where is an open subscheme of and we can conclude the proof in a couple of ways.
We can specialize to a rational value and proceed as before or alternatively, we can end with the slick observation that if a field is integral over a ring , then , then is itself a field. However, in our case and this is certainly not a field!
Remark: The idea here is that any transcendental extension over will necessarily contain . This is equivalent to the first step of building a transcendental basis.
However, once we know that it contains , we can spread out as before and obtain a map . Since finiteness can be preserved by spreading out, we can use our induction hypothesis to establish finiteness and then proceed as before.
This proof seems really magical because we seem to be getting the finiteness hypothesis for free. However, considering the first non trivial case of and one polynomial equation, we see that this proof is really somehow the obvious thing to do. Treat one of the variables as constant and solve for the other variable! Then, figure out an appropriate specialization so that a solution exists. The appropriate specialization is equivalent to finding a point in such that does not vanish…