This is going to be the first post in a series I’d like to call “Easy theorems, hard proofs”. The idea is simple. Take a classic theorem from some field, maybe whose proof is somewhat messy, and realize it as a corollary of a much more sophisticated theorem. Rarely will the proof be “better”, in the sense that it’s the proof one should initially learn, but hopefully the proof is neat enough, or elegant enough, to warrant attention.
I can think of no better way to start this series than with one of my favorite proofs of all time: the algebraic geometry proof of the Nullstellensatz.
The Nullstellensatz: Preamble
This may be worth skipping if you are already familiar with the various forms of the Nullstellensatz.
The Nullstellensatz (“zeroes theorem”) is a famous result, known under many guises. Perhaps the most well-known phrasing is the following:
Theorem(Classic Nullstellensatz): The maximal ideals of the polynomial ring are those of the form for .
This is the first theorem one learns in algebraic geometry. It gives one a glimpse of the bridge between algebra and geometry, by providing a natural bijection .
Perhaps the next most common phrasing is in terms of the bijection between closed subsets of (in the Zariski topology) and radical ideals of –an extension of the bijection given in the previous paragraph. Namely, let us define to be the set of closed subsets of in the Zariski topology, and the set of radical ideals of . We will define maps and that will be inverses of one another. Namely, we define and as follows, where is any subset of and is any ideal of :
Colloquially, is the “vanishing set” of an ideal, and is “ideal of vanishing” of a set. One would expect that these should be inverse to one another. It’s trivial to check that and are maps and respectively (i.e. their images land in the right set). It’s also easy to see that for all . What is not obvious, is that for .
More specifically, this says that if is an ideal in , and if is such that whenever for all , then for some , and for some one has that
To make it even more intuitive, if , this says that if whenever , then for some , and some . This seems intuitive, just because you can’t think of how else one would obtain such an , but it is far from “obvious”.
The proof follows from the more general version of the previous Nullstellensatz:
Theorem(Strong Classical Nullstellensatz): For any ideal the following equality holds:
where denotes the radical. Of course, the fact that for then follows, since is radical.
To see how the Classical Nullstellensatz follows from the Strong Classical Nullstellensatz, suppose that is a maximal ideal of . As one can easily check, and are inclusion reversing. We claim that this implies that is a minimal non-empty subset of . Indeed, if , then , for if then
But, since , this then implies that . But, if , then . Thus, must be empty.
But, since every set in contains a point, it follows that the minimal elements of are points, and so
for some .
But, by inspection, one sees that
Since the strong Nullstellensatz implies that and are inverses, and so in particular is injective, we may conclude that as desired.
The next version of the Nullstellensatz comes from using the Classic Nullstellensatz to reinterpret what the Strong Classic Nullstellensatz is saying. Let us first note that given , the ideal is just . Indeed, it’s clear that is contained in , and since is maximal we have equality. But, note that by definition, if then
And, since, by inspection, if and only if , we see that
If we then assume the Classic Nullstellensatz, the above shows that
Thus, the Strong Classic Nullstellensatz says precisely that the radical of an ideal is the intersection of all maximal ideals containing . Now, it’s a classic fact of algebra that the radical is the intersection of all prime ideals containing . Thus, the strong Nullstellensatz is saying that we need to intersect many fewer ideals to obtain the radical, only the maximal ones.
In general, a ring is called Jacobson if the radical of any ideal is the intersection of all maximal ideals containing it. Thus, the Strong Classical Nullstellensatz, assuming the Classical Nullstellensatz, says precisely that is a Jacobson ring. This leads us to make the following claim:
Theorem(Jacobson Classical Nullstellensatz): The ring is Jacobson.
The above analysis then shows that the Jacobson Classical Nullstellensatz together with the Classical Nullstellensatz imply the Strong Classical Nullstellensatz.
In fact, as is well-known, we can replace in all of the above examples with an algebraically closed field . We then arrive at the Semi-Classical versions of the above:
Theorem(Semi-Classic Nullstellensatz): Let be an algebraically closed field. The maximal ideals of the polynomial ring are those of the form for .
Theorem(Strong Semi-Classical Nullstellensatz): Let be an algebraically closed field. For any ideal the following equality holds:
Theorem(Jacobson Semi-Classical Nullstellensatz): Let be an algebraically closed field. The ring is Jacobson.
So, how can we prove all of these nice theorems? It turns out that the answer lies in a, ostensibly unrelated, but interesting theorem. To understand the content of the theorem, let us first recall a classic theorem from basic field theory.
Theorem 1: Let be a field extensions. Then, is algebraic over if and only if .
In other words, the ring , the ring of all -polynomial’s evaluated at , is a field if and only if is algebraic. But, since is algebraic over if and only if is finite, we can phrase this theorem as follows:
Theorem 2: Let be a field extensions. Then, is such that is finite if and only if .
The proof of this theorem was easy. We have a surjection defined by . If is algebraic, we know that this map is non-injective with kernel for some . But, since , and is a domain, we know that is irreducible. But, since is a PID, this implies that, in fact, is maximal and so is a field. Conversely, if is not algebraic, then the surjection is injective, and so which implies that is not a field.
The next natural thing one might ask then is if there is a generalization of Theorem 2, to non-simple extensions. Namely, one might conjecture the following:
Conjecture: Let be a field extensions. Then, are such that is finite if and only if .
Of course, we attempt to argue as we did in the case of . Namely, we start by inspecting the surjection
But, we then immediately see the difficulty. Whereas the ideal structure of is simple, it’s a PID, the ideal structure of is difficult–to prove this conjecture, we’d need to understand the ideal structure of better. From this, we see the connection to the Nullstellensatz.
Our conjecture turns out to be true, and can easily be proven by the result commonly known as Zariski’s lemma:
Theorem(Zariski’s Lemma): Let be a field extensions. Then, is a finite type -algebra, if and only if it is a finitely generated -module.
Said differently, Zariski’s lemma says that an ideal is maximal, if and only if is a finite extensions of .
Before we prove Zariski’s lemma, let us show explicitly how it lays waste to all the previously discussed results:
Proof(Semi-Classic Nullstellensatz): Let be a maximal ideal of . Then, the extension is a finite type -algebra, which is also a field. It follows from Zariski’s lemma that it’s a finite extensions. But, since is algebraically closed, this implies that it is a degree one extension. Thus, by definition, the composition of the following maps
is an isomorphism. In particular, for each , there exists such that
Thus, , and so . Since is maximal, this implies that as desired.
Proof(Jacobson Semi-Classical Nullstellensatz): We are trying to show that for any ideal of , the following holds:
where ranges over maximal ideals containing , and over prime ideals containing . Evidently we have the inclusion
To show the other inclusion, we need to show that if is such that there is a prime with , then there is some maximal with . To see this, consider the localization of . Since , we know that is prime. Take a maximal ideal of which contains . Let be the contraction of to (i.e. the ideal of numerators).
By the classic correspondence between primes of a rings and its localization, we know that is a prime ideal of not containing –the correspondence does not say that maximals correspond to maximals. We claim that in this case though is maximal. Indeed, we have an injection
Note though that is a finite type -algebra (localizing at is equivalent to appending ), and so by Zariski’s lemma, is a finite -module. But, the above injection (of -algebras!) then implies that is a finite -module.
But, since it is a domain (since is prime), it follows from the basic theory that is a field (i.e. the quotient is a dimension domain). If this parenthetical statement is meaningless to you, we can be more direct. Explicitly, given we know that the set is an infinite set, and so must be linearly dependent over . In particular, there exists some minimal , and some such that
If , then we could divide through by (since we’re in a domain), contradicting the minimality of . Thus, we see that
Since is a unit, this implies that is a unit. Since was arbitrary, it follows that is a field.
This proof was taken largely from Pete L. Clark’s commutative algebra notes (proposition 11.3). It is also worth noting that, in general, a finite type extension of a Jacobson ring is Jacobson. Since a field is Jacobson, this also implies the Jacobson Semi-Classical Nullstellensatz.
Proof(Strong Semi-Classical Nullstellensatz): Just as in the classical case, the Semi-Classical Nullstellensatz together with the Jacobson Semi-Classical Nullstellensatz gives you the Strong Semi-Classical Nullstellensatz.
Proof(Conjecture): If is is finite, then by the basic theory (check the end of the last proof!) we know that is a field. Conversely, if we know that , then is a finite type -algebra which is a field. By Zariski’s lemma this implies that is finite as desired.
Proof of Zariski’s Lemma
After all of this work, which is relatively elementary, we have reduced everything we could ever want to know about the Nullstellensatz, all its various forms, to Zariski’s lemma. There are several proofs of Zariski’s lemma, with varying degrees of sophistication/messiness. My absolute favorite, and by far the quickest (if not the most elementary), is by means of a fun theorem from algebraic geometry:
Theorem(Chevalley): Let and be Noetherian schemes, and let be finite type. Then, for any constructible (finite union of locally closed) subset , the image is constructible.
The proof of this is slightly cumbersome, but is something any student of algebraic geometry has worked through at least once. It tells you that while morphisms needn’t be open or closed in general, they have at least some amount of topological tameness.
How in the world is this useful in proving Zariksi’s lemma though? The key lemma, which ties it all together, is the following:
Lemma: Let be a field. Then, the generic point of is not constructible.
Proof: Suppose that was constructible. Then, there exists open sets , and closed sets such that
Now, clearly each is all of since it contains the dense point . Thus, we see that this implies is open. But, all non-empty open subsets of are infinite, which is a contradiction.
With this simple lemma, we are now able to prove Zariski’s lemma:
Proof(Zariki’s Lemma): Since is finite type, to show it’s finite, it suffices to show it’s integral. Suppose not. Then, there is a -algebra injection . This then induces a morphism . This morphism is obviously dominant, and so the image contains the generic point. But, since is a point, it follows that the image of is exactly the generic point of . Since is a finite type morphism between Noetherian schemes, we know from Chevalley’s theorem that the image of is constructible. But, we have already noted that the image of is the generic point of which, by the lemma, is not constructible. This is a contradiction.
I am not sure who was the first to come up with this proof. I encountered it first in Vakil’s Foundations of Algebraic Geometry. If anyone knows the original author of this proof, I would love to know.