# The Nullstellensatz and Partitions of Unity

The Nullstellensatz, first proved by Hilbert, is the following theorem:

Let $k$ be an algebraically closed field and let $f_1$, $f_2$, …, $f_n$ be polynomials in $k[x_1, \ldots, x_m]$. Then exactly one of the following is true: either (1) the polynomials $f_1$, $f_2$, …, $f_n$ have a common root in $k^m$ or else (2) there are polynomials $g_1$, … $g_n$ such that $f_1 g_1 + f_2 g_2+\ldots+f_n g_n=1$.

It is clear that at most one of these can be true. If (1) held, so that we had a root of $f_1=f_2=\ldots=f_n=0$, then plugging this root into both sides of the relation $f_1 g_1 + f_2 g_2+\ldots+f_n g_n=1$ would give $0=1$, a contradiction. So the challenge is to show that at least one of these two claims is true. By the way, this is sometimes called the Weak Nullstellensatz. There is a variant called the Strong Nullstellensatz where one tries to solve the equations $f_1=f_2=\ldots=f_n=0$, $g \neq 0$. It is straightforward (not trivial) to deduce Strong from Weak by considering the system of equations $f_1=f_2=\ldots=f_n=1-yg=0$ in $m+1$ variables $(x_1, x_2, \ldots, x_m, y)$, so I will focus on the weak case.

Terry Tao has a nice post explaining how to prove the Nullstellensatz by bulldozing forward in the most straightforward manner. (Hopefully, he won’t find this description insulting; I consider the ability to prove results by hammering on them to be very important.) An anonymous commentor on that post points out a very slick three page proof by Arrondo. In this post, I want to discuss a way of understanding the Nullstellensatz which I think is probably well known to experts, but I haven’t seen presented in any elementary books.

Let me remind the reader of the classical theorem of existence of partitions of unity. Let $X$ be a smooth manifold and let $U_1$, …, $U_n$ be a covering of $X$ by open sets. Then there are smooth functions $h_i$ on $X$ such that $h_i$ is zero on the complement of $U_i$ and $\sum h_i=1$.

Now, let’s consider the case that our manifold $X$ is $\mathbb{C}^m$ and our open sets $U_i$ are of the form $f_i \neq 0$, for $f_1$, $f_2$, …, $f_n$ some set of polynomials. The assumption that the $U_i$ cover all of $\mathbb{C}^m$ precisely means that the $f_i$ do not have a common zero; in other words, condition (1) does not hold. By the theorem of existence of partitions of unity, there are smooth functions $h_1$, …, $h_n$ such that $h_i$ is zero whenever $f_i$ is zero and $\sum h_i=1$. In general, the proof of existence of partitions of unity is extremely nonconstructive; it does not give us a practical way to build the $h_i$. However, in the example where $X$ is $\mathbb{C}^m$ and $U_i=\{ f_i \neq 0 \}$, the Nullstellansatz gives us an explicit answer. Since condition (1) does not hold, condition (2) must. Take $h_i=f_i g_i$. Since $f_i$ divides $h_i$, we know that $h_i$ is zero whenever $f_i$ is zero and we have $\sum f_i g_i=\sum h_i=1$.

So, the Nullstellensatz is a version of partitions of unity that works over any algebraically closed field, and with no mention of convergence issues or topology. (There is a tool called the Zariski topology one can use to make this statement precise, although it isn’t quite as easy as you might think.) Whenever I think to myself “I’d like to use a partition of unity here”, I think “Could I use the Nullstellensatz instead?” Indeed, in my own thought, I often refer to the existence of partitions of unity as “the smooth Nullstellensatz”. Of course one major difference between the two results is that partitions of unity exist in $\mathbb{R}^n$, while the Nullstellensatz is not true in $\mathbb{R}^n$. (Consider the polynomials $f_1(t)=(1+t^2)(1-t)$ and $f_2(t)=(1+t^2)(1+t)$. They have no real roots in common, but nonetheless there is no pair of polynomials $(g_1(t), g_2(t))$ such that $f_1 g_1+f_2 g_2=1$.) The ring of polynomials can see complex zeroes, while the ring of smooth functions can not. (Excercise: what about real analytic functions on $\mathbb{R}$?)

I could talk about how to use partitions of unity in setting up sheaf theory, but I suspect the audience for that is small. Instead, a challenge to check your intuition. Let $M$ be a bounded metric space and let $L(M)$ be the ring of real valued Lipschitz functions on $M$. True or false: for every open cover of $M$, there is a partition of unity subordinate to that open cover using functions in $L(M)$? If not, what condition should I have imposed on $M$ to make this true?

## 10 thoughts on “The Nullstellensatz and Partitions of Unity”

1. No offense taken – bulldozing is what I do for a living :-)

There is also a version of partitions of unity for bounded holomorphic functions on, say, the unit disk, which is known (for historical reasons) as the corona theorem: if $f_1,\ldots,f_n$ are bounded holomorphic functions on the disk which do not all simultaneously vanish, then there exist bounded holomorphic functions $g_1,\ldots,g_n$ such that $f_1 g_1 + \ldots + f_n g_n = 1$. It has slick proofs these days (in particular one by Wolff), but it is certainly not trivial (the first proof was due to Carleson). [In case you’re wondering, the reason it is called the corona theorem is that it is equivalent to the statement that the space of maximal ideals of $H^\infty(D)$, the Banach algebra of bounded holomorphic functions on the disk, with the usual weak topology, contains the disk D inside it as a dense subspace (identifying each point z in the disk with the maximal ideal $\{ f: f(z) = 0 \}$), thus this space of maximal ideas forms a “corona” around the disk.]

2. Rishi says:

I have always wondered how Hilbert originally proved the Nullstellensatz. He did not have Lasker’s theorem on primary decomposition of ideals or Noether’s normalization theorem, which give straightforward proofs. Perhaps it was a version of the elimination theory proof given in the earlier editions of van der Waerden’s Modern Algebra (vol. 2). Does anyone know?

3. David,

I think there is great utility in having many models (proofs) of the same idea (theorem). One proof might lead to explicit computable bounds, but is difficult to understand. Another might not lead to bounds, but has strong ties to topology, and is easily visualizable. And then there is my favorite, the algebraic version which tries to prove the nullstellensatz over the most general commutative ring possible (but isn’t necessarily immediately useful in such generality-but often leads to more examples showing the limitations of the theorem). In all cases, different generalizations of the theorem often present themselves.

I have an algebraic proof (that Lam showed us at Berkeley) hidden in one of my many notebooks (which is different than the proof given in his “First Course” book). Maybe I’ll drag it out and find the reference.

4. Pace,

I agree with every bit of that and I’d be curious to see your reference. Is it like the proof of Zariski, where you reduce to the statement that, if L is a field which is finitely generated as an algebra over a subfield K, then L is finite dimensional as a vector space over K? That is my favorite non-computational proof.

5. Anonymous says:

Something I wonder about, as a consumer rather than a producer of algebraic geometry theorems, is how infrequently I notice the Nullstellensatz or other foundational results being used. For instance, I flatter myself that I have a not undetailed understanding of why something like Borel-Weil, or Grothendieck-Riemann-Roch, is true, but come to think of it I have no idea where or if the Nullstellensatz (or the going-up theorem or anything like that) appears.

6. David,

It is slightly different than that proof. Lam’s proof was based on Munshi’s proof, which uses a clever induction and what Kaplansky calls G-domains. You might google “Munshi nullstellensatz” to find out more.

7. David Speyer says:

Anonymous —

In my experience, one often doesn’t really need the Nullstellansatz because one can often substitute the following weaker, and much easier, result:

Let f_1, …, f_n be polynomials in k[x_1, …, x_m] such that there do not exist any polynomials g_1, …, g_n with sum f_i g_i=1. Then there is an extension field K of k and a point (a_1,…,a_m) in K^m which is a zero of f_1, …, f_n.

This result differs from the Nullstellansatz in that the Nullstellansatz makes the additional claim that we can take K to be a finite extension of k. (If k is algebraically closed, which is the setting in which the Nullstellansatz is usually stated, then the Nullstellansatz says that we can take k=K.)

This weaker result is used constantly in the foundations of algebraic geometry; I suspect that you can’t find any major result which doesn’t use it. Whenever you talk about having an open cover of Spec A, what you are saying is that you have a collection of basic opens Spec A[f_1^{-1}], …, Spec A[f_n^{-1}] for which the f_i have no common zero in any extension field K of k. Almost always, if you look at the commutative algebra underlying your algebraic geometry, the next step in the proof is to take a linear combination sum f_i g_i=1.

In pre-Grothendieck algebraic geometry, Spec A only had points for the homomorphisms from A to k. So, in those days, one constantly had to use the Nullstellansatz in order to deduce anything from the hypothesis of having an open cover. But today, Spec A has points for every homomorphism of A to a domain. So we can usually use the weaker result.

However, my intuition for algebraic geometry would be totally different if the complex points of a complex variety were not dense in that variety. I could no longer think of myself as studying the behavior of complex manifolds; almost all the interesting behavior would be hidden in exotic extensions of C.

I’d be curious to hear examples within algebraic geometry where it is truly necessary to know that we can take K to be a finite extension of k. I’m sure that this fact shows up whenever we want to relate algebraic geometry to complex manifold theory. It is also going to turn up all over number theory, where people take varieties defined over Q and need to know that every closed point is defined over a finite extension of Q, so they can use Galois techniques. Any other nominations?

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