# The Nullstellensatz

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 $\mathbb{C}[T_1,\ldots,T_n]$ are those of the form $(T_1-a_1,\ldots,T_n-a_n)$ for $a_1,\ldots,a_n\in\mathbb{C}$.

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 $\text{MaxSpec}(\mathbb{C}[T_1,\ldots,T_n])\to\mathbb{C}^n$.

Perhaps the next most common phrasing is in terms of the bijection between closed subsets of $\mathbb{C}^n$ (in the Zariski topology) and radical ideals of $\mathbb{C}[T_1,\ldots,T_n]$–an extension of the bijection given in the previous paragraph. Namely, let us define $\mathcal{S}$ to be the set of closed subsets of $\mathbb{C}^n$ in the Zariski topology, and $\mathcal{R}$ the set of radical ideals of $\mathbb{C}[T_1,\ldots,T_n]$. We will define maps $I:\mathcal{S}\to\mathcal{R}$ and $V:\mathcal{R}\to\mathcal{S}$ that will be inverses of one another. Namely, we define $I$ and $V$ as follows, where $S$ is any subset of $\mathbb{C}^n$ and $J$ is any ideal of $\mathbb{C}[T_1,\ldots,T_n]$:

$I(S)=\left\{f\in \mathbb{C}[T_1,\ldots,T_n]:f(a_1,\ldots,a_n)=0\text{ for all }(a_1,\ldots,a_n)\in S\right\}$

$V(J)=\left\{(a_1,\ldots,a_n):f(a_1,\ldots,a_n)=0\text{ for all }f\in J\right\}$

Colloquially, $V$ is the “vanishing set” of an ideal, and $I$ is “ideal of vanishing” of a set. One would expect that these should be inverse to one another. It’s trivial to check that $I$ and $J$ are maps $\mathcal{S}\to\mathcal{R}$ and $\mathcal{R}\to\mathcal{S}$ respectively (i.e. their images land in the right set). It’s also easy to see that $V(I(S))=S$ for all $S\in\mathcal{S}$. What is not obvious, is that $I(V(J))=J$ for $J\in\mathcal{R}$.

More specifically, this says that if $(f_1,\ldots,f_r)$ is an ideal in $\mathbb{C}[T_1,\ldots,T_n]$, and if $f\in\mathbb{C}[T_1,\ldots,T_n]$ is such that $f(a_1,\ldots,a_n)=0$ whenever $f_i(a_1,\ldots,a_n)=0$ for all $i$, then for some $m\geqslant 0$, and for some $g_1,\ldots,g_r\in\mathbb{C}[T_1,\ldots,T_n]$ one has that

$\displaystyle f^m=\sum_{i=1}^{r}g_i f_i$

To make it even more intuitive, if $r=1$, this says that if $f(a_1,\ldots,a_n)=0$ whenever $f_1(a_1,\ldots,a_n)=0$, then $f^m=g f_1$ for some $m\geqslant 0$, and some $g\in\mathbb{C}[T_1,\ldots,T_n]$. This seems intuitive, just because you can’t think of how else one would obtain such an $f$, 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 $J\in\mathbb{C}[T_1,\ldots,T_n]$ the following equality holds:

$I(V(J))=\sqrt{J}$

where $\sqrt{ }$ denotes the radical. Of course, the fact that $I(V(J))=J$ for $J\in\mathcal{R}$ then follows, since $J$ is radical.

To see how the Classical Nullstellensatz follows from the Strong Classical Nullstellensatz, suppose that $\mathfrak{m}$ is a maximal ideal of $\mathbb{C}[T_1,\ldots,T_n]$. As one can easily check, $I$ and $V$ are inclusion reversing. We claim that this implies that $V(\mathfrak{m})$ is a minimal non-empty subset of $\mathcal{S}$. Indeed, if $S\subsetneq V(\mathfrak{m})$, then $I(S)\supsetneq I(V(\mathfrak{m}))$, for if $I(S)=I(V(\mathfrak{m}))$ then

$S=V(I(S))=V(I(V(\mathfrak{m})))=V(\mathfrak{m})$

But, since $I(V(\mathfrak{m}))\supseteq\mathfrak{m}$, this then implies that $I(S)=\varnothing$. But, if $(a_1,\ldots,a_n)\in S$, then $T_1-a_1\in I(S)$. Thus, $S$ must be empty.

But, since every set in $\mathcal{S}$ contains a point, it follows that the minimal elements of $\mathcal{S}$ are points, and so

$V(\mathfrak{m})=\{(a_1,\ldots,a_n)\}$

for some $(a_1,\ldots,a_n)\in\mathbb{C}^n$.

But, by inspection, one sees that

$V((T_1-a_1,\ldots,T_n-a_n))=\{(a_1,\ldots,a_n)\}$

Since the strong Nullstellensatz implies that $I$ and $V$ are inverses, and so in particular $V$ is injective, we may conclude that $\mathfrak{m}=(T_1-a_1,\ldots,T_n-a_n)$ 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 $(a_1,\ldots,a_n)\in\mathbb{C}^n$, the ideal $I(\{(a_1,\ldots,a_n)\})$ is just $(T_1-a_1,\ldots,T_n-a_n)$. Indeed, it’s clear that $(T_1-a_1,\ldots,T_n-a_n)$ is contained in $I(\{(a_1,\ldots,a_n)\})$, and since $(T_1-a_1,\ldots,T_n-a_n)$ is maximal we have equality. But, note that by definition, if $S\subseteq\mathbb{C}^n$ then

$\displaystyle I(S)=\bigcap_{(a_1,\ldots,a_n)\in S}(T_1-a_1,\ldots,T_n-a_n)$

And, since, by inspection, $J_1\subseteq J_2$ if and only if $V(J_1)\supseteq V(J_2)$, we see that

\begin{aligned}I(V(J)) & = \bigcap_{(a_1,\ldots,a_n)\in V(J)}(T_1-a_1,\ldots,T_n-a_n)\\ &= \bigcap_{\stackrel{(a_1,\ldots,a_n)\text{ such that}}{V((T_1-a_1,\ldots,T_n-a_n))\subseteq V(J)}} (T_1-a_1,\ldots,T_n-a_n)\\ &= \bigcap_{\stackrel{(a_1,\ldots,a_n)\text{ such that}}{J\subseteq (T_1-a_1,\ldots,T_n-a_n)}}(T_1-a_1,\ldots,T_n-a_n)\end{aligned}

If we then assume the Classic Nullstellensatz, the above shows that

$\displaystyle I(V(J))=\bigcap_{\stackrel{\mathfrak{m}\in\text{MaxSpec}(\mathbb{C}[T_1,\ldots,T_n])}{\mathfrak{m}\supseteq J}}\mathfrak{m}$

Thus, the Strong Classic Nullstellensatz says precisely that the radical of an ideal $J\subseteq\mathbb{C}[T_1,\ldots,T_n]$ is the intersection of all maximal ideals containing $J$. Now, it’s a classic fact of algebra that the radical is the intersection of all prime ideals containing $J$. 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 $R$ 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 $\mathbb{C}[T_1,\ldots,T_n]$ is a Jacobson ring. This leads us to make the following claim:

Theorem(Jacobson Classical Nullstellensatz): The ring $\mathbb{C}[T_1,\ldots,T_n]$ 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 $\mathbb{C}$ in all of the above examples with an algebraically closed field $k$. We then arrive at the Semi-Classical versions of the above:

Theorem(Semi-Classic Nullstellensatz): Let $k$ be an algebraically closed field. The maximal ideals of the polynomial ring $k[T_1,\ldots,T_n]$ are those of the form $(T_1-a_1,\ldots,T_n-a_n)$ for $a_1,\ldots,a_n\in k$.

Theorem(Strong Semi-Classical Nullstellensatz): Let $k$ be an algebraically closed field. For any ideal $J\in k[T_1,\ldots,T_n]$ the following equality holds:

$I(V(J))=\sqrt{J}$

Theorem(Jacobson Semi-Classical Nullstellensatz): Let $k$ be an algebraically closed field. The ring $k[T_1,\ldots,T_n]$ 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 $L/K$ be a field extensions. Then, $\alpha\in L$ is algebraic over $K$ if and only if $K[\alpha]=K(\alpha)$.

In other words, the ring $K[\alpha]$, the ring of all $K$-polynomial’s evaluated at $\alpha$, is a field if and only if $\alpha$ is algebraic. But, since $\alpha\in L$ is algebraic over $K$ if and only if $K[\alpha]/K$ is finite, we can phrase this theorem as follows:

Theorem 2: Let $L/K$ be a field extensions. Then, $\alpha\in L$ is such that $K[\alpha]/K$ is finite if and only if $K[\alpha]=K(\alpha)$.

The proof of this theorem was easy. We have a surjection $K[T]\to K[\alpha]$ defined by $f(T)\mapsto f(\alpha)$. If $\alpha$ is algebraic, we know that this map is non-injective with kernel $(m(T))$ for some $m(T)\in K[T]$. But, since $K[T]/(m(T))\cong K[\alpha]$, and $K[\alpha]$ is a domain, we know that $m(T)$ is irreducible. But, since $K[T]$ is a PID, this implies that, in fact, $(m(T))$ is maximal and so $K[T]/(m(T))\cong K[\alpha]$ is a field. Conversely, if $\alpha$ is not algebraic, then the surjection $K[T]\to K[\alpha]$ is injective, and so $K[T]\cong K[\alpha]$ which implies that $K[\alpha]$ 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 $L/K$ be a field extensions. Then, $\alpha_1,\ldots,\alpha_n\in L$ are such that $K[\alpha_1,\ldots,\alpha_n]/K$ is finite if and only if $K[\alpha_1,\ldots,\alpha_n]=K(\alpha_1,\ldots,\alpha_n)$.

Of course, we attempt to argue as we did in the case of $n=1$. Namely, we start by inspecting the surjection

$K[T_1,\ldots,T_n]\to K[\alpha_1,\ldots,\alpha_n]: f(T_1\ldots,T_n)\mapsto f(\alpha_1,\ldots,\alpha_n)$

But, we then immediately see the difficulty. Whereas the ideal structure of $K[T]$ is simple, it’s a PID, the ideal structure of $K[T_1,\ldots,T_n]$ is difficult–to prove this conjecture, we’d need to understand the ideal structure of $K[T_1,\ldots,T_n]$ 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 $L/K$ be a field extensions. Then, $L$ is a finite type $K$-algebra, if and only if it is a finitely generated $K$-module.

Said differently, Zariski’s lemma says that an ideal $I\subseteq K[T_1,\ldots,T_n]$ is maximal, if and only if $K[T_1,\ldots,T_n]/I$ is a finite extensions of $K$.

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 $\mathfrak{m}$ be a maximal ideal of $k[T_1,\ldots,T_n]$. Then, the extension $k[T_1,\ldots,T_n]/\mathfrak{m}$ is a finite type $k$-algebra, which is also a field. It follows from Zariski’s lemma that it’s a finite extensions. But, since $k$ is algebraically closed, this implies that it is a degree one extension. Thus, by definition, the composition of the following maps

$k\to k[T_1,\ldots,T_n]\to k[T_1,\ldots,T_n]\to k[T_1,\ldots,T_n]/\mathfrak{m}$

is an isomorphism. In particular, for each $i=1,\ldots,n$, there exists $a_i\in\mathfrak{m}$ such that

$T_i=a_i\mod\mathfrak{m}$

Thus, $T_i-a_i\in\mathfrak{m}$, and so $(T_1-a_1,\ldots,T_n-a_n)\subseteq\mathfrak{m}$. Since $(T_1-a_1,\ldots,T_n-a_n)$ is maximal, this implies that $(T_1-a_1,\ldots,T_n-a_n)=\mathfrak{m}$ as desired. $\blacksquare$

Proof(Jacobson Semi-Classical Nullstellensatz): We are trying to show that for any ideal $I$ of $k[T_1,\ldots,T_n]$, the following holds:

$\displaystyle \bigcap_\mathfrak{m}\mathfrak{m}=\bigcap_\mathfrak{p}\mathfrak{p}$

where $\mathfrak{m}$ ranges over maximal ideals containing $I$, and $\mathfrak{p}$ over prime ideals containing $I$. Evidently we have the inclusion

$\displaystyle \bigcap_\mathfrak{m}\mathfrak{m}\supseteq \bigcap_\mathfrak{p}\mathfrak{p}$

To show the other inclusion, we need to show that if $f\in k[T_1,\ldots,T_n]$ is such that there is a prime $\mathfrak{p}\supseteq I$ with $f\notin \mathfrak{p}$, then there is some maximal $\mathfrak{m}\supseteq I$ with $f\notin \mathfrak{m}$. To see this, consider the localization of $k[T_1,\ldots,T_n]_f$. Since $a\notin\mathfrak{p}$, we know that $\mathfrak{p}k[T_1,\ldots,T_n]_f$ is prime. Take a maximal ideal $\mathfrak{m}'$ of $k[T_1,\ldots,T_n]_f$ which contains $\mathfrak{p} k[T_1,\ldots,T_n]_f$. Let $\mathfrak{m}$ be the contraction of $\mathfrak{m}'$ to $k[T_1,\ldots,T_n]$ (i.e. the ideal of numerators).

By the classic correspondence between primes of a rings and its localization, we know that $\mathfrak{m}$ is a prime ideal of $k[T_1,\ldots,T_n]$ not containing $a$–the correspondence does not say that maximals correspond to maximals. We claim that in this case though $\mathfrak{m}$ is maximal. Indeed, we have an injection

$k[T_1,\ldots,T_n]/\mathfrak{m}\to k[T_1,\ldots,T_n]_f/\mathfrak{m}'$

Note though that $k[T_1,\ldots,T_n]_f$ is a finite type $k$-algebra (localizing at $f$ is equivalent to appending $f^{-1}$), and so by Zariski’s lemma, $k[T_1,\ldots,T_n]_f/\mathfrak{m}'$ is a finite $k$-module. But, the above injection (of $k$-algebras!) then implies that $k[T_1,\ldots,T_n]/\mathfrak{m}$ is a finite $k$-module.

But, since it is a domain (since $\mathfrak{m}$ is prime), it follows from the basic theory that $k[T_1,\ldots,T_n]/\mathfrak{m}$ is a field (i.e. the quotient is a dimension $0$ domain). If this parenthetical statement is meaningless to you, we can be more direct. Explicitly, given $g+\mathfrak{m}\ne 0$ we know that the set $\{1+\mathfrak{m},\ldots,g^m+\mathfrak{m},\ldots\}$ is an infinite set, and so must be linearly dependent over $k$. In particular, there exists some minimal $\ell$, and some $a_0,\ldots,a_\ell\in k$ such that

$(a_0+\cdots+a_\ell g^\ell)=0\mod \mathfrak{m}$

If $a_0=0$, then we could divide through by $x$ (since we’re in a domain), contradicting the minimality of $\ell$. Thus, we see that

$g(a_1+\cdots+a_\ell g^{\ell-1})=-a_0\mod \mathfrak{m}$

Since $-a_0$ is a unit, this implies that $g+\mathfrak{m}$ is a unit. Since $g\in k[T_1,\ldots,T_n]+\mathfrak{m}$ was arbitrary, it follows that $k[T_1,\ldots,T_n]/\mathfrak{m}$ is a field. $\blacksquare$

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. $\blacksquare$

Proof(Conjecture): If $K[\alpha_1,\ldots,\alpha_n]/K$ is is finite, then by the basic theory (check the end of the last proof!) we know that $K[\alpha_1,\ldots,\alpha_n]$ is a field. Conversely, if we know that $K[\alpha_1,\ldots,\alpha_n]=K(\alpha_1,\ldots,\alpha_n)$, then $K[\alpha_1,\ldots,\alpha_n]$ is a finite type $K$-algebra which is a field. By Zariski’s lemma this implies that $K[\alpha_1,\ldots,\alpha]/K$ is finite as desired. $\blacksquare$

## 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 $X$ and $Y$ be Noetherian schemes, and let $f:X\to Y$ be finite type. Then, for any constructible (finite union of locally closed) subset $U\subseteq X$, the image $f(U)\subseteq Y$ 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 $k$ be a field. Then, the generic point $\eta$ of $\mathbb{A}^1_k$ is not constructible.

Proof: Suppose that $\{\eta\}$ was constructible. Then, there exists open sets $U_1,\ldots,U_n\subseteq\mathbb{A}^1_k$, and closed sets $C_1,\ldots,C_n\subseteq\mathbb{A}^1_k$ such that

$\{\eta\}=(U_1\cap C_1)\cup\cdots\cup (U_n\cap C_n)$

Now, clearly each $C_i$ is all of $\mathbb{A}^1_k$ since it contains the dense point $\eta$. Thus, we see that this implies $\{\eta\}$ is open. But, all non-empty open subsets of $\mathbb{A}^1_k$ are infinite, which is a contradiction. $\blacksquare$

With this simple lemma, we are now able to prove Zariski’s lemma:

Proof(Zariki’s Lemma): Since $L/K$ is finite type, to show it’s finite, it suffices to show it’s integral. Suppose not. Then, there is a $K$-algebra injection $K[T]\to L$. This then induces a morphism $\varphi:\text{Spec}(L)\to \mathbb{A}^1_K$. This morphism is obviously dominant, and so the image contains the generic point. But, since $\text{Spec}(L)$ is a point, it follows that the image of $\varphi$ is exactly the generic point of $\mathbb{A}^1_K$. Since $\varphi$ is a finite type morphism between Noetherian schemes, we know from Chevalley’s theorem that the image of $\varphi$ is constructible. But, we have already noted that the image of $\varphi$ is the generic point of $\mathbb{A}^1_k$ which, by the lemma, is not constructible. This is a contradiction. $\blacksquare$

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.

## One comment

1. filsdesam says:

Reblogged this on Samuel Cavazos and commented:
Wonderfully written piece on Hilbert’s Nullstellensatz.