Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Is there an easy proof of the Nullstellensatz that avoids the standard Noether-normalization techniques?

One proof I know proves first the 'weak' Nullstellensatz which ensures that maximal ideals correspond to points (using normalization), and then the stronger version by introducing another variable and the Rabinovich trick.

share|cite|improve this question
What's not elementary about Noether normalization? – Qiaochu Yuan Feb 14 '10 at 3:53
Dear Qiaochu, Noether normalization is not particularly difficult, but it has (to my mind) a certain "hard" feel to it that some other methods do not. (I'm not sure how meaningful this will be, but here I mean "hard" as opposed to "soft", rather than as opposed to "easy".) Perhaps this is because it is really a geometric statement, about the possibility of choosing a suitably generic projection. Related to this is the fact that it is a little finicky to prove over finite fields. – Emerton Feb 14 '10 at 4:13
Noether normalization has a generalization to (Noetherian?) commutative rings which requires only localization at one element. I'd definitely call it a theorem of commutative algebra, but as with so many theorems in commutative algebra, there's always a geometric form. – Harry Gindi Feb 14 '10 at 4:43
Isn't the proof of the weak Nullstellensatz in Atiyah-MacDonald normalization-free? I did not recall the details, but it seems to me that the route starting in their (unpleasant…) Proposition 7.8 and ending in Corollary 7.10 (the weak Nullstellensatz) avoids Noether normalization. – user2734 Feb 14 '10 at 7:49
There are many more elementary or direct proofs of the Nullstellensatz then using Noether's normalization lemma, but I think we shoud not forget that this lemma has its own interest and applications. – Qing Liu Feb 14 '10 at 22:18

12 Answers 12

up vote 12 down vote accepted

I once wrote up a proof (in the more general context of Jacobson rings) which uses basic commutative algebra techniques and avoids Noether normalization; it is available here.

This proof is very similar to the proof of Chevalley's theorem stating that the image of a constructible set, under a finite type morphism of Noetherian schemes, is again constructible. Indeed this result also implies the Nullstellensatz; see e.g. the discussion and proofs in section II.2 of the unpublished book of Mumford and Oda, where Chevalley's theorem is referred to as Chevalley's Nullstellensatz.

share|cite|improve this answer
I fixed a typo in the URL to your jacobson pdf. – Harry Gindi Feb 14 '10 at 3:45
Thank you, fpqc. – Emerton Feb 14 '10 at 6:36
This is not directly related to the question, but: this is a really nice development of Jacobson rings! – Ravi Vakil Aug 18 '11 at 3:52
Dear Ravi, Thanks! Best wishes, Matt – Emerton Aug 18 '11 at 4:33
(I fixed a link that was no longer working.) – Akhil Mathew Feb 16 '12 at 13:29

There is a cheap proof of the weak Nullstellensatz in Artin's Algebra which goes like this: suppose $m$ is a maximal ideal of $F[x_1, ... x_n]$ where $F$ is an uncountable algebraically closed field (note the additional hypothesis) which is not of the form "all functions vanishing at a particular point of $F^n$." Then $F[x_1, ... x_n]/m$ cannot be an algebraic extension of $F$ (since it would then be equal to $F$, and then there would exist $a_i \in F$ such that $x_i - a_i \in m$, which contradicts our hypothesis), so it must have transcendence degree at least one. But $F(t)$ has uncountable dimension over $F$, since, for example, the set $\{ \frac{1}{t -a}, a \in F \}$ is linearly independent, and $F[x_1, ... x_n]/m$ has countable dimension over $F$. Contradiction.

This proof works great if all you want to do is talk about complex varieties, but I imagine those of the Bourbaki tradition would cringe.

Edit, 2/28/10: See also question #15611, in which Brian Conrad gives an argument that one can reduce the general case to the uncountable case.

share|cite|improve this answer
The use of uncountability is not entirely unmotivated since the theory of algebraically closed fields (of a fixed characteristic) is uncountably categorical but not countably categorical. – François G. Dorais Feb 14 '10 at 4:41
That ain't no direct sum. That's a direct limit. – Kevin Buzzard Feb 14 '10 at 10:18
Mistake. I meant, union. – Abhishek Parab Feb 14 '10 at 14:45
@fpqc: I see no complex analysis in this proof. I think it is great! This is also called the "quick and dirty proof" of the Nullstellensatz in Eisenbud. I was going to post it if no one else had. I believe it can be done just slightly differently which generalizes a little bit. – Matt Mar 1 '10 at 1:08
Artin may have used complex analysis in the version of the proof presented in the first edition, but I don't see it anywhere in the version he used to teach his algebra course last year. The only analytic part of the argument is the part where he argues that the functions 1/(t - a) are linearly independent, but this can be done in a totally algebraic way just by clearing the denominators and substituting or by computing the "Wronskian." – Qiaochu Yuan Mar 1 '10 at 1:15

I have been thinking about the question "What is the best -- i.e., some combination of shortest, most natural, easiest -- proof of the Nullstellensatz?" recently on the eve of a commutative algebra course.

In my notes up to this point I had been following Kaplansky's treatment of Goldman domains and Hilbert-Jacobson rings. This places the problem in a more general context and allows for an attractively thorough analysis. At the end one comes out with the following results:

(1) The polynomial ring $k[t_1,\ldots,t_n]$ is a Jacobson ring -- i.e., every radical ideal is the intersection of the maximal ideals containing it.

(2) (Zariski's Lemma): If $\mathfrak{m}$ is a maximal ideal of $k[t_1,\ldots,t_n]$, then $k[t_1,\ldots,t_n]/\mathfrak{m}$ is a finite degree field extension of $k$.

(That Hilbert's Nullstellensatz follows from (1) and (2) is an easy, standard argument that I won't discuss here.)

But it is well-known that to prove the Nullstellensatz one needs only (2), because then (1) follows by a short and easy argument that everyone seems to like: Rabinowitsch's Trick. So perhaps this is a sign that developing the theory of (Hilbert-)Jacobson rings to prove the Nullstellensatz is overkill.

So the question seems to be: what is the best proof of Zariski's Lemma?

After looking around at various proofs, here is what I think the answer is now: it is an easy consequence of the following result.

Theorem (Artin-Tate Lemma): Let $R \subset T \subset S$ be a tower of rings such that
(i) $R$ is Noetherian,
(ii) $S$ is finitely generated as an $R$-algebra, and
(iii) $S$ is finitely generated as a $T$-module.
Then $T$ is finitely generated as an $R$-algebra.

Proof: Let $x_1,\ldots,x_n$ be a set of generators for $S$ as an $R$-algebra, and let $\omega_1,\ldots,\omega_m$ be a set of generators for $S$ as a $T$-module. For all $1 \leq i \leq n$, we may write \begin{equation} \label{ARTINTATEEQ1} x_i = \sum_j a_{ij} \omega_j, \ a_{ij} \in T. \end{equation} Similarly, for all $1 \leq i,j \leq m$, we may write \begin{equation} \label{ARTINTATEEQ2} \omega_i \omega_j = \sum_{k} b_{ijk} \omega_k, \ b_{ijk} \in T. \end{equation} Let $T_0$ be the $R$-subalgebra of $T$ generated by the $a_{ij}$ and $b_{ijk}$. Since $T_0$ is a finitely generated algebra over the Noetherian ring $R$, it is itself a Noetherian ring by the Hilbert Basis Theorem. \ \indent Now each element of $S$ may be expressed as a polynomial in the $x_i$'s with $R$-coefficients. Making substitutions using the two equations above shows that $S$ is a finitely generated $T_0$-module. Since $T_0$ is Noetherian, the submodule $T$ is also finitely generated as a $T_0$-module. This immediately implies that $T$ is finitely generated as a $T_0$-algebra and then in turn that $T$ is finitely generated as an $R$-algebra, qed!

Proof that Artin-Tate implies Zariski's Lemma:

It suffices to prove the following: let $K/k$ be a field extension which is finitely generated as a $k$-algebra. Then $K/k$ is algebraic. Indeed, suppose otherwise: let $x_1,\ldots,x_n$ be a transcendence basis for $K/k$ (where $n \geq 1$ since $K/k$ is transcendental), put $k(x) = k(x_1,\ldots,x_n)$ and consider the tower of rings

$k \subset k(x) \subset K$.

To be sure, we recall the definition of a transcendence basis: the elements $x_i$ are algebraically independent over $k$ and $K/k(x)$ is algebraic. But since $K$ is a finitely generated $k$-algebra, it is certainly a finitely generated $k(x)$-algebra and thus $K/k(x)$ is a finite degree field extension. Thus the Artin-Tate Lemma applies to our tower: we conclude that $k(x)/k$ is a finitely generated $k$-algebra. But this is absurd. It implies the much weaker statement that $k(x) = k(x_1,\ldots,x_{n-1})(x_n)$ is finitely generated as a $k(x_1,\ldots,x_{n-1})[x_n]$-algebra, or weaker yet, that there exists some field $F$ such that $F(t)$ is finitely generated as an $F[t]$-algebra: i.e., there exist finitely many rational functions $\{r_i(t) = \frac{p_i(t)}{q_i(t)} \}_{i=1}^N$ such that every rational function is a polynomial in the $r_i$'s with $k$-coefficients. But $F[t]$ is a PID with infinitely many nonassociate nonzero prime elements (e.g. adapt Euclid's argument of the infinitude of the primes), so we may choose a nonzero prime element $q$ which does not divide $q_i(t)$ for any $i$. It is then clear that $\frac{1}{q}$ cannot be a polynomial in the $r_i(t)$'s: for instance, evaluation at a root of $q$ in $\overline{F}$ leads to a contradiction.

Note that this is almost all exactly as in Artin-Tate's paper, except for the endgame above, which has been made a little more explicit and simplified: their conclusion seems to depend upon unique factorization in $k[t_1,\ldots,t_n]$, which does not come until later on in my notes.

Further comments:

(i) The proof is essentially a reduction to Noether's normalization in the case of field extensions, which becomes the familiar result about existence of transcendence bases. Thus it is not so far away from the most traditional proof of the Nullstellensatz. But I think the Artin-Tate Lemma is easier than Noether Normalization.

(ii) Speaking of Noether: the proof of the Artin-Tate Lemma is embedded in the standard textbook proof that if $R$ is a finitely generated $k$-algebra and $G$ is a finite group acting on $R$ by ring automorphisms, then $R^G$ is a finitely generated $k$-algebra. In fact I had already typed this proof up elsewhere in my notes. Realizing that the Artin-Tate Lemma is something I was implicitly proving in the course of another result anyway was part of what convinced me that this was an efficient route to the Nullstellensatz. Note that the paper of Artin and Tate doesn't make any connection with Noether's theorem and conversely the textbooks on invariant theory that prove Noether's theorem don't seem to mention Artin-Tate. (However, googling -- Artin-Tate Lemma, Noether -- finds several research papers which allude to the connection in a way which suggests it is common knowledge among the cognoscenti.)

Added: It turns out this is the proof of Zariski's Lemma given in Chapter 7 of Atiyah-Macdonald. I had missed this because (i) they give another (nice) proof using valuation rings in Chapter 5 and (ii) they do not attribute the Artin-Tate Lemma to Artin and Tate, although their treatment of it is even closer to the Artin-Tate paper than mine is above. (In the introduction of their book, they state cheerfully that they have not attributed any results. I think this is a drawback of their otherwise excellent text.)

share|cite|improve this answer
+1; this is exactly the proof I remember from commutative algebra, and it feels the most natural of all of these (perhaps due to familiarity). – Daniel Litt Jan 1 '11 at 0:18
Dear Pete, 4 years late a tiny typo: at the end of the line beginning "To be sure...", the word "equivalent" is presumably meant to be "independent". – user5117 Jan 13 at 13:13

Well, easy is a relative term, but Terry Tao has a blog post about a computational proof of the Nullstellensatz he discovered which requires little more than high school algebra. The proof itself may be longer than the one(s) you are used to, but I'm willing to bet the tools needed are easier. The post is available here.

share|cite|improve this answer

From Serge Lang's Algebra:

Theorem 9.1.1

Let $k$ be a field, and let $k[x]:=k[x_1,...,x_n]$ be a finitely generated $k$-algebra. Let $\phi:k\to L$ be an embedding of k into an algebraically closed field $L$. Then there exists an extension of $\phi$ to a homomorphism $\bar{\phi}: k[x] \to L$.

Note: The $x_i$ are not indeterminates.

Corollary 9.1.2 (Zariski)

Let $k$ be a field, and let $k[x]:=k[x_1,...,x_n]$ be a finitely generated $k$-algebra. If $k[x]$ is a field, then $k[x]$ is algebraic over $k$.

Corollary 9.1.3

Let $k$ be a field, and let $k[x]:=k[x_1,...,x_n]$ be a finitely generated $k$-algebra. Fix a finite family of elements $(y_i)_{i=1}^m$ of $k[x]$. If $k[x]$ is an integral domain, there exists a homomorphism $\psi:k[x]\to k^a$, where $k^a$ is the algebraic closure of $k$ such that $\psi(y_i)\neq 0$ for all $1\leq i\leq m$.

Theorem 9.1.4 (Weak Nullstellensatz)

Let $k$ be a field, and let $k[X]:=k[X_1,...,X_n]$ be the polynomial ring in $n$ indeterminates over $k$.

Let $\mathfrak{q}$ be an ideal of $k[X]$. Then either $\mathfrak{q}$ is the unit ideal, or $\mathfrak{q}$ has a zero in $k^a$.

Theorem 9.1.5 (Hilbert's Nullstellensatz)

Let $k$ be a field, and let $k[X]:=k[X_1,...,X_n]$ be the polynomial ring in $n$ indeterminates over $k$.

Let $\mathfrak{q}$ be an ideal of $k[X]$. Let $f\in k[X]$ be a polynomial vanishing on every zero of $\mathfrak{q}$ in $k^a$. Then there exists $m>0$ such that $f^m\in \mathfrak{q}$.

The proof of 9.1.5 follows from the Rabinowitsch trick and 9.1.4, which in turn follows directly from 9.1.2, which is a straighforward application of 9.1.1.

There is an advantage to this proof because it allows us not only to extend the definition of a variety to non-algebraically closed fields, but also to define an algebraic space, which is defined as a functor that takes field extensions of the basefield $k$ to the zero set of the ideal in that extension, with some nice properties. (This is not standard terminology, but I believe that every functor arising this way is in fact an algebraic space.)

In Eisenbud's commutative algebra book, the Nullstellensatz is generalized further from fields to Jacobson rings, which are rings for which any prime ideal is an intersection of some family of maximal ideals (This is a theorem of Bourbaki).

None of these proofs uses Noether normalization.

Corollary 9.1.2 is a lemma of Zariski that he introduced to prove the Nullstellensatz. I believe Lang's proof of Theorem 9.1.1 is similar to Zariski's proof of 9.1.2.

This is Zariski's paper where he introduced the method used by Lang. If you read the introduction, it's really interesting, because all of the previous proofs had been somewhat nontrivial. This was somewhat groundbreaking for proofs of the Nullstellensatz.

Additionally this lemma of Zariski is a special case of Zariski's main theorem for commutative rings. The Nullstellensatz follows with basically no effort. However, Zariski's main theorem is highly nontrivial.

share|cite|improve this answer

I found this "geodesic" proof by Munshi, written by May (I like the very last paragraph (:).

share|cite|improve this answer
Unfortunately I don't remember (: – Hailong Dao Feb 14 '10 at 4:09

There is the snappy model-theoretic proof of the Nullstellensatz, which goes by first proving that ACF_p, the theory of algebraically closed fields of some fixed characteristic p, is model complete. Once you know this it is easy:

Let k be an algebraically closed field, and let m be a maximal ideal in k[x1,...,xn]. Let L be the algebraic closure of k[x1,...,xn]/m.

We are trying to prove that there exists $a_1,...,a_n \in k$ which is a common zero of some set of generators for m. Note that this assertion is a first-order sentence using the language of fields and symbols in k. So model completeness of ACF_p tells us that this sentence is true in k if and only if it is true in some (and then every) algebraically closed extension field of k. But we know that this sentence is true in the algebraically closed extension L by construction, so we are done.

share|cite|improve this answer
I feel like all of the real work is hidden proving that ACFields are model-complete. – Harry Gindi Feb 14 '10 at 10:15
Well, sure. But the proof that ACF_p is model complete is not that hard -- it's (probably) easier than showing the Nullstellensatz. – Dan Petersen Feb 14 '10 at 10:26
Lang proves the nullstellensatz in roughly a page and a half (If you remove the text between proofs. Not also that Corollary 9.1.3 is not actually used in the proof of the Nullstellensatz. If we include all of the irrelevant stuff, it's two and a half pages (springer GTM sized). – Harry Gindi Feb 14 '10 at 10:32
Also, Zariski's paper contains two full proofs of the Nullstellensatz with example applications and a history of earlier proofs running a total of 6 pages, each proof taking probably 2. – Harry Gindi Feb 14 '10 at 10:34
David Marker's notes on the model theory of fields proves quantifier eleminiation for ACF_p (which easily implies model completeness) in less than a page. – Dan Petersen Feb 18 '10 at 9:06

I have a feeling that some modern proofs hiding from the students the intuition of what is really going on... I am now teaching an undergraduate course "Introduction to algebraic geom. and comm. alg.". I am speaking a lot about Grobner bases and resultant -- they look natural and algorithmic. For the course I am going to use a proof of weak Nullstellensatz using the Groebner bases only.

Let $k$ be a field, $a\in k$. Let ev$_{a}:k[x_1,x_2,\dots,x_n]\to k[x_2,\dots,x_n]$ denote the evaluation homomorphism ev$_{a}:f(x_1,x_2,\dots,x_n)\to f(a,x_2,...,x_n)$. The proof is based on the following lemmas.


Let $k$ be an algebraically closed field, $I\subseteq k[x_1,\dots,x_n]$ be an ideal, such that $I\cap k[x_1]=\langle p \rangle$ and $p\in k[x]\setminus k$. Then there exists $a\in k$, $p(a)=0$ such that ev$_{a}(I)\neq k[x_2,\dots,x_n]$

Lemma2 (It is valid for any field):

Let $k$ be a field, $I\subseteq k[x_1,\dots,x_n]$ be an ideal, such that $I\cap k[x_1]=\{0\}$. Then there exist non-zero polynomial $q\in k[x_1]$ such that ev$_{a}(I)\neq k[x_2,\dots,x_n]$ for any $a\in k$, $q(a)\neq 0$.


Let $k$ be an infinite field, $I\subseteq k[x_1,\dots,x_n]$ be an ideal, such that $I\cap k[x_1]=\{0\}$. Then ev$_{a}(I)\neq k[x_2,\dots,x_n]$ for some $a\in k$.

The weak Nullstellensatz follows by induction...

Let me show how Groebner bases can be used to prove, say, Lemma 2. Let $I\subset k[x_1,\dots,x_n]$ be an ideal. Let $\langle I\rangle_{k(x_1)}$ denote the ideal generated by $I$ in $k(x_1)[x_2,\dots,x_n]$. Lemma 2 follows from the

1: $\langle I \rangle _{k(x_1)}=k(x_1)[x_2,\dots,x_n]$ iff $I\cap k[x_1]\neq \{0\}$.

2: Let $\Gamma=\{g_1,\dots,g_2\}$ be a Groebener basis for $\langle I\rangle_{k(x_1)}$. Suppose, that leading coefficients of $g_i$ are $1$. Let $q\in k[x_1]$ be the common denominator of all $g_i$
(i.e. $qg_i\in k[x_1,...x_n]$ for any $i$). Then ev$_{a}(\Gamma)$ is a Groebner basis for ev$_{a}(I)$ for any $a\in k$, $q(a)\neq 0$.

To show (2) it is suffices to note that during reduction of $f\in k[x_1,x_2,\dots,x_n]$ by $\Gamma$ all denominators divide a power of $q$.

share|cite|improve this answer

Richard Swan's proof (which is based on Munshi's ideas):

share|cite|improve this answer

I like Daniel Allcock's proof (which still uses the Rabinowitsch trick).

share|cite|improve this answer

Here is another proof I posted on math.SE.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.