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In his answer to a question about simple proofs of the Nullstellensatz (Elementary / Interesting proofs of the Nullstellensatz), Qiaochu Yuan referred to a really simple proof for the case of an uncountable algebraically closed field.

Googling, I found this construction also in Exercise 10 of a 2008 homework assignment from a course of J. Bernstein (see the last page of http://www.math.tau.ac.il/~bernstei/courses/2008%20spring/D-Modules_and_applications/pr/pr2.pdf). Interestingly, this exercise ends with the following (asterisked, hard) question:

(*) Reduce the case of arbitrary field $k$ to the case of an uncountable field.

After some tries to prove it myself, I gave up and returned to googling. I found several references to the proof provided by Qiaochu Yuan, but no answer to exercise (*) above.

So, my question is: To prove the Nullstellensatz, how can the general case of an arbitrary algebraically closed field be reduced to the easily-proved case of an uncountable algebraically closed field?

The exercise is from a course of Bernstein called 'D-modules and their applications.' One possibility is that the answer arises somehow when learning D-modules, but unfortunately I know nothing of D-modules. Hence, proofs avoiding D-modules would be particularly helpful.

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It seems natural to try to use the model completeness of the theory of algebraically closed fields. But if you're going to use model theory, it seems to me that you might as well prove the Nullstellensatz outright, which is possible: see the accepted answer to mathoverflow.net/questions/9667/…. –  Pete L. Clark Feb 18 '10 at 3:01
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It's possible that Bernstein had in mind a more direct reduction, although I can't imagine what it would look like. –  Qiaochu Yuan Feb 18 '10 at 3:24
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Is there a non-model-theory approach? –  Harry Gindi Feb 18 '10 at 4:44
    
@PLC: Thank you very much for your comment. Given the context of the question in the homework assignment, I tend to believe (or at least to hope) that there is a proof from commutative algebra. Clearly, this should not be an obvious proof, but I am still hoping that someone familiar with Bernstein's work in other fields will come up with the proof. Less ambitiously, perhaps a student from that course will reveal the secret... –  user2734 Feb 18 '10 at 6:27
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Also, +1 for the long but extremely informative title. It's good for people to realize that nothing is gained by making their titles shorter. –  Qiaochu Yuan Feb 18 '10 at 7:15
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5 Answers

up vote 21 down vote accepted

These logic/ZFC/model theory arguments seem out of proportion to the task at hand. Let $k$ be a field and $A$ a finitely generated $k$-algebra over a field $k$. We want to prove that there is a $k$-algebra map from $A$ to a finite extension of $k$. Pick an algebraically closed extension field $k'/k$ (e.g., algebraic closure of a massive transcendental extension, or whatever), and we want to show that if the result is known in general over $k'$ then it holds over $k$. We just need some very basic commutative algebra, as follows.

Proof: We may replace $k$ with its algebraic closure $\overline{k}$ in $k'$ and $A$ with a quotient $\overline{A}$ of $A \otimes_k \overline{k}$ by a maximal ideal (since if the latter equals $\overline{k}$ then $A$ maps to an algebraic extension of $k$, with the image in a finite extension of $k$ since $A$ is finitely generated over $k$). All that matters is that now $k$ is perfect and infinite.

By the hypothesis over $k'$, there is a $k'$-algebra homomorphism $$A' := k' \otimes_k A \rightarrow k',$$ or equivalently a $k$-algebra homomorphism $A \rightarrow k'$. By expressing $k'$ as a direct limit of finitely generated extension fields of $k$ such an algebra homomorphism lands in such a field (since $A$ is finitely generated over $k$). That is, there is a finitely generated extension field $k'/k$ such that the above kind of map exists. Now since $k$ is perfect, there is a separating transcendence basis $x_1, \dots, x_n$, so $k' = K[t]/(f)$ for a rational function field $K/k$ (in several variables) and a monic (separable) $f \in K[t]$ with positive degree. Considering coefficients of $f$ in $K$ as rational functions over $k$, there is a localization $$R = k[x_1,\dots,x_n][1/h]$$ so that $f \in R[t]$. By expressing $k'$ as the limit of such $R$ we get such an $R$ so that there is a $k$-algebra map $$A \rightarrow R[t]/(f).$$ But $k$ is infinite, so there are many $c \in k^n$ such that $h(c) \ne 0$. Pass to the quotient by $x_i \mapsto c_i$. QED

I think the main point is twofold: (i) the principle of proving a result over a field by reduction to the case of an extension field with more properties (e.g., algebraically closed), and (ii) spreading out (descending through direct limits) and specialization are very useful for carrying out (i).

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+1: This works nicely. –  Pete L. Clark Feb 18 '10 at 6:05
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@Brian: when you edit a post significantly, it is nice to give some indication of what you have changed. Was there something wrong with your previous argument? –  Pete L. Clark Feb 18 '10 at 6:22
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The previous post had an integrality argument that didn't apply when k'/k is not algebraic. The ironic thing is that my immediate reaction upon seeing the question was "Oh, it's just the old spread out and specialize business", and while typing that I thought I found an even "slicker" argument (the original post) which I realized was not right about 2 seconds after I posted it. So I went back to my original idea, which is correct. Better to follow one's instincts and not try to be too slick. :) –  BCnrd Feb 18 '10 at 6:27
    
Wow! this looks like what I am looking for. It will take me some time to process this proof, though. –  user2734 Feb 18 '10 at 7:53
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This is a comment on Brian's answer, which is however a bit long to fit into the comment box. I wanted to remark that Brian's argument is ulimately not so different from the Noether normalization argument, nor is it so different to the argument linked to here, or to the argument in II.2 of Mumford--Oda using Chevalley's theorem. What they all have in common is the fact that any finite type variety can be projected to affine space with generically finite fibres and big image. On affine space (at least over an infinite field) we can find lots of points, and by the generic finiteness and big image assumptions we can even find such a point lying in the image of the original affine variety with finite fibres. Finding a point on this fibre then involves solving a finite degree polynomial, which we can do over the algebraic closure. Hence our original finite-type variety has a point.

Here is a rewrite of Brian's argument which illustrates this: Following his reduction, we may assume that $k$ is infinite and perfect. We are given a non-zero finite type $k$-algebra $A$, and we want to show that Spec $A$ has a $\bar{k}$-point, i.e. that we can find a $k$-algebra homomorphism $A \to \bar{k}$. For this, we may as well replace $A$ by a quotient by a maximal ideal, and thus assume that $A$ is a field.
As Brian notes, the theory of finitely generated field extensions allows us to write $A = k(X_1,\ldots,X_d)[t]/f(t)$ (because $k$ is perfect). We then observe that since $A$ is finite type over $k$, its generators involve only finitely many denominators, as do the coefficients of $f$, and so in fact $A = k[X_1,\ldots,X_d][1/h][t]/f(t)$ for some well-chosen non-zero $h$.

Now because $k$ is infinite, $h$ is not identically zero on $k^d$, and so we are done: we choose a point $c_i$ where $h$ is non-zero, then solve $f(c_1,\ldots,c_d,t) = 0$ in $\bar{k}$.

So one sees that the role of the theory of finitely generated field extensions is simply to provide a weaker version of the Noether normalization, with generic finiteness replacing finiteness. As I already wrote, the other "soft" arguments for the Nullstellensatz proceed along essentially the same lines.

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Very helpful, +1 –  Hailong Dao Feb 18 '10 at 16:04
    
I fixed your first link. –  Harry Gindi Feb 18 '10 at 16:07
    
Thank you, fpqc –  Emerton Feb 18 '10 at 16:36
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Matt's formulation is more elegant, but I was trying to set up it to look more like a deduction from the case over a big extension field (hence my focus on the map to k' without changing A too much after the initial reduction step) since otherwise we'e just giving a direct proof of Nullstellensatz over the ground field, which seems to violate the spirit of the question. That said, in some sense the whole question is kind of pointless because we have so many nice direct proofs of Null. over any field. The above is yet another. :) –  BCnrd Feb 18 '10 at 17:20
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Dear Brian, Yes, I realized that you were trying to conform to the spirit of the question, which I happily disregarded! :) –  Emerton Feb 18 '10 at 18:55
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I know a way to do this, but it involves some very heavy machinery...

The first component are effective bounds on the degrees of the polynomials in the conclusion of the Weak Nullstellensatz. Such bounds are not that easy to get and there has been a lot of literature on the Effective Nullstellensatz. Perhaps the earliest effective bounds were found by Grete Hermann Die Frage der endlich vielen Schritte in der Theorie der Polynomideale (Mathematische Annalen 95, 1926), but there has been a lot of work on improving these bounds and also obtaining lower bounds over the years. [E.g., D. W. Brownawell, Bounds for the degrees in the Nullstellensatz, Ann. of Math. (2) 126 (1987), 577-591] It's interesting to read these papers, but I will only use the fact that effective bounds do exist.

Using these bounds it is possible to find a sequence of first-order sentences $\phi_{n,k,r}$, which together are equivalent to the Weak Nullstellensatz; the sentence $\phi_{n,k,r}$ is a first order rendition of the following statement.

If $p_1(\bar{x}),\dots,p_k(\bar{x})$ ($\bar{x} = x_1,\ldots,x_r$) are polynomials of degree at most $n$ without common zeros, then there are polynomials $q_1(\bar{x}),\dots,q_k(\bar{x})$ of degree at most $b(n,k,r)$ such that $p_1(\bar{x})q_1(\bar{x})+\cdots+p_k(\bar{x})q_k(\bar{x}) = 1$.

The bounds $n$ and $b(n,k,r)$ are necessary so that the $p_i(\bar{x})$ and $q_i(\bar{x})$ have a bounded number of coefficients. Otherwise, we could not use a fixed number of variables for these coefficients.

That said, the other piece of heavy machinery is the fact that the theory of algebraically closed fields of a given characteristic is complete, i.e. every first-order sentence is decided by the axioms. Therefore, if the above sentences $\phi_{n,k,r}$ are true in any algebraically closed field of a given characteristic, then they must be true in all algebraically closed fields of the same characteristic. In particular, the Weak Nullstellensatz for $\mathbb{C}$ implies the Weak Nullstellensatz for all algebraically closed fields of characteristic zero.

From here, you can use the Rabinowitsch trick to get the Strong Nullstellensatz...

PS: You do not need the Nullstellensatz to prove that the theory of algebraically closed fields of a given characteristic is complete. You implicitly need the Nullstellensatz to prove the effective upper bounds, but you only need them for the one field and you can think of them as wild guesses that turn out to be right.

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After seeing Pete's comment, a simpler approach is to first prove quantifier elimination and use model completeness. (Well, I don't know which is easiest between getting very crude effective bounds and proving quantifier elimination.) However, there is a small benefit of my brute force approach, namely that the Nullstellensatz is actually expressible in first-order logic. –  François G. Dorais Feb 18 '10 at 3:37
    
Thank you very much for your answer. While I am hoping for a "trick" using only commutative algebra, this is still very interesting! –  user2734 Feb 18 '10 at 6:27
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The easiest way to reduce to the uncountable case may be as follows. Let $I$ be an ideal of $k[X_1,...,X_d]$ which does not contain $1$. Let $P_1,\dots,P_r$ be a generating family of $I$.

Let $A=k^{\mathbf N}$ and let $m$ be a maximal ideal of $A$ which contains the ideal $N=k^{(\mathbf N)}$ of $A$. Then $K=A/m$ is an algebraically closed field which is has at least the power of the continuum. (Alternative description: let $K$ be an ultrapower of $k$, with respect to a non-principal ultrafilter.)

Lemma. For $i\in\{1,\dots,r\}$, let $a_i=(a_{i,n})\in A$. Assume that $(\bar a_1,\dots,\bar a_r)=0$ in $K^r$. Then the set of $n\in\mathbf N$ such that $(a_{1,n},\dots,a_{r,n})=0$ is infinite.

Proof. Assume otherwise. For every $n$ such that $(a_{1,n},...,a_{r,n}) \neq 0$, choose $(b_{1,n},\dots,b_{r,n})$ such that $\sum a_{i,n}b_{i,n}=1 $, and let $b_i=(b_{i,n})_n\in A$. Then $\sum a_i b_i - 1 $ belongs to $N^r$, hence $\sum \bar a_i \bar b_i=1$. Contradiction.

Thanks to the lemma, one proves easily that the ideal $I_K$ of $K[X_1,...,X_d]$ generated by $I$ does not contain $1$. By the uncountable case, there exists $x=(x_1,...,x_d)\in K^d$ such that $P_j(x_1,...,x_d)=0$ for every $j$. For every $i$, let $a=(a_{n})\in A^d$ be such that $\bar a=x$. By the lemma again the set of integers $n$ such that $P_j(a_n) \neq 0$ for some $j$ is finite. In particular, there exists a point $y\in k^d$ such that $P_j(y)=0$ for every $j$.

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I like this proof, it is completely elementary (and remark that also Brian's proof contains a choice of a maximal ideal). –  Martin Brandenburg Jan 6 '13 at 18:14
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Well, this is the opposite of what you asked, but there is an easy reduction in the other direction. Namely, if the result is true for countable fields, then it is true for all fields. I can give two totally different proofs of this, both very soft, using elementary methods from logic. While we wait for a solution in the requested direction, let me describe these two proofs.

Proof 1. Suppose k is any algebraically closed field, and J is an ideal in the polynomial ring k[x1,...,xn]. Consider the structure (k[x1,...,xn],k,J,+,.), which is the polynomial ring k[x1,...,xn], together with a predicate for the field k and for the ideal J. By the downward Loweheim-Skolem theorem, there is a countable elementary substructure, which must have the form (F[x1,...,xn],F,I,+,.), where F is a countable subfield of k, and I is a proper ideal in F[x1,...,xn]. The "elementarity" part means that any statement expressible in this language that is true in the subring is also true in the original structure. In particular, I is a proper ideal in F[x1,...,xn] and F is algebraically closed. Thus, by assumption, there is a1,...,an in F making all polynomials in I zero simultaneously. This is a fact about a1,...,an that is expressible in the smaller structure, and so it is also true in the upper structure. That is, every polynomial in J is zero at a1,...,an, as desired.

Proof 2. The second proof is much quicker, for it falls right out of simple considerations in set theory. Suppose that we can prove (in ZFC) that the theorem holds for countable fields. Now, suppose that k is any field and that J is a proper ideal in the ring k[x1,...,xn]. If V is the set-theoretic universe, let V[G] be a forcing extension where k has become countable. (It is a remarkable fact about forcing that any set at all can become countable in a forcing extension.) We may consider k and k[x1,...,xn] and J inside the forcing extension V[G]. Moving to the forcing extension does not affect any of our assumptions about k or k[x1,...,xn] or J, except that now, in the forcing extension, k has become countable. Thus, by our assumption, there is a1,...,an in kn making all polynomials in J zero. This fact was true in V[G], but since the elements of k and J are the same in V and V[G], and the evaluations of polynonmials is the same, it follows that this same solution works back in V. So the theorem is true for k in V, as desired.

But I know, it was the wrong reduction, since I am reducing from the uncountable to the countable, instead of from the countable to the uncountable, as you requested...

Nevertheless, I suppose that both of these arguments could be considered as alternative very soft short proofs of the uncountable case (assuming one has a proof of the countable case).

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I am sure this is a naive question, but: In your introduction your hypothesis was "if the result is true for all countable fields", but then in Proof 2 the hypothesis is "suppose we can prove in ZFC that the theorem holds for countable fields". Does the former imply the latter? I guess it does, by the completeness of the theory of algebraically closed fields of each characteristic. But can you conclude this without appealing to the completeness of this particular theory? –  Tom Church Feb 18 '10 at 3:00
    
It's not a naive question. In the forcing argument, one needs the theorem for countable fields to be true in V[G], rather than V. So if you assumed only that it was true (i.e. true in V), then the argument wouldn't quite work. If you assume it is provable in ZFC, then we get to use it in any model of ZFC, including V[G]. –  Joel David Hamkins Feb 18 '10 at 3:09
    
I've realized that the assertion that the claim is true for countable fields has complexity only Pi^1_1, and so if it is true in V, it will also be true in V[G] by the Schoenfield Absoluteness theorem. So there is no need for me to have assumed that the claim was provable, but rather only that it was true. –  Joel David Hamkins Feb 18 '10 at 14:31
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