Comments on: Hilbert’s nullstellensatz http://terrytao.wordpress.com/2007/11/26/hilberts-nullstellensatz/ Updates on my research and expository papers, discussion of open problems, and other maths-related topics. By Terence Tao Thu, 07 Aug 2008 21:45:00 +0000 http://wordpress.org/?v=MU By: Anonymous http://terrytao.wordpress.com/2007/11/26/hilberts-nullstellensatz/#comment-31827 Anonymous Mon, 04 Aug 2008 21:07:06 +0000 http://terrytao.wordpress.com/2007/11/26/hilberts-nullstellensatz/#comment-31827 This is late, but you might find Munshi's proof of the Nullstellensatz interesting. It's at www.math.uchicago.edu/~may/PAPERS/MunshiFinal2.pdf It avoids harder results in commutative algebra such as Noether's normalization theorem, and instead introduces the idea of a "flow" to prove the intersection of a maximal ideal in R[x_1, ..., x_n] and R is nonempty. The proof of the weak nullstellensatz follows easily this and the strong form follows from the standard Rabinowitsch argument. This is late, but you might find Munshi’s proof of the Nullstellensatz interesting. It’s at http://www.math.uchicago.edu/~may/PAPERS/MunshiFinal2.pdf

It avoids harder results in commutative algebra such as Noether’s normalization theorem, and instead introduces the idea of a “flow” to prove the intersection of a maximal ideal in R[x_1, ..., x_n] and R is nonempty. The proof of the weak nullstellensatz follows easily this and the strong form follows from the standard Rabinowitsch argument.

]]> By: My Conjecture « Reasonable Deviations http://terrytao.wordpress.com/2007/11/26/hilberts-nullstellensatz/#comment-29882 My Conjecture « Reasonable Deviations Mon, 19 May 2008 04:57:31 +0000 http://terrytao.wordpress.com/2007/11/26/hilberts-nullstellensatz/#comment-29882 [...] Hilbert’s nullstellensatz Possibly related posts: (automatically generated)Existence of God [...] [...] Hilbert’s nullstellensatz Possibly related posts: (automatically generated)Existence of God [...]

]]> By: SriRam http://terrytao.wordpress.com/2007/11/26/hilberts-nullstellensatz/#comment-27180 SriRam Wed, 12 Mar 2008 09:14:40 +0000 http://terrytao.wordpress.com/2007/11/26/hilberts-nullstellensatz/#comment-27180 “The presence of the exponent r in conclusion (2) is a little annoying; to get rid of it, one needs to generalise the notion of an algebraic variety to that of a scheme” Continuing on the above topic (by Jens) I would be very interested in having the reference where $latex R = \sum_{i=1}^m P_i(\mathbf{x}) Q_i(\mathbf{x}) $ can be proved using schemes. Here is the context: I have a bunch of cubic polynomials $latex F=\{f_1, f_2 ... f_k \} f_{k+1}$. I want to prove that $latex f_{k+1}$ is independent of $latex F$. Also, in my problem setting I can check at as many points, $latex \mathbf{x}$ as possible (at say W points, for some natural number W) that $latex \hbox{rank}([\nabla f_1 \nabla f_2 ... \nabla f_{k+1}]) < (k+1)$. For that, I would like to prove: If $latex f_{k+1}$ is dependent on $latex F$ then $latex \hbox{rank}([\nabla f_1 \nabla f_2 ... \nabla f_{k+1}]) < (k+1)$. In this proof by contradiction, suppose $latex f_{k+1}$ is dependent. Then $latex f_{k+1}$ vanishes wherever $latex F$ is zero (mutual zero) (Note that this is an assumption I make because of "dependence". And the fact that experimentally at a lot of points it is true). By Hilbert's Nullstellensatz, $latex \exists r, P_i(x)$ s.t. $latex f_{k+1}^r = \sum_{i=1}^{m} P_i(x) f_i(x)$ . Differentiating on both sides we get, $latex r f_{k+1}^{r-1} \nabla f_{k+1} = \sum_{i=1}^k (P_i(x) \nabla f_i(x) + f_i(x) \nabla P_i(x))$. Evaluating it at points $latex F=0$, I get $latex r f_{k+1}^{r-1} \nabla f_{k+1} = \sum_{i=1}^k P_i(x) \nabla f_i(x)$, which would have nicely proved the result $latex \hbox{rank}([\nabla f_1 \nabla f_2 ... \nabla f_{k+1}]) < (k+1)$, except that we are not sure if $latex P_i(x) = 0 \forall i$ between 1 and k. In which case I dont get any result as it becomes $latex 0 = \sum_{i=1}^k (0 \nabla f_i)$. However, in case the exponent is not there, I will get $latex \nabla f_{k+1} = \sum_{i=1}^k P_i(x) f_i$ which will give my result. Any reference to prove / solve the above issue would be great. Thanks in advance. “The presence of the exponent r in conclusion (2) is a little annoying; to get rid of it, one needs to generalise the notion of an algebraic variety to that of a scheme”

Continuing on the above topic (by Jens) I would be very interested in having the reference where R = \sum_{i=1}^m P_i(\mathbf{x}) Q_i(\mathbf{x}) can be proved using schemes.

Here is the context: I have a bunch of cubic polynomials F=\{f_1, f_2 ... f_k \} f_{k+1}. I want to prove that f_{k+1} is independent of F. Also, in my problem setting I can check at as many points, \mathbf{x} as possible (at say W points, for some natural number W) that \hbox{rank}([\nabla f_1 \nabla f_2 ... \nabla f_{k+1}]) < (k+1).

For that, I would like to prove: If f_{k+1} is dependent on F then \hbox{rank}([\nabla f_1 \nabla f_2 ... \nabla f_{k+1}]) < (k+1).

In this proof by contradiction, suppose f_{k+1} is dependent. Then f_{k+1} vanishes wherever F is zero (mutual zero) (Note that this is an assumption I make because of “dependence”. And the fact that experimentally at a lot of points it is true). By Hilbert’s Nullstellensatz, \exists r, P_i(x) s.t. f_{k+1}^r = \sum_{i=1}^{m} P_i(x) f_i(x) .
Differentiating on both sides we get, r f_{k+1}^{r-1} \nabla f_{k+1} = \sum_{i=1}^k (P_i(x) \nabla f_i(x) + f_i(x) \nabla P_i(x)).

Evaluating it at points F=0, I get r f_{k+1}^{r-1} \nabla f_{k+1} = \sum_{i=1}^k P_i(x) \nabla f_i(x), which would have nicely proved the result \hbox{rank}([\nabla f_1 \nabla f_2 ... \nabla f_{k+1}]) < (k+1), except that we are not sure if P_i(x) = 0 \forall i between 1 and k.

In which case I dont get any result as it becomes 0 = \sum_{i=1}^k (0 \nabla f_i).

However, in case the exponent is not there, I will get
\nabla f_{k+1} = \sum_{i=1}^k P_i(x) f_i which will give my result.

Any reference to prove / solve the above issue would be great.

Thanks in advance.

]]> By: Affine Varieties « Rigorous Trivialities http://terrytao.wordpress.com/2007/11/26/hilberts-nullstellensatz/#comment-20842 Affine Varieties « Rigorous Trivialities Tue, 18 Dec 2007 03:06:36 +0000 http://terrytao.wordpress.com/2007/11/26/hilberts-nullstellensatz/#comment-20842 [...] two facts which seem fairly straightforward, are in fact rather difficult to prove (here is a proof that Tao wrote up), and are essential to algebraic geometry. The first tells us that we [...] [...] two facts which seem fairly straightforward, are in fact rather difficult to prove (here is a proof that Tao wrote up), and are essential to algebraic geometry. The first tells us that we [...]

]]> By: The Nullstellensatz and Partitions of Unity « Secret Blogging Seminar http://terrytao.wordpress.com/2007/11/26/hilberts-nullstellensatz/#comment-20466 The Nullstellensatz and Partitions of Unity « Secret Blogging Seminar Fri, 14 Dec 2007 19:32:47 +0000 http://terrytao.wordpress.com/2007/11/26/hilberts-nullstellensatz/#comment-20466 [...] Tao has a nice post explaining how to prove the Nullstellensatz by bulldozing forward in the most straightforward [...] [...] Tao has a nice post explaining how to prove the Nullstellensatz by bulldozing forward in the most straightforward [...]

]]> By: Emmanuel Kowalski http://terrytao.wordpress.com/2007/11/26/hilberts-nullstellensatz/#comment-18807 Emmanuel Kowalski Tue, 04 Dec 2007 14:07:27 +0000 http://terrytao.wordpress.com/2007/11/26/hilberts-nullstellensatz/#comment-18807 One of the simple ways of looking at infinitesimals in the scheme framework over a general field (or any commutative ring), and why they can be useful, is the following : Let $latex D$ be the ring $latex D=k[\epsilon]/(\epsilon^2)$, where $latex \epsilon$ is seen as an indeterminate. This is often called the ring of "dual numbers" over $latex k$. It has the property that $latex \epsilon$ is non-zero, but $latex \epsilon^2=0$, so $latex \epsilon$ is "infinitesimal of the first order". Now take any $latex k$-algebra $latex A$, say $latex A=k[X_1,....,X_r]/I$ for concreteness, where $latex I$ is some ideal. Then look at $latex k$-linear ring homomorphisms $latex d:A\rightarrow D$. Writing $latex d(a)=x(a)+\epsilon t(a)$, for uniquely defined $latex x(a)$ and $latex t(a)$, we find by writing out the conditions that (1) $latex d(I)=0$, and (2) $latex d(ab)=d(a)d(b)$, that to give $latex d$ is equivalent to giving a pair $latex (x,t)$ where (1) $latex x$ is a ring homomorphism $latex A\rightarrow k$, which means nothing else than an assignation of variables $latex X_i\mapsto x_i$ so that $latex (x_1,...,x_r)\in k^r$ is a solution of the equations defined by the ideal $latex I$; (2) $latex t$ is a $latex k$-linear map $latex A\rightarrow k$, with $latex t(I)=0$, such that $latex t(ab)=x(b)t(a)+x(a)t(b)$ or in other words, $latex t$ is a derivation "at the point $latex x$". Intuitively, $latex t$ is a vector tangent (at the point $latex x$) to the zero set $latex V$ in $latex k^r$ defined by the polynomial equations in the ideal $latex I$. This means that the set $latex T(A)$ of all homomorphisms $latex d:A\rightarrow D$ can be seen naturally as being the whole tangent "bundle" of $latex V$ (quotes because there might be singularities where this bundle is not as nicely behaved as the differential geometry analogy would suggest). Intuitively, it is the same to give a point on $latex V$ with coordinates in the ring of dual numbers, or to give a tangent vector to $latex V$. As hint that this is useful, one may note that if we have a $latex k$-linear ring homomorphism $latex A_1\rightarrow A_2$, we immediately obtain by composition an induced map $latex T(A_2)\rightarrow T(A_1)$, where $latex T(A)$ is $latex Hom(A,D)$, and this corresponds geometrically to the association of the tangent map $latex T(f): T(V)\rightarrow T(W)$ to a (polynomial) map $latex f:V\rightarrow W$ between their zero sets (in the reverse direction because $latex A$ corresponds to functions on $latex V$). One of the simple ways of looking at infinitesimals in the scheme framework over a general field (or any commutative ring), and why they can be useful, is the following :

Let D be the ring D=k[\epsilon]/(\epsilon^2), where \epsilon is seen as an indeterminate. This is often called the ring of “dual numbers” over k. It has the property that \epsilon is non-zero, but \epsilon^2=0, so \epsilon is “infinitesimal of the first order”.

Now take any k-algebra A, say A=k[X_1,....,X_r]/I for concreteness, where I is some ideal.

Then look at k-linear ring homomorphisms d:A\rightarrow D.

Writing d(a)=x(a)+\epsilon t(a), for uniquely defined x(a) and t(a), we find by writing out the conditions that (1) d(I)=0, and (2) d(ab)=d(a)d(b), that to give d is equivalent to giving a pair (x,t) where

(1) x is a ring homomorphism A\rightarrow k, which means nothing else than an assignation of variables X_i\mapsto x_i so that (x_1,...,x_r)\in k^r is a solution of the equations defined by the ideal I;

(2) t is a k-linear map A\rightarrow k, with t(I)=0, such that

t(ab)=x(b)t(a)+x(a)t(b)

or in other words, t is a derivation “at the point x“. Intuitively, t is a vector tangent (at the point x) to the zero set V in k^r defined by the polynomial equations in the ideal I.

This means that the set T(A) of all homomorphisms d:A\rightarrow D can be seen naturally as being the whole tangent “bundle” of V (quotes because there might be singularities where this bundle is not as nicely behaved as the differential geometry analogy would suggest). Intuitively, it is the same to give a point on V with coordinates in the ring of dual numbers, or to give a tangent vector to V.

As hint that this is useful, one may note that if we have a k-linear ring homomorphism A_1\rightarrow A_2, we immediately obtain by composition an induced map T(A_2)\rightarrow T(A_1), where T(A) is Hom(A,D), and this corresponds geometrically to the association of the tangent map T(f): T(V)\rightarrow T(W) to a (polynomial) map f:V\rightarrow W between their zero sets (in the reverse direction because A corresponds to functions on V).

]]> By: Terence Tao http://terrytao.wordpress.com/2007/11/26/hilberts-nullstellensatz/#comment-18733 Terence Tao Tue, 04 Dec 2007 05:43:59 +0000 http://terrytao.wordpress.com/2007/11/26/hilberts-nullstellensatz/#comment-18733 Dear Jens: I don't work with schemes on a regular basis, so I am not well placed to answer your question. But one viewpoint that I learned from a friend, which appeals to my background in analysis, is to think of schemes (in, say, $latex {\Bbb R}^n$, for sake of discussion, though I guess one has to take a little care here as $latex {\Bbb R}$ is not algebraically closed) as "infinitesimally thickened" varieties. An (affine) variety is the intersection of various zero loci $latex \{ x \in {\Bbb R}^n: P(x) = 0 \}$ of polynomials; you can think of an (affine) scheme as the intersection of near-zero loci $latex \{ x \in {\Bbb R}^n: |P(x)| \leq \varepsilon \}$, where $latex \varepsilon$ is an infinitesimal. Instead of looking at the functions which vanish on the zero locus, one then looks at functions which are of size $latex O(\varepsilon)$ (locally, at least) on the near-zero locus. For instance, the scheme $latex \{ x \in {\Bbb R}: |x^2| \leq \varepsilon \}$ is an infinitesimal thickening of the origin $latex \{0\}$ which is "fatter" than than the scheme $latex \{ x \in {\Bbb R}: |x| \leq \varepsilon \}$; the polynomial $latex x$ "vanishes" on the latter but not the former, while $latex x^2$ vanishes on both schemes. So schemes carry some information about multiplicity of vanishing which algebraic varieties do not; in particular they can distinguish between the vanishing of a polynomial $latex Q$ and the vanishing of a power $latex Q^r$ of that polynomial. This kind of viewpoint is sort of specific to fields such as $latex {\Bbb R}$ and $latex {\Bbb C}$ though, which have good notions of infinitesimals; I don't know how to interpret schemes like this over, say, fields of finite characteristic. Dear Anonymous: thanks for the reference! It is a nice argument; it proceeds by induction on dimension, as with my own argument, but with the neat additional trick of randomly rotating (or skewing) the variety first to prepare it for the projection to one lower dimension. I'm quite fond of random projection arguments in combinatorics, and it's nice to see one come up in algebra too. Dear Jens: I don’t work with schemes on a regular basis, so I am not well placed to answer your question. But one viewpoint that I learned from a friend, which appeals to my background in analysis, is to think of schemes (in, say, {\Bbb R}^n, for sake of discussion, though I guess one has to take a little care here as {\Bbb R} is not algebraically closed) as “infinitesimally thickened” varieties. An (affine) variety is the intersection of various zero loci \{ x \in {\Bbb R}^n: P(x) = 0 \} of polynomials; you can think of an (affine) scheme as the intersection of near-zero loci \{ x \in {\Bbb R}^n: |P(x)| \leq \varepsilon \}, where \varepsilon is an infinitesimal. Instead of looking at the functions which vanish on the zero locus, one then looks at functions which are of size O(\varepsilon) (locally, at least) on the near-zero locus. For instance, the scheme \{ x \in {\Bbb R}: |x^2| \leq \varepsilon \} is an infinitesimal thickening of the origin \{0\} which is “fatter” than than the scheme \{ x \in {\Bbb R}: |x| \leq \varepsilon \}; the polynomial x “vanishes” on the latter but not the former, while x^2 vanishes on both schemes. So schemes carry some information about multiplicity of vanishing which algebraic varieties do not; in particular they can distinguish between the vanishing of a polynomial Q and the vanishing of a power Q^r of that polynomial. This kind of viewpoint is sort of specific to fields such as {\Bbb R} and {\Bbb C} though, which have good notions of infinitesimals; I don’t know how to interpret schemes like this over, say, fields of finite characteristic.

Dear Anonymous: thanks for the reference! It is a nice argument; it proceeds by induction on dimension, as with my own argument, but with the neat additional trick of randomly rotating (or skewing) the variety first to prepare it for the projection to one lower dimension. I’m quite fond of random projection arguments in combinatorics, and it’s nice to see one come up in algebra too.

]]> By: Anonymous http://terrytao.wordpress.com/2007/11/26/hilberts-nullstellensatz/#comment-18695 Anonymous Mon, 03 Dec 2007 23:25:28 +0000 http://terrytao.wordpress.com/2007/11/26/hilberts-nullstellensatz/#comment-18695 That's funny, because only a month ago I was looking at precisely the same problem: there must be a reasonably elementary proof of the (weak) Nullstellensatz (I am teaching an undergraduate-level algebraic geometry class, so I certainly wanted to avoid too much algebra). I soon realized that the problem was not quite as elementary as I'd hoped, but there was a very nice proof by Arrondo published in the Monthly in 2006, that uses only resultants and Noether normalization. It's only two pages long, I think it's about as easy as it gets. I'm happy to share the reference with your readers. http://www.mat.ucm.es/~arrondo/publications.html That’s funny, because only a month ago I was looking at precisely the same problem: there must be a reasonably elementary proof of the (weak) Nullstellensatz (I am teaching an undergraduate-level algebraic geometry class, so I certainly wanted to avoid too much algebra).

I soon realized that the problem was not quite as elementary as I’d hoped, but there was a very nice proof by Arrondo published in the Monthly in 2006, that uses only resultants and Noether normalization. It’s only two pages long, I think it’s about as easy as it gets. I’m happy to share the reference with your readers.

http://www.mat.ucm.es/~arrondo/publications.html

]]> By: Jens Gottschalk http://terrytao.wordpress.com/2007/11/26/hilberts-nullstellensatz/#comment-18371 Jens Gottschalk Sun, 02 Dec 2007 11:38:14 +0000 http://terrytao.wordpress.com/2007/11/26/hilberts-nullstellensatz/#comment-18371 Hi Terence! "The presence of the exponent r in conclusion (2) is a little annoying; to get rid of it, one needs to generalise the notion of an algebraic variety to that of a scheme" This sounds interesting! I'm not very familiar with schemes, but I would be very glad if you or somebody else could explain the connection between this exponent and the notion of schemes. You wrote that this could be a whole story itself, but maybe somebody wants to write one or two words about it here. Thank you in advance and the best regards to you and all the other readers of this great blog! Jens Hi Terence!

“The presence of the exponent r in conclusion (2) is a little annoying; to get rid of it, one needs to generalise the notion of an algebraic variety to that of a scheme”

This sounds interesting! I’m not very familiar with schemes, but I would be very glad if you or somebody else could explain the connection between this exponent and the notion of schemes.

You wrote that this could be a whole story itself, but maybe somebody wants to write one or two words about it here.

Thank you in advance and the best regards to you and all the other readers of this great blog!

Jens

]]> By: The Hahn-Banach theorem, Menger’s theorem, and Helly’s theorem « What’s new http://terrytao.wordpress.com/2007/11/26/hilberts-nullstellensatz/#comment-18347 The Hahn-Banach theorem, Menger’s theorem, and Helly’s theorem « What’s new Sun, 02 Dec 2007 06:36:30 +0000 http://terrytao.wordpress.com/2007/11/26/hilberts-nullstellensatz/#comment-18347 [...] Helly's theorem, linear programming, Menger's theorem, minimax theorem, separation theorem In the previous post, I discussed how an induction on dimension approach could establish Hilbert’s [...] [...] Helly’s theorem, linear programming, Menger’s theorem, minimax theorem, separation theorem In the previous post, I discussed how an induction on dimension approach could establish Hilbert’s [...]

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