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| '''Hilbert's Nullstellensatz''' (German for "theorem of zeros," or more literally, "zero-locus-theorem" – see ''[[Satz (disambiguation)|Satz]]'') is a theorem which establishes a fundamental relationship between [[geometry]] and [[algebra]]. This relationship is the basis of [[algebraic geometry]], an important branch of [[mathematics]]. It relates [[algebraic set]]s to [[ideal (ring theory)|ideals]] in [[polynomial ring]]s over [[algebraically closed field]]s. This relationship was discovered by [[David Hilbert]] who proved the Nullstellensatz and several other important related theorems named after him (like [[Hilbert's basis theorem]]).
| | Asbsestos was a typical building content back the delayed '40s and early '50s. It had been used thus typically as a result of proven fact that it absolutely was cheap to make and an easy task to install. Asbestos has-been found in complexes throughout the planet, including London, and merely currently are we just starting to truly understand just why it was this type of bad idea. Asbestos, as it works out, is quite dangerous when broken down and breathed in. The allergens in it can cause some significant dilemmas for the respiratory system of the people that breathe it--including mesothelioma. So that it must clearly be eliminated when feasible, but merely with a trained professional.<br><br>Absestos Treatment in London<br><br>If you have asbestos that really needs to be eliminated then you definitely have to contact one of the most skilled business that London has to offer. The simple truth is that asbestos is most harmful when it is cracking, failing, or elsewhere being changed. This is obviously not really a job for any frequent Joe to try and execute. A professional firm can come to trigger the removal so when they are doing so they really will soon be donning the correct breathing device to help make the occupation secure along with the people around it protected also. The asbestos is then removed and damaged in a secure method.<br><br>While you is able to see there are lots of reasons that you might want to acquire gone asbestos sooner than later. The content, while in one piece, does not symbolize a lot of a security danger nevertheless the time it's changed it may become a problem. Do not wait till it becomes a challenge, once you know that asbestos is in your house--get it out! Take a look at [http://www.aulamusicapoetica.com/component/content/article/501-programa-del-congreso.html click]. |
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| == Formulation ==
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| Let ''k'' be a field (such as the [[rational number]]s) and ''K'' be an algebraically closed field extension (such as the [[complex number]]s), consider the [[polynomial ring]] ''k''[''X''<sub>1</sub>,''X''<sub>2</sub>,..., ''X''<sub>''n''</sub>] and let ''I'' be an [[Ideal (ring theory)|ideal]] in this ring. The [[algebraic set]] V(''I'') defined by this ideal consists of all ''n''-tuples '''x''' = (''x''<sub>1</sub>,...,''x''<sub>''n''</sub>) in ''K''<sup>''n''</sup> such that ''f''('''x''') = 0 for all ''f'' in ''I''. Hilbert's Nullstellensatz states that if ''p'' is some polynomial in ''k''[''X''<sub>1</sub>,''X''<sub>2</sub>,..., ''X''<sub>''n''</sub>] which vanishes on the algebraic set V(''I''), i.e. ''p''('''x''') = 0 for all '''x''' in ''V''(''I''), then there exists a [[natural number]] ''r'' such that ''p''<sup>''r''</sup> is in ''I''.
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| An immediate corollary is the "weak Nullstellensatz": The ideal ''I'' in ''k''[''X''<sub>1</sub>,''X''<sub>2</sub>,..., ''X''<sub>''n''</sub>] contains 1 if and only if the polynomials in ''I'' do not have any common zeros in ''K''<sup>''n''</sup>. It may also be formulated as follows:
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| if ''I'' is a proper ideal in ''k''[''X''<sub>1</sub>,''X''<sub>2</sub>,..., ''X''<sub>''n''</sub>], then V(''I'') cannot be [[empty set|empty]], i.e. there exists a common zero for all the polynomials in the ideal in every algebraically closed extension of ''k''. This is the reason for the name of the theorem, which can be proved easily from the 'weak' form using the [[Rabinowitsch trick]]. The assumption of considering common zeros in an algebraically closed field is essential here; for example, the elements of the proper ideal (''X''<sup>2</sup> + 1) in '''R'''[''X''] do not have a common zero in '''R'''.
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| With the notation common in algebraic geometry, the Nullstellensatz can also be formulated as
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| :<math>\hbox{I}(\hbox{V}(J))=\sqrt{J}</math>
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| for every ideal ''J''. Here, <math>\sqrt{J}</math> denotes the [[radical of an ideal|radical]] of ''J'' and I(''U'') is the ideal of all polynomials which vanish on the set ''U''.
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| In this way, we obtain an order-reversing [[bijective]] correspondence between the algebraic sets in ''K''<sup>''n''</sup> and the [[radical ideal]]s of ''K''[''X''<sub>1</sub>,''X''<sub>2</sub>,..., ''X''<sub>''n''</sub>]. In fact, more generally, one has a [[Galois connection]] between subsets of the space and subsets of the algebra, where "[[Zariski closure]]" and "radical of the ideal generated" are the [[closure operator]]s.
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| As a particular example, consider a point <math>P = (a_1, \cdots, a_n) \in K^n</math>. Then <math>I(P) = (X_1 - a_1, \cdots, X_n - a_n)</math>. More generally,
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| :<math>\sqrt{I} = \bigcap_{P \in V(I)} (X_1 - a_1, \cdots, X_n - a_n), \quad P = (a_1, \cdots, a_n).</math>
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| As another example, an algebraic subset ''W'' in ''K''<sup>''n''</sup> is irreducible (in the Zariski topology) if and only if <math>I(W)</math> is a prime ideal.
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| == Proof and generalization ==
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| There are many known proofs of the theorem. One proof <!-- due to Zariski? --> is the following:
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| # Note that it is enough to prove [[Zariski's lemma]]: a finitely generated algebra over a field ''k'' that is a field is a finite field extension of ''k''.
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| # Prove Zariski's lemma.
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| The proof of Step 1 is elementary. Step 2 is deeper. It follows, for example, from the [[Noether normalization lemma]]. See [[Zariski's lemma]] for more. Here we sketch the proof of Step 1.<ref>{{harvnb|Atiyah-MacDonald|1969|loc=Ch. 7}}</ref> Let <math>A = k[t_1, ..., t_n]</math> (''k'' algebraically closed field), ''I'' an ideal of ''A'' and ''V'' the common zeros of ''I'' in <math>k^n</math>. Clearly, <math>\sqrt{I} \subseteq I(V)</math>. Let <math>f \not\in \sqrt{I}</math>. Then <math>f \not\in \mathfrak{p}</math> for some prime ideal <math>\mathfrak{p}\supseteq I</math> in ''A''. Let <math>R = (A/\mathfrak{p}) [f^{-1}]</math> and <math>\mathfrak{m}</math> a maximal ideal in <math>R</math>. By Zariski's lemma, <math>R/\mathfrak{m}</math> is a finite extension of ''k''; thus, is ''k'' since ''k'' is algebraically closed. Let <math>x_i</math> be the images of <math>t_i</math> under the natural map <math>A \to k</math>. It follows that <math>x = (x_1, ..., x_n) \in V</math> and <math>f(x) \ne 0</math>.
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| The Nullstellensatz will also follow trivially once one systematically developed the theory of a [[Jacobson ring]], a ring in which a radical ideal is an intersection of maximal ideals. Let <math>R</math> be a [[Jacobson ring]]. If <math>S</math> is a finitely generated [[associative algebra|''R''-algebra]], then <math>S</math> is a Jacobson ring. Further, if <math>\mathfrak{n}\subset S</math> is a maximal ideal, then <math>\mathfrak{m} := \mathfrak{n} \cap R</math> is a maximal ideal of R, and <math>S/\mathfrak{n}</math> is a finite extension field of <math>R/\mathfrak{m}</math>.
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| Another generalization states that a faithfully flat morphism <math>f: Y \to X</math> locally of finite type with ''X'' quasi-compact has a ''quasi-section'', i.e. there exists <math>X'</math> affine and faithfully flat and quasi-finite over ''X'' together with an ''X''-morphism <math>X' \to Y</math>
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| == Effective Nullstellensatz ==
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| In all of its variants, Hilbert's Nullstellensatz asserts that some polynomial <math>g</math> belongs or not to an ideal generated, say, by <math>f_1,\dots,f_k</math>; we have <math>g=f^r</math> in the strong version, <math>g=1</math> in the weak form. This means the existence or the non existence of polynomials <math>g_1,\dots,g_k</math> such that <math>g=f_1g_1+\cdots +f_kg_k.</math> The usual proofs of the Nullstellensatz are non effective in the sense that they do not give any way to compute the <math>g_i</math>.
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| It is thus a rather natural question to ask if there is an effective way to compute the <math>g_i</math> (and the exponent <math>r</math> in the strong form) or to prove that they do not exist. To solve this problem, it suffices to provide an upper bound on the total degree of the <math>g_i</math>: such a bound reduces the problem to a finite [[system of linear equations]] that may be solved by usual [[linear algebra]] techniques.
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| Any such upper bound is called an '''effective Nullstellensatz'''.
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| A related problem is the '''ideal membership problem''', which consists in testing if a polynomial belongs to an ideal. For this problem also, a solution is provided by an upper bound on the degree of the <math>g_i.</math> A general solution of the ideal membership problem provides an effective Nullstellensatz, at least for the weak form.
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| In 1925, [[Grete Hermann]] gave an upper bound for ideal membership problem that is doubly exponential in the number of variables. In 1982 Mayr and Meyer gave an example where the <math>g_i</math> have a degree which is at least double exponential, showing that every general upper bound for the ideal membership problem is doubly exponential in the number of variables.
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| Until 1987, nobody had the idea that effective Nullstellensatz was easier than ideal membership, when [[Brownawell]] gave an upperbound for the effective Nullstellensatz which is simply exponential in the number of variables. Brownawell proof uses calculus techniques and thus is valid only in characteristic 0. Soon after, in 1988, [[János Kollár]] gave a purely algebraic proof valid in any characteristic, leading to a better bound.
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| In the case of the weak Nullstellensatz, Kollár's bound is the following:<ref>{{citation|first=János|last=Kollár|title=Sharp Effective Nullstellensatz|journal=Journal of the American Mathematical Society|volume= 1|issue=4|date=October 1988|pages=963–975|url=http://www.math.ucdavis.edu/~deloera/MISC/BIBLIOTECA/trunk/Kollar/kollarnullstellen.pdf}}</ref>
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| :Let <math>f_1,\ldots, f_s</math> be polynomials in ''n''≥2 variables, of total degree <math>d_1\ge \cdots \ge d_s.</math> If there exist polynomials <math>g_i</math> such that <math>f_1g_1+\cdots +f_sg_s=1,</math> then they can be chosen such that <math>\deg(f_ig_i) \le \max(d_s,3)\prod_{j=1}^{\min(n,s)-1}\max(d_j,3).</math> This bound is optimal if all the degrees are greater than 2.
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| If ''d'' is the maximum of the degrees of the <math>f_i</math>, this bound may be simplified to <math>\max(3,d)^{\min(n,s)}.</math>
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| Kollár's result has been improved by several authors. M. Sombra has provided the best improvement, up to date, giving the bound<ref>{{citation|first=Martín|last=Sombra|title=A Sparse Effective Nullstellensatz|journal=Advances in Applied Mathematics|volume= 22|issue=2|date=February 1999|pages=271–295|url=http://arxiv.org/pdf/alg-geom/9710003.pdf}}</ref> <math>\deg(f_ig_i) \le 2d_s\prod_{j=1}^{\min(n,s)-1}d_j.</math>. His bound is better than Kollár's as soon as at least two of the degrees that are involved are lower than 3..
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| ==Projective Nullstellensatz==
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| We can formulate a certain correspondence between homogeneous ideals of polynomials and algebraic subsets of a projective space, called the '''projective Nullstellensatz''', that is analogous to the affine one. To do that, we introduce some notations. Let <math>R = k[t_0, ..., t_n].</math> The homogeneous ideal <math>R_+ = \bigoplus_{d \ge 1} R_d</math> is called the ''maximal homogeneous ideal'' (see also [[irrelevant ideal]]). As in the affine case, we let: for a subset <math>S \subseteq \mathbb{P}^n</math> and a homogeneous ideal ''I'' of ''R'',
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| :<math>\begin{align}
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| \operatorname{I}_{\mathbb{P}^n}(S) &= \{ f \in R_+ | f = 0 \text{ on } S \}, \\
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| \operatorname{V}_{\mathbb{P}^n}(I) &= \{ x \in \mathbb{P}^n | f(x) = 0 \text{ for all }f \in I \}.
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| \end{align}
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| </math>
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| By <math>f = 0 \text{ on } S</math> we mean: for every homogeneous coordinates <math>(a_0 : \cdots : a_n)</math> of a point of ''S'' we have <math>f(a_0,\ldots, a_n)=0</math>. This implies that the homogeneous components of ''f'' are also zero on ''S'' and thus that <math>\operatorname{I}_{\mathbb{P}^n}(S)</math> is a homogeneous ideal. Equivalently, <math>\operatorname{I}_{\mathbb{P}^n}(S)</math> is the homogeneous ideal generated by homogeneous polynomials ''f'' that vanish on ''S''. Now, for any homogeneous ideal <math>I \subseteq R_+</math>, by the usual Nullstellensatz, we have:
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| :<math>\sqrt{I} = \operatorname{I}_{\mathbb{P}^n}(\operatorname{V}_{\mathbb{P}^n}(I)),</math>
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| and so, like in the affine case, we have:<ref>This formulation comes from Milne, Algebraic geometry [http://www.jmilne.org/math/CourseNotes/ag.html] and differs from {{harvnb|Hartshorne|1977|loc=Ch. I, Exercise 2.4}}</ref>
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| :There exists an order-reversing one-to-one correspondence between proper homogeneous radical ideals of ''R'' and subsets of <math>\mathbb{P}^n</math> of the form <math>\operatorname{V}_{\mathbb{P}^n}(I).</math> The correspondence is given by <math>\operatorname{I}_{\mathbb{P}^n}</math> and <math>\operatorname{V}_{\mathbb{P}^n}.</math>
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| ==See also==
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| *[[Stengle's Positivstellensatz]]
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| *[[Differential Nullstellensatz]]
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| *[[Combinatorial Nullstellensatz]]
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| ==Notes==
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| {{reflist}}
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| ==References==
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| *[[Michael Atiyah|M. Atiyah]], [[Ian G. Macdonald|I.G. Macdonald]], ''Introduction to Commutative Algebra'', [[Addison–Wesley]], 1994. ISBN 0-201-40751-5
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| * {{cite book | author=Shigeru Mukai | authorlink=Shigeru Mukai | coauthors=William Oxbury (translator) | title=An Introduction to Invariants and Moduli | series=Cambridge studies in advanced mathematics | volume=81 | year=2003 | isbn=0-521-80906-1 | page=82 }}
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| * [[David Eisenbud]], ''Commutative Algebra With a View Toward Algebraic Geometry'', New York : Springer-Verlag, 1999.
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| *{{Hartshorne AG}}
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| [[Category:Polynomials]]
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| [[Category:Theorems in algebraic geometry]]
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Asbsestos was a typical building content back the delayed '40s and early '50s. It had been used thus typically as a result of proven fact that it absolutely was cheap to make and an easy task to install. Asbestos has-been found in complexes throughout the planet, including London, and merely currently are we just starting to truly understand just why it was this type of bad idea. Asbestos, as it works out, is quite dangerous when broken down and breathed in. The allergens in it can cause some significant dilemmas for the respiratory system of the people that breathe it--including mesothelioma. So that it must clearly be eliminated when feasible, but merely with a trained professional.
Absestos Treatment in London
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