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# Hilbert's Nullstellensatz

Hilbert's Nullstellensatz (German for "theorem of zeros," or more literally, "zero-locus-theorem"—see Satz) is a theorem that establishes a fundamental relationship between geometry and algebra. This relationship is the basis of algebraic geometry, a branch of mathematics. It relates algebraic sets to ideals in polynomial rings over algebraically closed fields. 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).

## Formulation

Let k be a field (such as therational numbers) and K be an algebraically closed field extension (such as thecomplex numbers), consider thepolynomial ringand let I be anidealin this ring. Thealgebraic setV(I) defined by this ideal consists of all n-tuples x = (x1,...,xn) in *Kn
• such that f(x) = 0 for all f in I. Hilbert's Nullstellensatz states that if p is some polynomial in
that vanishes on the algebraic set V(I), i.e. p(x) = 0 for all x in V(I), then there exists anatural numberr such that pris in I.
An immediate corollary is the "weak Nullstellensatz": The idealcontains 1 if and only if the polynomials in I do not have any common zeros in *Kn*. It may also be formulated as follows: if I is a proper ideal inthen V(I) cannot beempty, 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 theRabinowitsch trick. The assumption of considering common zeros in an algebraically closed field is essential here; for example, the elements of the proper ideal (X2
1. in
do not have a common zero in

With the notation common in algebraic geometry, the Nullstellensatz can also be formulated as

for every ideal J. Here,denotes theradicalof J and I(U) is the ideal of all polynomials that vanish on the set U.
In this way, we obtain an order-reversingbijectivecorrespondence between the algebraic sets in Knand theradical idealsofIn fact, more generally, one has aGalois connectionbetween subsets of the space and subsets of the algebra, where "Zariski closure" and "radical of the ideal generated" are theclosure operators.
As a particular example, consider a point. Then. More generally,
Conversely, everymaximal idealof the polynomial ring(note thatis algebraically closed) is of the formfor some.
As another example, an algebraic subset W in Knis irreducible (in the Zariski topology) if and only ifis a prime ideal.

## Proof and generalization

There are many known proofs of the theorem. One proof uses Zariski's lemma, which asserts that, if a field is finitely generated as an associative algebra over a field k, then it is a finite field extension of k (that is, it is also finitely generated as a vector space). Here is a sketch of this proof.[1]

Let(k algebraically closed field), I an ideal of A and V the common zeros of I in. Clearly,. Let. Thenfor some prime idealin A. Letanda maximal ideal in. By Zariski's lemma,is a finite extension of k; thus, is k since k is algebraically closed. Letbe the images ofunder the natural map. It follows thatand.
The Nullstellensatz will also follow trivially once one systematically developed the theory of aJacobson ring, a ring in which a radical ideal is an intersection of maximal ideals. Letbe a Jacobson ring. Ifis a finitely generatedR-algebra, thenis a Jacobson ring. Further, ifis a maximal ideal, thenis a maximal ideal of R, andis a finite extension field of.
Another generalization states that a faithfully flat morphism of schemeslocally of finite type with X quasi-compact has a quasi-section, i.e. there existsaffine and faithfully flat and quasi-finite over X together with an X-morphism

## Effective Nullstellensatz

In all of its variants, Hilbert's Nullstellensatz asserts that some polynomial g belongs or not to an ideal generated, say, by f1, ..., fk; we have g = f r in the strong version, g = 1 in the weak form. This means the existence or the non-existence of polynomials g1, ..., gk such that g = f1g1 + ... + fkgk. The usual proofs of the Nullstellensatz are not constructive, non-effective, in the sense that they do not give any way to compute the gi.

It is thus a rather natural question to ask if there is an effective way to compute the gi (and the exponent r 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 gi: such a bound reduces the problem to a finite system of linear equations that may be solved by usual linear algebra techniques. Any such upper bound is called an effective Nullstellensatz.

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 gi. A general solution of the ideal membership problem provides an effective Nullstellensatz, at least for the weak form.

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 gi have a degree that is at least double exponential, showing that every general upper bound for the ideal membership problem is doubly exponential in the number of variables.

Since most mathematicians at the time assumed the effective Nullstellensatz was at least as hard as ideal membership, few mathematicians sought a bound better than double-exponential. In 1987, however, W. Dale Brownawell gave an upper bound for the effective Nullstellensatz that is simply exponential in the number of variables.[2] Brownawell's proof relied on analytic techniques valid only in characteristic 0, but, one year later, János Kollár gave a purely algebraic proof, valid in any characteristic, of a slightly better bound.

In the case of the weak Nullstellensatz, Kollár's bound is the following:[3]

Letf1, ..., *fs
be polynomials inn ≥ 2variables, of total degreed1≥ ... ≥ *ds
. If there exist polynomials*gi
such thatf1g1
• ... + *f
sgs
• = 1
, then they can be chosen such that
This bound is optimal if all the degrees are greater than 2.

If d is the maximum of the degrees of the fi, this bound may be simplified to

Kollár's result has been improved by several authors. As of 14 October 2012, the best improvement, due to M. Sombra is[4]

His bound improves Kollár's as soon as at least two of the degrees that are involved are lower than 3.

## Projective Nullstellensatz

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. LetThe homogeneous ideal,
is called the maximal homogeneous ideal (see alsoirrelevant ideal). As in the affine case, we let: for a subsetand a homogeneous ideal I of R,
Bywe mean: for every homogeneous coordinatesof a point of S we have. This implies that the homogeneous components of f are also zero on S and thus thatis a homogeneous ideal. Equivalently,is the homogeneous ideal generated by homogeneous polynomials f that vanish on S. Now, for any homogeneous ideal, by the usual Nullstellensatz, we have:

and so, like in the affine case, we have:[5]

There exists an order-reversing one-to-one correspondence between proper homogeneous radical ideals of R and subsets ofof the formThe correspondence is given byand

## Analytic Nullstellensatz

The Nullstellensatz also holds for the germs of holomorphic functions at a point of complex n-spacePrecisely, for each open subsetletdenote the ring of holomorphic functions on U; thenis a *sheaf
• on
The stalkat, say, the origin can be shown to be aNoetherianlocal ringthat is aunique factorization domain.
Ifis a germ represented by a holomorphic function, then letbe the equivalence class of the set
where two subsetsare considered equivalent iffor some neighborhood U of 0. Noteis independent of a choice of the representativeFor each idealletdenotefor some generatorsof I. It is well-defined; i.e., is independent of a choice of the generators.
For each subset, let
It is easy to see thatis an ideal ofand thatifin the sense discussed above.
The analytic Nullstellensatz then states:[6] for each ideal,

where the left-hand side is the radical of I.

• Stengle's Positivstellensatz

• Differential Nullstellensatz

• Combinatorial Nullstellensatz

• Artin–Tate lemma

## References

[1]
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[2]
Citation Link//www.ams.org/mathscinet-getitem?mr=0916719Brownawell, W. Dale (1987), "Bounds for the degrees in the Nullstellensatz", Ann. of Math., 126 (3): 577–591, doi:10.2307/1971361, MR 0916719
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[3]
Citation Link//www.ams.org/mathscinet-getitem?mr=0944576Kollár, János (1988), "Sharp Effective Nullstellensatz" (PDF), Journal of the American Mathematical Society, 1 (4): 963–975, doi:10.2307/1990996, MR 0944576
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[4]
Citation Link//www.ams.org/mathscinet-getitem?mr=1659402Sombra, Martín (1999), "A Sparse Effective Nullstellensatz", Advances in Applied Mathematics, 22 (2): 271–295, arXiv:alg-geom/9710003, doi:10.1006/aama.1998.0633, MR 1659402
Oct 1, 2019, 9:11 PM
[5]
Citation Linkopenlibrary.orgThis formulation comes from Milne, Algebraic geometry [1] and differs from Hartshorne, Robin (1977), Algebraic Geometry, Graduate Texts in Mathematics, 52, New York: Springer-Verlag, ISBN 978-0-387-90244-9, MR 0463157, Ch. I, Exercise 2.4
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