Hilbert's Nullstellensatz

Let k be a field (such as the rational numbers) and K be an algebraically closed field extension (such as the complex numbers). Consider the polynomial ring ${\displaystyle k[X_{1},\ldots ,X_{n}]}$ and let I be an ideal in this ring. The algebraic set V(I) defined by this ideal consists of all n-tuples x = (x₁,...,x_n) in Kⁿ such that f(x) = 0 for all f in I. Hilbert's Nullstellensatz states that if p is some polynomial in ${\displaystyle k[X_{1},\ldots ,X_{n}]}$ that 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^r is in I.[1]

An immediate corollary is the weak Nullstellensatz: The ideal ${\displaystyle I\subset k[X_{1},\ldots ,X_{n}]}$ contains 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 in ${\displaystyle k[X_{1},\ldots ,X_{n}],}$ then V(I) cannot be 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 (X² + 1) in ${\displaystyle \mathbb {R} [X]}$ do not have a common zero in ${\displaystyle \mathbb {R} .}$

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

${\displaystyle {\hbox{I}}({\hbox{V}}(J))={\sqrt {J}}}$

for every ideal J. Here, ${\displaystyle {\sqrt {J}}}$ denotes the radical of J and I(U) is the ideal of all polynomials that vanish on the set U.

In this way, taking ${\displaystyle k=K}$ we obtain an order-reversing bijective correspondence between the algebraic sets in Kⁿ and the radical ideals of ${\displaystyle K[X_{1},\ldots ,X_{n}].}$ 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 operators.

As a particular example, consider a point ${\displaystyle P=(a_{1},\dots ,a_{n})\in K^{n}}$. Then ${\displaystyle I(P)=(X_{1}-a_{1},\ldots ,X_{n}-a_{n})}$. More generally,

${\displaystyle {\sqrt {I}}=\bigcap _{(a_{1},\dots ,a_{n})\in V(I)}(X_{1}-a_{1},\dots ,X_{n}-a_{n}).}$

Conversely, every maximal ideal of the polynomial ring ${\displaystyle K[X_{1},\ldots ,X_{n}]}$ (note that ${\displaystyle K}$ is algebraically closed) is of the form ${\displaystyle (X_{1}-a_{1},\ldots ,X_{n}-a_{n})}$ for some ${\displaystyle a_{1},\ldots ,a_{n}\in K}$.

As another example, an algebraic subset W in Kⁿ is irreducible (in the Zariski topology) if and only if ${\displaystyle I(W)}$ is a prime ideal.

There are many known proofs of the theorem. Some are non-constructive, such as the first one. Others are constructive, as based on algorithms for expressing 1 or p^r as a linear combination of the generators of the ideal.

The Nullstellensatz is subsumed by a systematic development of the theory of Jacobson rings, which are those rings in which every radical ideal is an intersection of maximal ideals. Given Zariski's lemma, proving the Nullstellensatz amounts to showing that if k is a field, then every finitely generated k-algebra R (necessarily of the form ${\textstyle R=k[t_{1},\cdots ,t_{n}]/I}$) is Jacobson. More generally, one has the following theorem:

Let ${\displaystyle R}$ be a Jacobson ring. If ${\displaystyle S}$ is a finitely generated R-algebra, then ${\displaystyle S}$ is a Jacobson ring. Furthermore, if ${\displaystyle {\mathfrak {n}}\subset S}$ is a maximal ideal, then ${\displaystyle {\mathfrak {m}}:={\mathfrak {n}}\cap R}$ is a maximal ideal of ${\textstyle R}$, and ${\displaystyle S/{\mathfrak {n}}}$ is a finite extension of ${\displaystyle R/{\mathfrak {m}}}$.[3]

Other generalizations proceed from viewing the Nullstellensatz in scheme-theoretic terms as saying that for any field k and nonzero finitely generated k-algebra R, the morphism ${\textstyle \mathrm {Spec} \,R\to \mathrm {Spec} \,k}$ admits a section étale-locally (equivalently, after base change along ${\textstyle \mathrm {Spec} \,L\to \mathrm {Spec} \,k}$ for some finite field extension ${\textstyle L/k}$). In this vein, one has the following theorem:

Any faithfully flat morphism of schemes ${\textstyle f:Y\to X}$ locally of finite presentation admits a quasi-section, in the sense that there exists a faithfully flat and locally quasi-finite morphism ${\textstyle g:X'\to X}$ locally of finite presentation such that the base change ${\textstyle f':Y\times _{X}X'\to X'}$ of ${\textstyle f}$ along ${\textstyle g}$ admits a section.[4] Moreover, if ${\textstyle X}$ is quasi-compact (resp. quasi-compact and quasi-separated), then one may take ${\textstyle X'}$ to be affine (resp. ${\textstyle X'}$ affine and ${\textstyle g}$ quasi-finite), and if ${\textstyle f}$ is smooth surjective, then one may take ${\textstyle g}$ to be étale.[5]

Serge Lang gave an extension of the Nullstellensatz to the case of infinitely many generators:

Let ${\textstyle \kappa }$ be an infinite cardinal and let ${\textstyle K}$ be an algebraically closed field whose transcendence degree over its prime subfield is strictly greater than ${\displaystyle \kappa }$. Then for any set ${\textstyle S}$ of cardinality ${\textstyle \kappa }$, the polynomial ring ${\textstyle A=K[x_{i}]_{i\in S}}$ satisfies the Nullstellensatz, i.e., for any ideal ${\textstyle J\subset A}$ we have that ${\displaystyle {\sqrt {J}}={\hbox{I}}({\hbox{V}}(J))}$.[6]

In all of its variants, Hilbert's Nullstellensatz asserts that some polynomial g belongs or not to an ideal generated, say, by f₁, ..., f_k; 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, ..., g_k such that g = f₁g₁ + ... + f_kg_k. The usual proofs of the Nullstellensatz are not constructive, non-effective, in the sense that they do not give any way to compute the g_i.

It is thus a rather natural question to ask if there is an effective way to compute the g_i (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 g_i: 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 g_i. 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 g_i 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.[7] 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:[8]

Let f₁, ..., f_s be polynomials in n ≥ 2 variables, of total degree d₁ ≥ ... ≥ d_s. If there exist polynomials g_i such that f₁g₁ + ... + f_sg_s = 1, then they can be chosen such that

${\displaystyle \deg(f_{i}g_{i})\leq \max(d_{s},3)\prod _{j=1}^{\min(n,s)-1}\max(d_{j},3).}$

This bound is optimal if all the degrees are greater than 2.

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

${\displaystyle \max(3,d)^{\min(n,s)}.}$

Kollár's result has been improved by several authors. As of 14 October 2012, the best improvement, due to M. Sombra is[9]

${\displaystyle \deg(f_{i}g_{i})\leq 2d_{s}\prod _{j=1}^{\min(n,s)-1}d_{j}.}$

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

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 ${\displaystyle R=k[t_{0},\ldots ,t_{n}].}$ The homogeneous ideal,

${\displaystyle R_{+}=\bigoplus _{d\geqslant 1}R_{d}}$

is called the maximal homogeneous ideal (see also irrelevant ideal). As in the affine case, we let: for a subset ${\displaystyle S\subseteq \mathbb {P} ^{n}}$ and a homogeneous ideal I of R,

${\displaystyle {\begin{aligned}\operatorname {I} _{\mathbb {P} ^{n}}(S)&=\{f\in R_{+}\mid f=0{\text{ on }}S\},\\\operatorname {V} _{\mathbb {P} ^{n}}(I)&=\{x\in \mathbb {P} ^{n}\mid f(x)=0{\text{ for all }}f\in I\}.\end{aligned}}}$

By ${\displaystyle f=0{\text{ on }}S}$ we mean: for every homogeneous coordinates ${\displaystyle (a_{0}:\cdots :a_{n})}$ of a point of S we have ${\displaystyle f(a_{0},\ldots ,a_{n})=0}$. This implies that the homogeneous components of f are also zero on S and thus that ${\displaystyle \operatorname {I} _{\mathbb {P} ^{n}}(S)}$ is a homogeneous ideal. Equivalently, ${\displaystyle \operatorname {I} _{\mathbb {P} ^{n}}(S)}$ is the homogeneous ideal generated by homogeneous polynomials f that vanish on S. Now, for any homogeneous ideal ${\displaystyle I\subseteq R_{+}}$, by the usual Nullstellensatz, we have:

${\displaystyle {\sqrt {I}}=\operatorname {I} _{\mathbb {P} ^{n}}(\operatorname {V} _{\mathbb {P} ^{n}}(I)),}$

and so, like in the affine case, we have:[10]

There exists an order-reversing one-to-one correspondence between proper homogeneous radical ideals of R and subsets of ${\displaystyle \mathbb {P} ^{n}}$ of the form ${\displaystyle \operatorname {V} _{\mathbb {P} ^{n}}(I).}$ The correspondence is given by ${\displaystyle \operatorname {I} _{\mathbb {P} ^{n}}}$ and ${\displaystyle \operatorname {V} _{\mathbb {P} ^{n}}.}$

The Nullstellensatz also holds for the germs of holomorphic functions at a point of complex n-space ${\displaystyle \mathbb {C} ^{n}.}$ Precisely, for each open subset ${\displaystyle U\subset \mathbb {C} ^{n},}$ let ${\displaystyle {\mathcal {O}}_{\mathbb {C} ^{n}}(U)}$ denote the ring of holomorphic functions on U; then ${\displaystyle {\mathcal {O}}_{\mathbb {C} ^{n}}}$ is a sheaf on ${\displaystyle \mathbb {C} ^{n}.}$ The stalk ${\displaystyle {\mathcal {O}}_{\mathbb {C} ^{n},0}}$ at, say, the origin can be shown to be a Noetherian local ring that is a unique factorization domain.

If ${\displaystyle f\in {\mathcal {O}}_{\mathbb {C} ^{n},0}}$ is a germ represented by a holomorphic function ${\displaystyle {\widetilde {f}}:U\to \mathbb {C} }$, then let ${\displaystyle V_{0}(f)}$ be the equivalence class of the set

${\displaystyle \left\{z\in U\mid {\widetilde {f}}(z)=0\right\},}$

where two subsets ${\displaystyle X,Y\subset \mathbb {C} ^{n}}$ are considered equivalent if ${\displaystyle X\cap U=Y\cap U}$ for some neighborhood U of 0. Note ${\displaystyle V_{0}(f)}$ is independent of a choice of the representative ${\displaystyle {\widetilde {f}}.}$ For each ideal ${\displaystyle I\subset {\mathcal {O}}_{\mathbb {C} ^{n},0},}$ let ${\displaystyle V_{0}(I)}$ denote ${\displaystyle V_{0}(f_{1})\cap \dots \cap V_{0}(f_{r})}$ for some generators ${\displaystyle f_{1},\ldots ,f_{r}}$ of I. It is well-defined; i.e., is independent of a choice of the generators.

For each subset ${\displaystyle X\subset \mathbb {C} ^{n}}$, let

${\displaystyle I_{0}(X)=\left\{f\in {\mathcal {O}}_{\mathbb {C} ^{n},0}\mid V_{0}(f)\supset X\right\}.}$

It is easy to see that ${\displaystyle I_{0}(X)}$ is an ideal of ${\displaystyle {\mathcal {O}}_{\mathbb {C} ^{n},0}}$ and that ${\displaystyle I_{0}(X)=I_{0}(Y)}$ if ${\displaystyle X\sim Y}$ in the sense discussed above.

The analytic Nullstellensatz then states:[11] for each ideal ${\displaystyle I\subset {\mathcal {O}}_{\mathbb {C} ^{n},0}}$,

${\displaystyle {\sqrt {I}}=I_{0}(V_{0}(I))}$

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

• Stengle's Positivstellensatz
• Differential Nullstellensatz
• Combinatorial Nullstellensatz
• Artin–Tate lemma
• Real radical
• Restricted power series#Tate algebra, an analog of Hilbert's nullstellensatz holds for Tate algebras.

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