Hilbert's Nullstellensatz

Hilbert's Nullstellensatz (German for "theorem of zeros," or more literally, "zero-locus-theorem" – see 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 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 the rational numbers) and K be an algebraically closed field extension (such as the complex numbers), consider the polynomial ring k[X₁,X₂,..., 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 k[X₁,X₂,..., X_n] 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^r is in I.

An immediate corollary is the "weak Nullstellensatz": The ideal I in k[X₁,X₂,..., X_n] contains 1 if and only if the polynomials in I do not have any common zeros in Kⁿ.

When k=K the "weak Nullstellensatz" may also be stated as follows: if I is a proper ideal in K[X₁,X₂,..., X_n], then V(I) cannot be empty, i.e. there exists a common zero for all the polynomials in the ideal. 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 that K be algebraically closed is essential here; the elements of the proper ideal (X² + 1) in R[X] do not have a common zero. With the notation common in algebraic geometry, the Nullstellensatz can also be formulated as

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

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

In this way, we obtain an order-reversing bijective correspondence between the algebraic sets in Kⁿ and the radical ideals of K[X₁,X₂,..., 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 $P = (a_1, \cdots, a_n) \in K^n$. Then $I(P) = (X_1 - a_1, \cdots, X_n - a_n)$. More generally,

$\sqrt{I} = \bigcap_{P \in V(I)} (X_1 - a_1, \cdots, X_n - a_n), \quad P = (a_1, \cdots, a_n).$

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

Proof and generalization

There are many known proofs of the theorem. One proof is the following:

1. 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.
2. Prove Zariski's lemma.

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. Let $A = k[t_1, ..., t_n]$ (k algebraically closed field), I an ideal of A and V the common zeros of I in $k^n$. Clearly, $\sqrt{I} \subseteq I(V)$. Let $f \not\in \sqrt{I}$. Then $f \not\in \mathfrak{p}$ for some prime ideal $\mathfrak{p}\supseteq I$ in A. Let $R = (A/\mathfrak{p}) [f^{-1}]$ and $\mathfrak{m}$ a maximal ideal in $R$. By Zariski's lemma, $R/\mathfrak{m}$ is a finite extension of k; thus, is k since k is algebraically closed. Let $x_i$ be the images of $t_i$ under the natural map $A \to k$. It follows that $x = (x_1, ..., x_n) \in V$ and $f(x) \ne 0$.

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. This idea is due to Bourbaki.^{[citation needed]} Let $R$ be a Jacobson ring. If $S$ is a finitely generated R-algebra, then $S$ is a Jacobson ring. Further, if $\mathfrak{n}\subset S$ is a maximal ideal, then $\mathfrak{m} := \mathfrak{n} \cap R$ is a maximal ideal of R, and $S/\mathfrak{n}$ is a finite extension field of $R/\mathfrak{m}$.

Another generalization states that a faithfully flat morphism $f: Y \to X$ locally of finite type with X quasi-compact has a quasi-section, i.e. there exists $X'$ affine and faithfully flat and quasi-finite over X together with an X-morphism $X' \to Y$

Effective Nullstellensatz

In all of its variants, Hilbert's Nullstellensatz asserts that some polynomial $g$ belongs or not to an ideal generated, say, by $f_1,\dots,f_k$; 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 $g_1,\dots,g_k$ such that $g=f_1g_1+\cdots +f_kg_k.$ The usual proofs of the Nullstellensatz are non effective in the sense that they do not give any way to compute the $g_i$.

This is thus a rather natural question to ask if there 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 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.

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.

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

Let $f_1,\ldots, f_s$ be polynomials in n≥2 variables, of total degree $d_1\ge \cdots \ge d_s.$ If there exist polynomials $g_i$ such that $f_1g_1+\cdots +f_sg_s=1,$ then they can be chosen such that $\deg(f_ig_i) \le \max(d_s,3)\prod_{j=1}^{\min(n,s)-1}\max(d_j,3).$ 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 $\max(3,d)^{\min(n,s)}.$

Kollár's result has been improved by several authors. M. Sombra has provided the best improvement, up to date, giving the bound $\deg(f_ig_i) \le 2d_s\prod_{j=1}^{\min(n,s)-1}d_j.$. His bound is better than 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. Let $R = k[t_0, ..., t_n].$ The homogeneous ideal $R_+ = \bigoplus_{d \ge 1} R_d$ is called the maximal homogeneous ideal (see also irrelevant ideal). As in the affine case, we let: for a subset $S \subseteq \mathbb{P}^n$ and a homogeneous ideal I of R,

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

By $f = 0 \text{ on } S$ we mean: for every homogeneous coordinates $(a_0 : \cdots : a_n)$ of a point of S we have $f(a_0,\ldots, a_n)=0$. This implies that the homogeneous components of f are also zero on S and thus that $\operatorname{I}_{\mathbb{P}^n}(S)$ is a homogeneous ideal. Equivalently, $\operatorname{I}_{\mathbb{P}^n}(S)$ is the homogeneous ideal generated by homogeneous polynomials f that vanish on S. Now, for any homogeneous ideal $I \subseteq R_+$, by the usual Nullstellensatz, we have:

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

and so, like in the affine case, we have:

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

Applications

Commuting matrices

The fact that commuting matrices have a common eigenvector – and hence by induction stabilize a common flag and are simultaneously triangularizable – can be interpreted as a result of the weak Nullstellensatz, as follows: commuting matrices form a commutative algebra

$K[A_1,\ldots,A_k]$ over $K[x_1,\ldots,x_k];$

the matrices satisfy various polynomials such as their minimal polynomials, which form a proper ideal (because they are not all zero, in which case the result is trivial); one might call this the characteristic ideal, by analogy with the characteristic polynomial.

One then defines an eigenvector for a commutative algebra as a vector v such that for all $x \in A$ one has $x(v) = \lambda(x)\cdot v$ for a linear functional

$\lambda\colon A \to K.$

This simply linearizes the definition of an eigenvalue, and is the correct definition for a common eigenvector, as if v is a common eigenvector, meaning $A_i(v)=\lambda_i v,$ then the functional is defined as

$\textstyle{\lambda(c_0 I + c_1 A_1 + \cdots c_k A_k) := c_0 + \sum c_i \lambda_i}$

(treating scalars as multiples of the identity matrix $A_0 := I$, which has eigenvalue 1 for all vectors), and conversely an eigenvector for such a functional $\lambda$ is a common eigenvector. Geometrically, the eigenvalue corresponds to the point in affine k-space with coordinates $(\lambda_1,\ldots,\lambda_k)$ with respect to the basis given by $A_i.$

Then the existence of an eigenvalue $\lambda$ is equivalent to the ideal generated by (the relations satisfied by) $A_i$ being non-empty, which exactly generalizes the usual proof of existence of an eigenvalue existing for a single matrix over an algebraically closed field by showing that the characteristic polynomial has a zero.

See also

• Stengle's Positivstellensatz
• Differential Nullstellensatz
• Combinatorial Nullstellensatz

References

• M. Atiyah, I.G. Macdonald, Introduction to Commutative Algebra, Addison–Wesley, 1994. ISBN 0-201-40751-5
• Shigeru Mukai; William Oxbury (translator) (2003). An Introduction to Invariants and Moduli. Cambridge studies in advanced mathematics 81. p. 82. ISBN 0-521-80906-1. 
• David Eisenbud, Commutative Algebra With a View Toward Algebraic Geometry, New York : Springer-Verlag, 1999.
• Hartshorne, Robin (1977), Algebraic Geometry, Graduate Texts in Mathematics 52, New York: Springer-Verlag, ISBN 978-0-387-90244-9, MR0463157