Hilbert's Nullstellensatz

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Hilbert's Nullstellensatz (German for "theorem of zeros," or more literally, "zero-locus-theorem"—see ''
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'') 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 ideal (ring theory), 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 $k\left[X_1, \ldots, X_n\right]$ 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''1,...,''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 $k\left[X_1, \ldots, X_n\right]$ 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''. An immediate corollary is the ''weak Nullstellensatz'': The ideal $I \subset k\left[X_1, \ldots, X_n\right]$ 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 $k\left[X_1, \ldots, X_n\right],$ 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''2 + 1) in $\R\left[X\right]$ do not have a common zero in $\R.$ With the notation common in algebraic geometry, the Nullstellensatz can also be formulated as :$\hbox\left(\hbox\left(J\right)\right)=\sqrt$ for every ideal ''J''. Here, $\sqrt$ denotes the radical of an ideal, radical of ''J'' and I(''U'') is the ideal of all polynomials that vanish on the set ''U''. In this way, we obtain an order-reversing bijective correspondence between the algebraic sets in ''K''''n'' and the radical ideals of $K\left[X_1, \ldots, X_n\right].$ 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 $P = \left(a_1, \dots, a_n\right) \in K^n$. Then $I\left(P\right) = \left(X_1 - a_1, \ldots, X_n - a_n\right)$. More generally, :$\sqrt = \bigcap_ \left(X_1 - a_1, \dots, X_n - a_n\right).$ Conversely, every maximal ideal of the polynomial ring $K\left[X_1,\ldots,X_n\right]$ (note that $K$ is algebraically closed) is of the form $\left(X_1 - a_1, \ldots, X_n - a_n\right)$ for some $a_1,\ldots,a_n \in K$. As another example, an algebraic subset ''W'' in ''K''''n'' is irreducible (in the Zariski topology) if and only if $I\left(W\right)$ is 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 algebra, 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. Let $A = k\left[t_1, \ldots, t_n\right]$ (''k'' algebraically closed field), ''I'' an ideal of ''A,'' and ''V'' the common zeros of ''I'' in $k^n$. Clearly, $\sqrt \subseteq I\left(V\right)$. Let $f \not\in \sqrt$. Then $f \not\in \mathfrak$ for some prime ideal $\mathfrak\supseteq I$ in ''A''. Let $R = \left(A/\mathfrak\right) \left[f^\right]$ and $\mathfrak$ a maximal ideal in $R$. By Zariski's lemma, $R/\mathfrak$ 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 $x = \left(x_1, \ldots, x_n\right) \in V$ and $f\left(x\right) \ne 0$. The Nullstellensatz also follows trivially from a systematic development of Jacobson rings, in which a radical ideal is an intersection of maximal ideals. Let $R$ be a Jacobson ring. If $S$ is a finitely generated associative algebra, ''R''-algebra, then $S$ is a Jacobson ring. Further, if $\mathfrak\subset S$ is a maximal ideal, then $\mathfrak := \mathfrak \cap R$ is a maximal ideal of R, and $S/\mathfrak$ is a finite extension field of $R/\mathfrak$. Another generalization states that a faithfully flat morphism of schemes $f: Y \to X$ locally of finite type with ''X'' quasi-compact has a ''quasi-section'', i.e. there exists $X\text{'}$ affine and faithfully flat and quasi-finite over ''X'' together with an ''X''-morphism $X\text{'} \to Y$

# Effective Nullstellensatz

In all of its variants, Hilbert's Nullstellensatz asserts that some polynomial belongs or not to an ideal generated, say, by ; we have in the strong version, in the weak form. This means the existence or the non-existence of polynomials such that . The usual proofs of the Nullstellensatz are not constructive, non-effective, in the sense that they do not give any way to compute the . It is thus a rather natural question to ask if there is an effective way to compute the (and the exponent 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 : 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 . 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 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. 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: :Let be polynomials in variables, of total degree . If there exist polynomials such that , then they can be chosen such that ::$\deg\left(f_ig_i\right) \le \max\left(d_s,3\right)\prod_^\max\left(d_j,3\right).$ :This bound is optimal if all the degrees are greater than 2. If is the maximum of the degrees of the , this bound may be simplified to :$\max\left(3,d\right)^.$ Kollár's result has been improved by several authors. , the best improvement, due to M. Sombra is :$\deg\left(f_ig_i\right) \le 2d_s\prod_^d_j.$ 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. Let $R = k\left[t_0, \ldots, t_n\right].$ The homogeneous ideal, :$R_+ = \bigoplus_ R_d$ is called the ''maximal homogeneous ideal'' (see also irrelevant ideal). As in the affine case, we let: for a subset $S \subseteq \mathbb^n$ and a homogeneous ideal ''I'' of ''R'', :$\begin \operatorname_\left(S\right) &= \, \\ \operatorname_\left(I\right) &= \. \end$ By $f = 0 \text S$ we mean: for every homogeneous coordinates $\left(a_0 : \cdots : a_n\right)$ of a point of ''S'' we have $f\left(a_0,\ldots, a_n\right)=0$. This implies that the homogeneous components of ''f'' are also zero on ''S'' and thus that $\operatorname_\left(S\right)$ is a homogeneous ideal. Equivalently, $\operatorname_\left(S\right)$ 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: :$\sqrt = \operatorname_\left(\operatorname_\left(I\right)\right),$ 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 $\mathbb^n$ of the form $\operatorname_\left(I\right).$ The correspondence is given by $\operatorname_$ and $\operatorname_.$

# Analytic Nullstellensatz (Rückert’s Nullstellensatz)

The Nullstellensatz also holds for the germs of holomorphic functions at a point of complex ''n''-space $\Complex^n.$ Precisely, for each open subset $U \subset \Complex^n,$ let $\mathcal_\left(U\right)$ denote the ring of holomorphic functions on ''U''; then $\mathcal_$ is a ''sheaf (mathematics), sheaf'' on $\Complex^n.$ The stalk $\mathcal_$ at, say, the origin can be shown to be a Noetherian ring, Noetherian local ring that is a unique factorization domain. If $f \in \mathcal_$ is a germ represented by a holomorphic function $\widetilde: U \to \Complex$, then let $V_0\left(f\right)$ be the equivalence class of the set :$\left \,$ where two subsets $X, Y \subset \Complex^n$ are considered equivalent if $X \cap U = Y \cap U$ for some neighborhood ''U'' of 0. Note $V_0\left(f\right)$ is independent of a choice of the representative $\widetilde.$ For each ideal $I \subset \mathcal_,$ let $V_0\left(I\right)$ denote $V_0\left(f_1\right) \cap \dots \cap V_0\left(f_r\right)$ for some generators $f_1, \ldots, f_r$ of ''I''. It is well-defined; i.e., is independent of a choice of the generators. For each subset $X \subset \Complex ^n$, let :$I_0\left(X\right) = \left \.$ It is easy to see that $I_0\left(X\right)$ is an ideal of $\mathcal_$ and that $I_0\left(X\right) = I_0\left(Y\right)$ if $X \sim Y$ in the sense discussed above. The analytic Nullstellensatz then states: for each ideal $I \subset \mathcal_$, :$\sqrt = I_0\left(V_0\left(I\right)\right)$ where the left-hand side is the radical of an ideal, radical of ''I''.