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Nullstelle means zero locus. Hilbert’s Nullstellensatz (theorem about zero loci) characterizes the joint zero loci of an ideal of functions in a polynomial ring. This is a foundational result in algebraic geometry at the heart of the duality by which rings are dually interpreted as spaces (varieties).
In classical algebraic geometry over an algebraically closed field $k$, the Nullstellensatz concerns the fixed points of a standard Galois connection between ideals $I$ of the polynomial ring $k[x_1, \ldots, x_n]$ and subsets $V \subseteq k^n$. The Galois connection is induced by a relation $I \perp V$ iff $f(x) = 0$ for all $f \in I, x \in V$. Accordingly, letting $Idl(k[x_1, \ldots, x_n])$ be the set of ideals ordered by inclusion and $P(k^n)$ the set of subsets of $k^n$ ordered by inclusion, there are contravariant maps $Ideal: P(k^n) \to Idl(k[x_1, \ldots, x_n])$ and $Var: Idl(k[x_1, \ldots, x_n]) \to P(k^n)$ defined by
which defines the Galois connection.
The fixed points of $Var \circ Ideal: P(k^n) \to P(k^n)$ are, by definition, the closed sets of the Zariski topology on the affine space $k^n$ (“affine space” here in the sense of algebraic geometry). These are the closed (not necessarily reduced) subvarieties? of $k^n$.
On the other side, the fixed points of $Ideal \circ Var$ is what the classical Nullstellensatz is about:
(Strong Nullstellensatz) The fixed points of $Ideal \circ Var$ are precisely the radical ideals $I$ of $k[x_1, \ldots, x_n]$, i.e., ideals $I$ such that for any $r \in \mathbb{N}$, the condition $f^r \in I$ implies $f \in I$.
Thus there is a Galois correspondence between closed subvarieties and radical ideals.
Typically the way that the strong Nullstellensatz is proved is by reduction to the so-called “weak Nullstellensatz” by means of the “Rabinowitsch trick”. (The “weak” here may be misleading, as the weak Nullstellensatz may be considered the core result and the strong Nullstellensatz a corollary.) We now turn to these.
One weak version of the Nullstellensatz says the following (e.g. Theorem 3.99.1 here: pdf)
For $k$ an algebraically closed field and $I$ a proper ideal in the polynomial ring $k[X_1, \cdots, X_n]$, the set $V(I)$ (of $n$-tuples $x = (x_i) \in k^n$ such that all polynomials in $I$ vanish when evaluated on these $x$) is an inhabited set.
Before giving a proof, we remark (in view of more abstract formulations to come later) that an element of $V(I)$ is just a $k$-algebra homomorphism of the form
Dually this is a morphism of affine schemes (ring spectra) of the form
Moreover, since $Spec(k)$ is the terminal object in this context, such a map is the same as a “point”, a global element of $Spec(k[X_1, \cdots, X_n]/I)$.
Hence in this form the Nullstellensatz simply says that (for $k$ algebraically closed) affine schemes have points. This formulation of the Nullstellensatz leads one to a more general abstract formulation.
We turn now to the proof of the weak Nullstellensatz.
If $k$ is an uncountable algebraically closed field, then for any proper ideal $I \subset k[X_1, \ldots, X_n]$ there is $(a_1, \ldots, a_n) \in k^n$ such that $f(a_1, \ldots, a_n) = 0$ for all $f \in I$.
By the axiom of choice, $I$ is contained in a maximal ideal $M$. So it suffices to prove it in the case where $I = M$ is maximal, where we prove there is an isomorphism $\phi: k[X_1, \ldots, X_n]/M \cong k$. Then the composite
sends each $X_i$ to a value $a_i$, so we have an inclusion of maximal ideals
which must be an equality. In that case $(a_1, \ldots, a_n)$ is the desired point.
Since $k$ is algebraically closed, it suffices to prove that the extension field $F = k[X_1, \ldots, X_n]/M$ over $k$ is algebraic over $k$. Clearly $F$ as a vector space over $k$ has countable dimension. Supposing to the contrary there is an element $t \in F$ that is transcendental over $k$, we have a subquotient vector space of $F$,
where the isomorphism is by partial fraction decomposition (ultimately by the Chinese remainder theorem?), but the vector space on the right clearly has uncountable dimension because there are uncountably many $a \in k$, and this leads to a contradiction.
Now we reduce to the case where $k$ is uncountable by employing a model-theoretic argument:
(Weak Nullstellensatz) If $k$ is an algebraically closed field, then for any proper ideal $I \subset k[X_1, \ldots, X_n]$ there is $(a_1, \ldots, a_n) \in k^n$ such that $f(a_1, \ldots, a_n) = 0$ for all $f \in I$.
Let $U$ be a non-principal ultrafilter on $\mathbb{N}$, and let $K$ be the ultrapower $k^\mathbb{N}/U$. By Łoś's ultraproduct theorem?, $K$ is an algebraically closed field. Also $K$ is uncountable (Lemma below).
First we claim $K \otimes_k I$ is a proper ideal of $K \otimes_k k[X_1, \ldots, X_n] \cong K[X_1,\ldots, X_n]$. For suppose we have $1 = \sum_{i = 1}^s f_i p_i$ for $p_1, \ldots, p_s \in I$ and $f_i \in K[X_1, \ldots, X_n]$. Writing the $f_i$ as $K$-linear combinations of monomials $X^\alpha$, there are finitely many $q_1, \ldots, q_t \in I$, each of the form $X^\alpha p_i$, so as to enable us to write $1 = \sum_{i=1}^t a_i q_i$ for $a_i = [(a_{i, m})] \in K$ and $q_i \in I$. By Łoś‘s theorem,
this set is nonempty, and so there is at least one $m$ rendering the equation true. This contradicts $1 \notin I$.
By Lemma , there exists $(a_1, \ldots, a_n) \in K^n$ belonging to $Var(K \otimes_k I)$. By Hilbert's basis theorem, $I \subset k[X_1, \ldots, X_n]$ is generated by finitely many polynomials $p_1, \ldots, p_s$, so $Var(K \otimes_k I) \subseteq K^n$ is defined by finitely many formulas $p_1(X) = 0, \ldots, p_s(X) = 0$ and thus by Łoś‘s theorem again,
and so this set is nonempty and therefore there is a point $(a_{1, m}, \ldots, a_{n, m}) \in k^n$ that belongs to $Var(I)$.
To prove $K$ uncountable, we use the fact that an algebraically closed field is at least infinite. Then the missing piece is supplied by the following result.
If $X$ is an infinite set and $U$ is a non-principal ultrafilter on $\mathbb{N}$, then the ultrapower $X^\mathbb{N}/U$ has cardinality ${|X|}^{\aleph_0}$ (so at least $2^{\aleph_0}$).
For $j \in \mathbb{N}$, put $[j] = \{i \in \mathbb{N}: i \lt j\}$. If $X$ is infinite, then there is a mono $i_j: X^{[j]} \to X$ for each $j \geq 0$. For any $h: \mathbb{N} \to X$, let $h|_{[j]}$ denote the composite $[j] \hookrightarrow \mathbb{N} \stackrel{h}{\to} X$, and define $h': \mathbb{N} \to X$ by $h'(j) = i_j(h|_{[j]})$. Let $[h']_U$ denote the value of $h'$ under the quotient map $X^\mathbb{N} \to X^\mathbb{N}/U$.
Claim: The map $X^\mathbb{N} \to X^\mathbb{N}/U$ that takes $h$ to $[h']_U$ is injective. For, if $g, h: \mathbb{N} \to X$ are distinct, then they differ at some $i \in \mathbb{N}$. Then $g|_{[j]} \neq h|_{[j]}$ whenever $i \lt j$, which in turn implies $i_j(g|_{[j]}) \neq i_j(h|_{[j]})$ i.e. $g'(j) \neq h'(j)$ for all $j$ such that $j \gt i$. (That is, “for almost all $j$”, belonging to a cofinite set that belongs to $U$.) Thus $[g']_U \neq [h']_U$, which establishes the claim. The lemma follows immediately from the claim.
Rabinowitsch, who later renamed himself Rainich (see the story here), showed how to deduce the Strong Nullstellensatz from the Weak Nullstellensatz. Often this is described as Rabinowitsch’s “trick” (i.e., something involving a stroke of cleverness), but see the remark below which places the trick conceptually within the general method of localization.
Let $k$ be algebraically closed and let $I \subseteq k[x_1, \ldots, x_n]$ be an ideal. To show that $Ideal(Var(I)) = \sqrt{I}$, suppose that $f \in k[x_1, \ldots, x_n]$ vanishes at any point where all of a given set of polynomials $f_1, \ldots, f_s \in k[x_1, \ldots, x_n]$ vanish. It suffices to show that for some $r$ we have that $f^r$ belongs to the ideal generated by the $f_i$.
Introduce a fresh variable $x_0$, and observe that there is no point $(a_0, a_1, \ldots, a_n) \in k^{n+1}$ where the polynomials $1 - x_0 f, f_1, \ldots, f_s$ simultaneously vanish. By the Weak Nullstellensatz (Theorem ), these polynomials must generate the unit ideal, say
where $x = (x_0, x_1, \ldots, x_n)$. There is a $k$-algebra map $k[x_0, x_1, \ldots, x_n] \to k(x_1, \ldots, x_n)$ sending $x_0 \mapsto 1/f$ and $x_i \mapsto x_i$ for $1 \leq i \leq n$, where the above identity yields
which, if $r$ is the highest degree of $x_0$ appearing in a $p_i$, may be rewritten in the form $f^r = \sum_{i=1}^s g_i f_i$ in $k[x_1, \ldots, x_n]$ by clearing denominators.
Proving that $f \in \sqrt{I}$ means we prove that $f$ is nilpotent in $k[x_1, \ldots, x_n]/I$, in other words that the localization of $k[x_1, \ldots, x_n]/I$ with respect to the (image of the) multiplicative system $\{1, f, \ldots, f^n, \ldots\}$ vanishes. Clearly we localize by inverting (the image of) $f$, by constructing
Clearly $Var(I, 1 - x_0 f) = \emptyset$, and so by the weak Nullstellensatz the ideal $(I, 1 - x_0 f)$ is improper, i.e., the localization vanishes.
By the discussion above at Existence of points it follows that phrased in terms of the sheaf topos over the Zariski site, the weak Nullstellensatz says that “objects have global points”.
Motivated by this, in (Lawvere 07, def. 2) is a proposal for a general abstract formulation of what constitutes a Nullstellensatz, formalized in the context of cohesion.
A cohesive topos $\mathbf{H}$ over some base topos $\mathbf{S}$ is a relative topos such that the terminal geometric morphism extended to an adjoint quadruple $(\Pi \dashv Disc \dashv \Gamma \dashv coDisc)$ with $Disc, coDisc \colon \mathbf{S}\hookrightarrow \mathbf{H}$ full and faithful and $\Pi$ preserving finite products. The top adjoint triple here induces a canonical natural transformation
which deserves to be called the points-to-pieces transformation.
Consider the condition that $ptp$ is on every object $X$ of $\mathbf{H}$ an epimorphism. In terms of the geometric interpretation of cohesion this means that “all pieces of $X$ have at least one point”. (See at cohesive topos – Pieces have points) This statement is what in Lawvere 07, def. 2 c) is also referred to as Nullstellensatz. More comments on this are in (Lawvere 11). (In (Johnstone 11) this condition is called “punctual local connectedness”.)
For a detailed analysis of how this general abstract concept indeed captures the traditional meaning of Nullstellensatz, see (Lawvere 15, Menni), also (Tholen).
Wikipedia, Hilbert’s Nullstellensatz
David Hilbert. Ueber die vollen Invariantensysteme. Mathematische Annalen 42(3), pp. 313–373. 1893.
M. Bunge, On the transfer of an abstract Nullstellensatz , Comm. Alg. 10 (1982) pp.1891-1906.
W. A. MacCaull, Hilbert’s Nullstellensatz revisited , JPAA 54 (1988) pp.289-297.
V. Weispfenning, Nullstellensätze - a model theoretic framework , Z. Math. Logik Grundl. Math. 23 (1977) pp. 539-545.
Discussion of the Nullstellensatz in terms of cohesion:
William Lawvere, Axiomatic cohesion Theory and Applications of Categories, Vol. 19, No. 3 (2007) pp. 41–49. (pdf)
William Lawvere, MO comment on the general abstract Nullstellensatz (2011)
F. William Lawvere, Birkhoff’s Theorem from a geometric
perspective: A simple example_ , Categories and General Algebraic Structures with Applications, Vol. 4, No. 1 (2015) pp. 1–7. (pdf)
Peter Johnstone, Remarks on punctual local connectedness, Theory and Applications of Categories, Vol. 25, 2011, No. 3, pp 51-63 (TAC)
Marie La Palme Reyes, G. E. Reyes, H. Zolfaghari, Generic Figures and their Glueings , Polimetrica Milano 2004. (pp.206-224)
Matias Menni, The manifestation of Hilbert’s Nullstellensatz in Lawvere’s Axiomatic Cohesion (pdf slides pdf abstract)
Matías Menni, Sufficient Cohesion over Atomic Toposes , Cah.Top.Géom.Diff.Cat. LV (2014). (preprint)
Matías Menni, Continuous Cohesion over Sets , TAC 29 no.20 (2014) pp.542-568. (pdf)
Walter Tholen, Nullstellen and Subdirect Representation (pdf)
For a version of the Nullstellensatz in spectral algebraic geometry see
Last revised on April 19, 2023 at 15:53:01. See the history of this page for a list of all contributions to it.