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In 1946, Ulam, in response to a theorem of Anning and Erdös, posed the following problem:

Problem 1 (Erdös-Ulam problem) Let ${S \subset {\bf R}^2}$ be a set such that the distance between any two points in ${S}$ is rational. Is it true that ${S}$ cannot be (topologically) dense in ${{\bf R}^2}$?

The paper of Anning and Erdös addressed the case that all the distances between two points in ${S}$ were integer rather than rational in the affirmative.

The Erdös-Ulam problem remains open; it was discussed recently over at Gödel’s lost letter. It is in fact likely (as we shall see below) that the set ${S}$ in the above problem is not only forbidden to be topologically dense, but also cannot be Zariski dense either. If so, then the structure of ${S}$ is quite restricted; it was shown by Solymosi and de Zeeuw that if ${S}$ fails to be Zariski dense, then all but finitely many of the points of ${S}$ must lie on a single line, or a single circle. (Conversely, it is easy to construct examples of dense subsets of a line or circle in which all distances are rational, though in the latter case the square of the radius of the circle must also be rational.)

The main tool of the Solymosi-de Zeeuw analysis was Faltings’ celebrated theorem that every algebraic curve of genus at least two contains only finitely many rational points. The purpose of this post is to observe that an affirmative answer to the full Erdös-Ulam problem similarly follows from the conjectured analogue of Falting’s theorem for surfaces, namely the following conjecture of Bombieri and Lang:

Conjecture 2 (Bombieri-Lang conjecture) Let ${X}$ be a smooth projective irreducible algebraic surface defined over the rationals ${{\bf Q}}$ which is of general type. Then the set ${X({\bf Q})}$ of rational points of ${X}$ is not Zariski dense in ${X}$.

In fact, the Bombieri-Lang conjecture has been made for varieties of arbitrary dimension, and for more general number fields than the rationals, but the above special case of the conjecture is the only one needed for this application. We will review what “general type” means (for smooth projective complex varieties, at least) below the fold.

The Bombieri-Lang conjecture is considered to be extremely difficult, in particular being substantially harder than Faltings’ theorem, which is itself a highly non-trivial result. So this implication should not be viewed as a practical route to resolving the Erdös-Ulam problem unconditionally; rather, it is a demonstration of the power of the Bombieri-Lang conjecture. Still, it was an instructive algebraic geometry exercise for me to carry out the details of this implication, which quickly boils down to verifying that a certain quite explicit algebraic surface is of general type (Theorem 4 below). As I am not an expert in the subject, my computations here will be rather tedious and pedestrian; it is likely that they could be made much slicker by exploiting more of the machinery of modern algebraic geometry, and I would welcome any such streamlining by actual experts in this area. (For similar reasons, there may be more typos and errors than usual in this post; corrections are welcome as always.) My calculations here are based on a similar calculation of van Luijk, who used analogous arguments to show (assuming Bombieri-Lang) that the set of perfect cuboids is not Zariski-dense in its projective parameter space.

We also remark that in a recent paper of Makhul and Shaffaf, the Bombieri-Lang conjecture (or more precisely, a weaker consequence of that conjecture) was used to show that if ${S}$ is a subset of ${{\bf R}^2}$ with rational distances which intersects any line in only finitely many points, then there is a uniform bound on the cardinality of the intersection of ${S}$ with any line. I have also recently learned (private communication) that an unpublished work of Shaffaf has obtained a result similar to the one in this post, namely that the Erdös-Ulam conjecture follows from the Bombieri-Lang conjecture, plus an additional conjecture about the rational curves in a specific surface.

Let us now give the elementary reductions to the claim that a certain variety is of general type. For sake of contradiction, let ${S}$ be a dense set such that the distance between any two points is rational. Then ${S}$ certainly contains two points that are a rational distance apart. By applying a translation, rotation, and a (rational) dilation, we may assume that these two points are ${(0,0)}$ and ${(1,0)}$. As ${S}$ is dense, there is a third point of ${S}$ not on the ${x}$ axis, which after a reflection we can place in the upper half-plane; we will write it as ${(a,\sqrt{b})}$ with ${b>0}$.

Given any two points ${P, Q}$ in ${S}$, the quantities ${|P|^2, |Q|^2, |P-Q|^2}$ are rational, and so by the cosine rule the dot product ${P \cdot Q}$ is rational as well. Since ${(1,0) \in S}$, this implies that the ${x}$-component of every point ${P}$ in ${S}$ is rational; this in turn implies that the product of the ${y}$-coordinates of any two points ${P,Q}$ in ${S}$ is rational as well (since this differs from ${P \cdot Q}$ by a rational number). In particular, ${a}$ and ${b}$ are rational, and all of the points in ${S}$ now lie in the lattice ${\{ ( x, y\sqrt{b}): x, y \in {\bf Q} \}}$. (This fact appears to have first been observed in the 1988 habilitationschrift of Kemnitz.)

Now take four points ${(x_j,y_j \sqrt{b})}$, ${j=1,\dots,4}$ in ${S}$ in general position (so that the octuplet ${(x_1,y_1\sqrt{b},\dots,x_4,y_4\sqrt{b})}$ avoids any pre-specified hypersurface in ${{\bf C}^8}$); this can be done if ${S}$ is dense. (If one wished, one could re-use the three previous points ${(0,0), (1,0), (a,\sqrt{b})}$ to be three of these four points, although this ultimately makes little difference to the analysis.) If ${(x,y\sqrt{b})}$ is any point in ${S}$, then the distances ${r_j}$ from ${(x,y\sqrt{b})}$ to ${(x_j,y_j\sqrt{b})}$ are rationals that obey the equations

$\displaystyle (x - x_j)^2 + b (y-y_j)^2 = r_j^2$

for ${j=1,\dots,4}$, and thus determine a rational point in the affine complex variety ${V = V_{b,x_1,y_1,x_2,y_2,x_3,y_3,x_4,y_4} \subset {\bf C}^5}$ defined as

$\displaystyle V := \{ (x,y,r_1,r_2,r_3,r_4) \in {\bf C}^6:$

$\displaystyle (x - x_j)^2 + b (y-y_j)^2 = r_j^2 \hbox{ for } j=1,\dots,4 \}.$

By inspecting the projection ${(x,y,r_1,r_2,r_3,r_4) \rightarrow (x,y)}$ from ${V}$ to ${{\bf C}^2}$, we see that ${V}$ is a branched cover of ${{\bf C}^2}$, with the generic cover having ${2^4=16}$ points (coming from the different ways to form the square roots ${r_1,r_2,r_3,r_4}$); in particular, ${V}$ is a complex affine algebraic surface, defined over the rationals. By inspecting the monodromy around the four singular base points ${(x,y) = (x_i,y_i)}$ (which switch the sign of one of the roots ${r_i}$, while keeping the other three roots unchanged), we see that the variety ${V}$ is connected away from its singular set, and thus irreducible. As ${S}$ is topologically dense in ${{\bf R}^2}$, it is Zariski-dense in ${{\bf C}^2}$, and so ${S}$ generates a Zariski-dense set of rational points in ${V}$. To solve the Erdös-Ulam problem, it thus suffices to show that

Claim 3 For any non-zero rational ${b}$ and for rationals ${x_1,y_1,x_2,y_2,x_3,y_3,x_4,y_4}$ in general position, the rational points of the affine surface ${V = V_{b,x_1,y_1,x_2,y_2,x_3,y_3,x_4,y_4}}$ is not Zariski dense in ${V}$.

This is already very close to a claim that can be directly resolved by the Bombieri-Lang conjecture, but ${V}$ is affine rather than projective, and also contains some singularities. The first issue is easy to deal with, by working with the projectivisation

$\displaystyle \overline{V} := \{ [X,Y,Z,R_1,R_2,R_3,R_4] \in {\bf CP}^6: Q(X,Y,Z,R_1,R_2,R_3,R_4) = 0 \} \ \ \ \ \ (1)$

of ${V}$, where ${Q: {\bf C}^7 \rightarrow {\bf C}^4}$ is the homogeneous quadratic polynomial

$\displaystyle (X,Y,Z,R_1,R_2,R_3,R_4) := (Q_j(X,Y,Z,R_1,R_2,R_3,R_4) )_{j=1}^4$

with

$\displaystyle Q_j(X,Y,Z,R_1,R_2,R_3,R_4) := (X-x_j Z)^2 + b (Y-y_jZ)^2 - R_j^2$

and the projective complex space ${{\bf CP}^6}$ is the space of all equivalence classes ${[X,Y,Z,R_1,R_2,R_3,R_4]}$ of tuples ${(X,Y,Z,R_1,R_2,R_3,R_4) \in {\bf C}^7 \backslash \{0\}}$ up to projective equivalence ${(\lambda X, \lambda Y, \lambda Z, \lambda R_1, \lambda R_2, \lambda R_3, \lambda R_4) \sim (X,Y,Z,R_1,R_2,R_3,R_4)}$. By identifying the affine point ${(x,y,r_1,r_2,r_3,r_4)}$ with the projective point ${(X,Y,1,R_1,R_2,R_3,R_4)}$, we see that ${\overline{V}}$ consists of the affine variety ${V}$ together with the set ${\{ [X,Y,0,R_1,R_2,R_3,R_4]: X^2+bY^2=R^2; R_j = \pm R_1 \hbox{ for } j=2,3,4\}}$, which is the union of eight curves, each of which lies in the closure of ${V}$. Thus ${\overline{V}}$ is the projective closure of ${V}$, and is thus a complex irreducible projective surface, defined over the rationals. As ${\overline{V}}$ is cut out by four quadric equations in ${{\bf CP}^6}$ and has degree sixteen (as can be seen for instance by inspecting the intersection of ${\overline{V}}$ with a generic perturbation of a fibre over the generically defined projection ${[X,Y,Z,R_1,R_2,R_3,R_4] \mapsto [X,Y,Z]}$), it is also a complete intersection. To show (3), it then suffices to show that the rational points in ${\overline{V}}$ are not Zariski dense in ${\overline{V}}$.

Heuristically, the reason why we expect few rational points in ${\overline{V}}$ is as follows. First observe from the projective nature of (1) that every rational point is equivalent to an integer point. But for a septuple ${(X,Y,Z,R_1,R_2,R_3,R_4)}$ of integers of size ${O(N)}$, the quantity ${Q(X,Y,Z,R_1,R_2,R_3,R_4)}$ is an integer point of ${{\bf Z}^4}$ of size ${O(N^2)}$, and so should only vanish about ${O(N^{-8})}$ of the time. Hence the number of integer points ${(X,Y,Z,R_1,R_2,R_3,R_4) \in {\bf Z}^7}$ of height comparable to ${N}$ should be about

$\displaystyle O(N)^7 \times O(N^{-8}) = O(N^{-1});$

this is a convergent sum if ${N}$ ranges over (say) powers of two, and so from standard probabilistic heuristics (see this previous post) we in fact expect only finitely many solutions, in the absence of any special algebraic structure (e.g. the structure of an abelian variety, or a birational reduction to a simpler variety) that could produce an unusually large number of solutions.

The Bombieri-Lang conjecture, Conjecture 2, can be viewed as a formalisation of the above heuristics (roughly speaking, it is one of the most optimistic natural conjectures one could make that is compatible with these heuristics while also being invariant under birational equivalence).

Unfortunately, ${\overline{V}}$ contains some singular points. Being a complete intersection, this occurs when the Jacobian matrix of the map ${Q: {\bf C}^7 \rightarrow {\bf C}^4}$ has less than full rank, or equivalently that the gradient vectors

$\displaystyle \nabla Q_j = (2(X-x_j Z), 2(Y-y_j Z), -2x_j (X-x_j Z) - 2y_j (Y-y_j Z), \ \ \ \ \ (2)$

$\displaystyle 0, \dots, 0, -2R_j, 0, \dots, 0)$

for ${j=1,\dots,4}$ are linearly dependent, where the ${-2R_j}$ is in the coordinate position associated to ${R_j}$. One way in which this can occur is if one of the gradient vectors ${\nabla Q_j}$ vanish identically. This occurs at precisely ${4 \times 2^3 = 32}$ points, when ${[X,Y,Z]}$ is equal to ${[x_j,y_j,1]}$ for some ${j=1,\dots,4}$, and one has ${R_k = \pm ( (x_j - x_k)^2 + b (y_j - y_k)^2 )^{1/2}}$ for all ${k=1,\dots,4}$ (so in particular ${R_j=0}$). Let us refer to these as the obvious singularities; they arise from the geometrically evident fact that the distance function ${(x,y\sqrt{b}) \mapsto \sqrt{(x-x_j)^2 + b(y-y_j)^2}}$ is singular at ${(x_j,y_j\sqrt{b})}$.

The other way in which could occur is if a non-trivial linear combination of at least two of the gradient vectors vanishes. From (2), this can only occur if ${R_j=R_k=0}$ for some distinct ${j,k}$, which from (1) implies that

$\displaystyle (X - x_j Z) = \pm \sqrt{b} i (Y - y_j Z) \ \ \ \ \ (3)$

and

$\displaystyle (X - x_k Z) = \pm \sqrt{b} i (Y - y_k Z) \ \ \ \ \ (4)$

for two choices of sign ${\pm}$. If the signs are equal, then (as ${x_j, y_j, x_k, y_k}$ are in general position) this implies that ${Z=0}$, and then we have the singular point

$\displaystyle [X,Y,Z,R_1,R_2,R_3,R_4] = [\pm \sqrt{b} i, 1, 0, 0, 0, 0, 0]. \ \ \ \ \ (5)$

If the non-trivial linear combination involved three or more gradient vectors, then by the pigeonhole principle at least two of the signs involved must be equal, and so the only singular points are (5). So the only remaining possibility is when we have two gradient vectors ${\nabla Q_j, \nabla Q_k}$ that are parallel but non-zero, with the signs in (3), (4) opposing. But then (as ${x_j,y_j,x_k,y_k}$ are in general position) the vectors ${(X-x_j Z, Y-y_j Z), (X-x_k Z, Y-y_k Z)}$ are non-zero and non-parallel to each other, a contradiction. Thus, outside of the ${32}$ obvious singular points mentioned earlier, the only other singular points are the two points (5).

We will shortly show that the ${32}$ obvious singularities are ordinary double points; the surface ${\overline{V}}$ near any of these points is analytically equivalent to an ordinary cone ${\{ (x,y,z) \in {\bf C}^3: z^2 = x^2 + y^2 \}}$ near the origin, which is a cone over a smooth conic curve ${\{ (x,y) \in {\bf C}^2: x^2+y^2=1\}}$. The two non-obvious singularities (5) are slightly more complicated than ordinary double points, they are elliptic singularities, which approximately resemble a cone over an elliptic curve. (As far as I can tell, this resemblance is exact in the category of real smooth manifolds, but not in the category of algebraic varieties.) If one blows up each of the point singularities of ${\overline{V}}$ separately, no further singularities are created, and one obtains a smooth projective surface ${X}$ (using the Segre embedding as necessary to embed ${X}$ back into projective space, rather than in a product of projective spaces). Away from the singularities, the rational points of ${\overline{V}}$ lift up to rational points of ${X}$. Assuming the Bombieri-Lang conjecture, we thus are able to answer the Erdös-Ulam problem in the affirmative once we establish

Theorem 4 The blowup ${X}$ of ${\overline{V}}$ is of general type.

This will be done below the fold, by the pedestrian device of explicitly constructing global differential forms on ${X}$; I will also be working from a complex analysis viewpoint rather than an algebraic geometry viewpoint as I am more comfortable with the former approach. (As mentioned above, though, there may well be a quicker way to establish this result by using more sophisticated machinery.)

I thank Mark Green and David Gieseker for helpful conversations (and a crash course in varieties of general type!).

Remark 5 The above argument shows in fact (assuming Bombieri-Lang) that sets ${S \subset {\bf R}^2}$ with all distances rational cannot be Zariski-dense, and thus (by Solymosi-de Zeeuw) must lie on a single line or circle with only finitely many exceptions. Assuming a stronger version of Bombieri-Lang involving a general number field ${K}$, we obtain a similar conclusion with “rational” replaced by “lying in ${K}$” (one has to extend the Solymosi-de Zeeuw analysis to more general number fields, but this should be routine, using the analogue of Faltings’ theorem for such number fields).

[Note: the idea for this post originated before the recent preprint of Mochizuki on the abc conjecture was released, and is not intended as a commentary on that work, which offers a much more non-trivial perspective on scheme theory. -T.]

In classical algebraic geometry, the central object of study is an algebraic variety ${V}$ over a field ${k}$ (and the theory works best when this field ${k}$ is algebraically closed). One can talk about either affine or projective varieties; for sake of discussion, let us restrict attention to affine varieties. Such varieties can be viewed in at least four different ways:

• (Algebraic geometry) One can view a variety through the set ${V(k)}$ of points (over ${k}$) in that variety.
• (Commutative algebra) One can view a variety through the field of rational functions ${k(V)}$ on that variety, or the subring ${k[V]}$ of polynomial functions in that field.
• (Dual algebraic geometry) One can view a variety through a collection of polynomials ${P_1,\ldots,P_m}$ that cut out that variety.
• (Dual commutative algebra) One can view a variety through the ideal ${I(V)}$ of polynomials that vanish on that variety.

For instance, the unit circle over the reals can be thought of in each of these four different ways:

• (Algebraic geometry) The set of points ${\{ (x,y) \in {\bf R}^2: x^2+y^2 = 1 \}}$.
• (Commutative algebra) The quotient ${{\bf R}[x,y] / (x^2+y^2-1)}$ of the polynomial ring ${{\bf R}[x,y]}$ by the ideal generated by ${x^2+y^2-1}$ (or equivalently, the algebra generated by ${x,y}$ subject to the constraint ${x^2+y^2=1}$), or the fraction field of that quotient.
• (Dual algebraic geometry) The polynomial ${x^2+y^2-1}$.
• (Dual commutative algebra) The ideal ${(x^2+y^2-1)}$ generated by ${x^2+y^2-1}$.

The four viewpoints are almost equivalent to each other (particularly if the underlying field ${k}$ is algebraically closed), as there are obvious ways to pass from one viewpoint to another. For instance, starting with the set of points on a variety, one can form the space of rational functions on that variety, or the ideal of polynomials that vanish on that variety. Given a set of polynomials, one can cut out their zero locus, or form the ideal that they generate. Given an ideal in a polynomial ring, one can quotient out the ring by the ideal and then form the fraction field. Finally, given the ring of polynomials on a variety, one can form its spectrum (the space of prime ideals in the ring) to recover the set of points on that variety (together with the Zariski topology on that variety).

Because of the connections between these viewpoints, there are extensive “dictionaries” (most notably the ideal-variety dictionary) that convert basic concepts in one of these four perspectives into any of the other three. For instance, passing from a variety to a subvariety shrinks the set of points and the function field, but enlarges the set of polynomials needed to cut out the variety, as well as the associated ideal. Taking the intersection or union of two varieties corresponds to adding or multiplying together the two ideals respectively. The dimension of an (irreducible) algebraic variety can be defined as the transcendence degree of the function field, the maximal length of chains of subvarieties, or the Krull dimension of the ring of polynomials. And so on and so forth. Thanks to these dictionaries, it is now commonplace to think of commutative algebras geometrically, or conversely to approach algebraic geometry from the perspective of abstract algebra. There are however some very well known defects to these dictionaries, at least when viewed in the classical setting of algebraic varieties. The main one is that two different ideals (or two inequivalent sets of polynomials) can cut out the same set of points, particularly if the underlying field ${k}$ is not algebraically closed. For instance, if the underlying field is the real line ${{\bf R}}$, then the polynomial equations ${x^2+1=0}$ and ${1=0}$ cut out the same set of points, namely the empty set, but the ideal generated by ${x^2+1}$ in ${{\bf R}[x]}$ is certainly different from the ideal generated by ${1}$. This particular example does not work in an algebraically closed field such as ${{\bf C}}$, but in that case the polynomial equations ${x^2=0}$ and ${x=0}$ also cut out the same set of points (namely the origin), but again ${x^2}$ and ${x}$ generate different ideals in ${{\bf C}[x]}$. Thanks to Hilbert’s nullstellensatz, we can get around this problem (in the case when ${k}$ is algebraically closed) by always passing from an ideal to its radical, but this causes many aspects of the theory of algebraic varieties to become more complicated when the varieties involved develop singularities or multiplicities, as can already be seen with the simple example of Bezout’s theorem.

Nowadays, the standard way to deal with these issues is to replace the notion of an algebraic variety with the more general notion of a scheme. Roughly speaking, the way schemes are defined is to focus on the commutative algebra perspective as the primary one, and to allow the base field ${k}$ to be not algebraically closed, or even to just be a commutative ring instead of a field. (One could even consider non-commutative rings, leading to non-commutative geometry, but we will not discuss this extension of scheme theory further here.) Once one generalises to these more abstract rings, the notion of a rational function becomes more complicated (one has to work locally instead of globally, cutting out the points where the function becomes singular), but as a first approximation one can think of a scheme as basically being the same concept as a commutative ring. (In actuality, due to the need to localise, a scheme is defined as a sheaf of rings rather than a single ring, but these technicalities will not be important for the purposes of this discussion.) All the other concepts from algebraic geometry that might previously have been defined using one of the other three perspectives, are then redefined in terms of this ring (or sheaf of rings) in order to generalise them to schemes.

Thus, for instance, in scheme theory the rings ${{\bf R}[x]/(x^2)}$ and ${{\bf R}[x]/(x)}$ describe different schemes; from the classical perspective, they cut out the same locus, namely the point ${\{0\}}$, but the former scheme makes this point “fatter” than the latter scheme, giving it a degree (or multiplicity) of ${2}$ rather than ${1}$.

Because of this, it seems that the link between the commutative algebra perspective and the algebraic geometry perspective is still not quite perfect in scheme theory, unless one is willing to start “fattening” various varieties to correctly model multiplicity or singularity. But – and this is the trivial remark I wanted to make in this blog post – one can recover a tight connection between the two perspectives as long as one allows the freedom to arbitrarily extend the underlying base ring.

Here’s what I mean by this. Consider classical algebraic geometry over some commutative ring ${R}$ (not necessarily a field). Any set of polynomials ${P_1,\ldots,P_m \in R[x_1,\ldots,x_d]}$ in ${d}$ indeterminate variables ${x_1,\ldots,x_d}$ with coefficients in ${R}$ determines, on the one hand, an ideal

$\displaystyle I := (P_1,\ldots,P_m)$

$\displaystyle = \{P_1Q_1+\ldots+P_mQ_m: Q_1,\ldots,Q_m \in R[x_1,\ldots,x_d]\}$

in ${R[x_1,\ldots,x_d]}$, and also cuts out a zero locus

$\displaystyle V[R] := \{ (y_1,\ldots,y_d) \in R^d: P_1(y_1,\ldots,y_d) = \ldots = P_m(y_1,\ldots,y_d) = 0 \},$

since each of the polynomials ${P_1,\ldots,P_m}$ clearly make sense as maps from ${R^d}$ to ${R}$. Of course, one can also write ${V[R]}$ in terms of ${I}$:

$\displaystyle V[R] := \{ (y_1,\ldots,y_d) \in R^d: P(y_1,\ldots,y_d) = 0 \hbox{ for all } P \in I \}.$

Thus the ideal ${I}$ uniquely determines the zero locus ${V[R]}$, and we will emphasise this by writing ${V[R]}$ as ${V_I[R]}$. As the previous counterexamples illustrate, the converse is not true. However, whenever we have any extension ${R'}$ of the ring ${R}$ (i.e. a commutative ring ${R'}$ that contains ${R}$ as a subring), then we can also view the polynomials ${P_1,\ldots,P_m}$ as maps from ${(R')^d}$ to ${R'}$, and so one can also define the zero locus for all the extensions:

$\displaystyle V[R'] := \{ (y_1,\ldots,y_d) \in (R')^d:$

$\displaystyle P_1(y_1,\ldots,y_d) = \ldots = P_m(y_1,\ldots,y_d) = 0 \}.$

As before, ${V[R']}$ is determined by the ideal ${I}$:

$\displaystyle V[R'] = V_I[R'] := \{ (y_1,\ldots,y_d) \in (R')^d:$

$\displaystyle P(y_1,\ldots,y_d) = 0 \hbox{ for all } P \in I \}.$

The trivial remark is then that while a single zero locus ${V_I[R]}$ is insufficient to recover ${I}$, the collection of zero loci ${V_I[R']}$ for all extensions ${R'}$ of ${R}$ (or more precisely, the assignment map ${R' \mapsto V_I[R']}$, known as the functor of points of ${V_I}$) is sufficient to recover ${I}$, as long as at least one zero locus, say ${V_I[R_0]}$, is non-empty. Indeed, suppose we have two ideals ${I, I'}$ of ${R[x_1,\ldots,x_d]}$ that cut out the same non-empty zero locus for all extensions ${R'}$ of ${R}$, thus

$\displaystyle V_I[R'] = V_{I'}[R'] \neq \emptyset$

for all extensions ${R'}$ of ${R}$. We apply this with the extension ${R'}$ of ${R}$ given by ${R' := R_0[x_1,\ldots,x_d]/I}$. Note that the embedding of ${R}$ in ${R_0[x_1,\ldots,x_d]/I}$ is injective, since otherwise ${I}$ would cut out the empty set as the zero locus over ${R_0}$, and so ${R'}$ is indeed an extension of ${R}$. Tautologically, the point ${(x_1 \hbox{ mod } I, \ldots, x_d \hbox{ mod } I)}$ lies in ${V_I[R']}$, and thus necessarily lies in ${V_{I'}[R']}$ as well. Unpacking what this means, we conclude that ${P \in I}$ whenever ${P \in I'}$, that is to say that ${I' \subset I}$. By a symmetric argument, we also have ${I \subset I'}$, and thus ${I=I'}$ as claimed. (As pointed out in comments, this fact (and its proof) is essentially a special case of the Yoneda lemma. The connection is tighter if one allows ${R'}$ to be any ring with a (not necessarily injective) map from ${R}$ into it, rather than an extension of ${R}$, in which case one can also drop the hypothesis that ${V_I[R_0]}$ is non-empty for at least one ${R_0}$. For instance, ${V_{(2)}[R'] = V_{(3)}[R'] = \emptyset}$ for every extension ${R'}$ of the integers, but if one also allows quotients such as ${{\bf Z}/(2)}$ or ${{\bf Z}/(3)}$ instead, then ${V_{(2)}[R']}$ and ${V_{(3)}[R']}$ are no longer necessarily equal.)

Thus, as long as one thinks of a variety or scheme as cutting out points not just in the original base ring or field, but in all extensions of that base ring or field, one recovers an exact correspondence between the algebraic geometry perspective and the commutative algebra perspective. This is similar to the classical algebraic geometry position of viewing an algebraic variety as being defined simultaneously over all fields that contain the coefficients of the defining polynomials, but the crucial difference between scheme theory and classical algebraic geometry is that one also allows definition over commutative rings, and not just fields. In particular, one needs to allow extensions to rings that may contain nilpotent elements, otherwise one cannot distinguish an ideal from its radical.

There are of course many ways to extend a field into a ring, but as an analyst, one way to do so that appeals particularly to me is to introduce an epsilon parameter and work modulo errors of ${O(\varepsilon)}$. To formalise this algebraically, let’s say for sake of concreteness that the base field is the real line ${{\bf R}}$. Consider the ring ${\tilde R}$ of real-valued quantities ${x = x_\varepsilon}$ that depend on a parameter ${\varepsilon \geq 0}$ (i.e. functions from ${{\bf R}^+}$ to ${{\bf R}}$), which are locally bounded in the sense that ${x}$ is bounded whenever ${\varepsilon}$ is bounded. (One can, if one wishes, impose some further continuity or smoothness hypotheses on how ${x}$ depends on ${\varepsilon}$, but this turns out not to be relevant for the following discussion. Algebraists often prefer to use the ring of Puiseux series here in place of ${\tilde R}$, and a nonstandard analyst might instead use the hyperreals, but again this will not make too much difference for our purposes.) Inside this commutative ring, we can form the ideal ${I}$ of quantities ${x = x_\varepsilon}$ that are of size ${O(\varepsilon)}$ as ${\varepsilon \rightarrow 0}$, i.e. there exists a quantity ${C>0}$ independent of ${\varepsilon}$ such that ${|x| \leq C\varepsilon}$ for all sufficiently small ${\varepsilon}$. This can easily be seen to indeed be an ideal in ${\tilde R}$. We then form the quotient ring ${R' := \tilde R/I := \{ x \hbox{ mod } I: x \in \tilde R \}}$. Note that ${x = y \hbox{ mod } I}$ is equivalent to the assertion that ${x = y + O(\varepsilon)}$, so we are encoding the analyst’s notion of “equal up to errors of ${O(\varepsilon)}$” into algebraic terms.

Clearly, ${R'}$ is a commutative ring extending ${{\bf R}}$. Hence, any algebraic variety

$\displaystyle V[{\bf R}] = \{ (y_1,\ldots,y_d) \in {\bf R}^d: P_1(y_1,\ldots,y_d) = \ldots = P_m(y_1,\ldots,y_d) = 0 \}$

defined over the reals ${{\bf R}}$ (so the polynomials ${P_1,\ldots,P_m}$ have coefficients in ${{\bf R}}$), also is defined over ${R'}$:

$\displaystyle V[R'] = \{ (y_1,\ldots,y_d) \in (R')^d:$

$\displaystyle P_1(y_1,\ldots,y_d) = \ldots = P_m(y_1,\ldots,y_d) = 0 \}.$

In language that more closely resembles analysis, we have

$\displaystyle V[R'] = \{ (y_1,\ldots,y_d) \in \tilde R^d: P_1(y_1,\ldots,y_d), \ldots, P_m(y_1,\ldots,y_d) = O(\varepsilon) \}$

$\displaystyle \hbox{ mod } I^d.$

Thus we see that ${V[R']}$ is in some sense an “${\varepsilon}$-thickening” of ${V[{\bf R}]}$, and is thus one way to give rigorous meaning to the intuition that schemes can “thicken” varieties. For instance, the scheme associated to the ideal ${(x)}$, when interpreted over ${R'}$, becomes an ${O(\varepsilon)}$ neighbourhood of the origin

$\displaystyle V_{(x)}[R'] = \{ y \in \tilde R: y = O(\varepsilon) \} \hbox{ mod } I,$

but the scheme associated to the smaller ideal ${(x^2)}$, when interpreted over ${R'}$, becomes an ${O(\varepsilon^{1/2})}$-neighbourhood of the origin, thus being a much “fatter” point:

$\displaystyle V_{(x^2)}[R'] = \{ y \in \tilde R: y^2 = O(\varepsilon) \} \hbox{ mod } I$

$\displaystyle = \{ y \in \tilde R: y = O(\varepsilon^{1/2} ) \} \hbox{ mod } I.$

Once one introduces the analyst’s epsilon, one can see quite clearly that ${V_{(x^2)}[R']}$ is coming from a larger scheme than ${V_{(x)}[R']}$, with fewer polynomials vanishing on it; in particular, the polynomial ${x}$ vanishes to order ${O(\varepsilon)}$ on ${V_{(x)}[R']}$ but does not vanish to order ${O(\varepsilon)}$ on ${V_{(x^2)}[R']}$.

By working with this analyst’s extension of ${{\bf R}}$, one can already get a reasonably good first approximation of what schemes over ${{\bf R}}$ look like, which I found particularly helpful for getting some intuition on these objects. However, since this is only one extension of ${{\bf R}}$, and not a “universal” such extension, it cannot quite distinguish any two schemes from each other, although it does a better job of this than classical algebraic geometry. For instance, consider the scheme cut out by the polynomials ${x^2, y^2}$ in two dimensions. Over ${R'}$, this becomes

$\displaystyle V_{(x^2,y^2)}[R'] = \{ (x,y) \in \tilde R^2: x^2, y^2 = O(\varepsilon) \} \hbox{ mod } I^2$

$\displaystyle = \{ (x,y) \in \tilde R^2: x, y = O(\varepsilon^{1/2}) \} \hbox{ mod } I^2.$

Note that the polynomial ${xy}$ vanishes to order ${O(\varepsilon)}$ on this locus, but ${xy}$ fails to lie in the ideal ${(x^2,y^2)}$. Equivalently, we have ${V_{(x^2,y^2)}[R'] = V_{(x^2,y^2,xy)}[R']}$, despite ${(x^2,y^2)}$ and ${(x^2,y^2,xy)}$ being distinct ideals. Basically, the analogue of the nullstellensatz for ${R'}$ does not completely remove the need for performing a closure operation on the ideal ${I}$; it is less severe than taking the radical, but is instead more like taking a “convex hull” in that one needs to be able to “interpolate” between two polynomials in the ideal (such as ${x^2}$ and ${y^2}$ to arrive at intermediate polynomials (such as ${xy}$) that one then places in the ideal.

One can also view ideals (and hence, schemes), from a model-theoretic perspective. Let ${I}$ be an ideal of a polynomial ring ${R[x_1,\ldots,x_d]}$ generated by some polynomials ${P_1,\ldots,P_m \in R[x_1,\ldots,x_d]}$. Then, clearly, if ${Q}$ is another polynomial in the ideal ${I}$, then we can use the axioms of commutative algebra (which are basically the axioms of high school algebra) to obtain the syntactic deduction

$\displaystyle P_1(x_1,\ldots,x_d) = \ldots = P_m(x_1,\ldots,x_d) = 0 \vdash Q(x_1,\ldots,x_d) = 0$

(since ${Q}$ is just a sum of multiples of ${P_1,\ldots,P_m}$). In particular, we have the semantic deduction

$\displaystyle P_1(y_1,\ldots,y_d) = \ldots = P_m(y_1,\ldots,y_d) = 0 \implies Q(y_1,\ldots,y_d) = 0 \ \ \ \ \ (1)$

for any assignment of indeterminates ${y_1,\ldots,y_d}$ in ${R}$ (or in any extension ${R'}$ of ${R}$). If we restrict ${y_1,\ldots,y_d}$ to lie in ${R}$ only, then (even if ${R}$ is an algebraically closed field), the converse of the above statement is false; there can exist polynomials ${Q}$ outside of ${I}$ for which (1) holds for all assignments ${y_1,\ldots,y_d}$ in ${R}$. For instance, we have

$\displaystyle y^2 = 0 \implies y = 0$

for all ${y}$ in an algebraically closed field, despite ${x}$ not lying in the ideal ${(x^2)}$. Of course, the nullstellensatz again explains what is going on here; (1) holds whenever ${Q}$ lies in the radical of ${I}$, which can be larger than ${I}$ itself. But if one allows the indeterminates ${y_1,\ldots,y_d}$ to take values in arbitrary extensions ${R'}$ of ${R}$, then the truth of the converse is restored, thus giving a “completeness theorem” relating the syntactic deductions of commutative algebra to the semantic interpretations of such algebras over the extensions ${R'}$. For instance, since

$\displaystyle y^2 = O(\varepsilon) \not \implies y = O(\varepsilon)$

we no longer have a counterexample to the converse coming from ${x}$ and ${(x^2)}$ once we work in ${R'}$ instead of ${{\bf R}}$. On the other hand, we still have

$\displaystyle x^2, y^2 = O(\varepsilon) \implies xy = O(\varepsilon)$

so the extension ${R'}$ is not powerful enough to detect that ${xy}$ does not actually lie in ${(x^2,y^2)}$; a larger ring (which is less easy to assign an analytic interpretation to) is needed to achieve this.

I have blogged a number of times in the past about the relationship between finitary (or “hard”, or “quantitative”) analysis, and infinitary (or “soft”, or “qualitative”) analysis. One way to connect the two types of analysis is via compactness arguments (and more specifically, contradiction and compactness arguments); such arguments can convert qualitative properties (such as continuity) to quantitative properties (such as bounded), basically because of the fundamental fact that continuous functions on a compact space are bounded (or the closely related fact that sequentially continuous functions on a sequentially compact space are bounded).

A key stage in any such compactness argument is the following: one has a sequence ${X_n}$ of “quantitative” or “finitary” objects or spaces, and one has to somehow end up with a “qualitative” or “infinitary” limit object ${X}$ or limit space. One common way to achieve this is to embed everything inside some universal space and then use some weak compactness property of that space, such as the Banach-Alaoglu theorem (or its sequential counterpart). This is for instance the idea behind the Furstenberg correspondence principle relating ergodic theory to combinatorics; see for instance this post of mine on this topic.

However, there is a slightly different approach, which I will call ultralimit analysis, which proceeds via the machinery of ultrafilters and ultraproducts; typically, the limit objects ${X}$ one constructs are now the ultraproducts (or ultralimits) of the original objects ${X_\alpha}$. There are two main facts that make ultralimit analysis powerful. The first is that one can take ultralimits of arbitrary sequences of objects, as opposed to more traditional tools such as metric completions, which only allow one to take limits of Cauchy sequences of objects. The second fact is Los’s theorem, which tells us that ${X}$ is an elementary limit of the ${X_\alpha}$ (i.e. every sentence in first-order logic which is true for the ${X_\alpha}$ for ${\alpha}$ large enough, is true for ${X}$). This existence of elementary limits is a manifestation of the compactness theorem in logic; see this earlier blog post for more discussion. So we see that compactness methods and ultrafilter methods are closely intertwined. (See also my earlier class notes for a related connection between ultrafilters and compactness.)

Ultralimit analysis is very closely related to nonstandard analysis. I already discussed some aspects of this relationship in an earlier post, and will expand upon it at the bottom of this post. Roughly speaking, the relationship between ultralimit analysis and nonstandard analysis is analogous to the relationship between measure theory and probability theory.

To illustrate how ultralimit analysis is actually used in practice, I will show later in this post how to take a qualitative infinitary theory – in this case, basic algebraic geometry – and apply ultralimit analysis to then deduce a quantitative version of this theory, in which the complexity of the various algebraic sets and varieties that appear as outputs are controlled uniformly by the complexity of the inputs. The point of this exercise is to show how ultralimit analysis allows for a relatively painless conversion back and forth between the quantitative and qualitative worlds, though in some cases the quantitative translation of a qualitative result (or vice versa) may be somewhat unexpected. In an upcoming paper of myself, Ben Green, and Emmanuel Breuillard (announced in the previous blog post), we will rely on ultralimit analysis to reduce the messiness of various quantitative arguments by replacing them with a qualitative setting in which the theory becomes significantly cleaner.

For sake of completeness, I also redo some earlier instances of the correspondence principle via ultralimit analysis, namely the deduction of the quantitative Gromov theorem from the qualitative one, and of Szemerédi’s theorem from the Furstenberg recurrence theorem, to illustrate how close the two techniques are to each other.