Much as group theory is the study of groups, or graph theory is the study of graphs, model theory is the study of models (also known as structures) of some language (which, in this post, will always be a single-sorted, first-order language). A structure is a set
, equipped with one or more operations, constants, and relations. This is of course an extremely general type of mathematical object, but (quite remarkably) one can still say a substantial number of interesting things about very broad classes of structures.
We will observe the common abuse of notation of using the set as a metonym for the entire structure, much as we usually refer to a group
simply as
, a vector space
simply as
, and so forth. Following another common bending of the rules, we also allow some operations on structures (such as the multiplicative inverse operation on a group or field) to only be partially defined, and we allow use of the usual simplifying conventions for mathematical formulas (e.g. writing
instead of
or
, in cases where associativity is known). We will also deviate slightly from the usual practice in logic by emphasising individual structures, rather than the theory of general classes of structures; for instance, we will talk about the theory of a single field such as
or
, rather than the theory of all fields of a certain type (e.g. real closed fields or algebraically closed fields).
Once one has a structure , one can introduce the notion of a definable subset of
, or more generally of a Cartesian power
of
, defined as a set
of the form
for some formula in the language
with
free variables and any number of constants from
(that is,
is a well-formed formula built up from a finite number of constants
in
, the relations and operations on
, logical connectives such as
,
,
, and the quantifiers
). Thus, for instance, in the theory of the arithmetic of the natural numbers
, the set of primes
is a definable set, since we have
In the theory of the field of reals , the unit circle
is an example of a definable set,
but so is the the complement of the circle,
and the interval :
Due to the unlimited use of constants, any finite subset of a power of any structure
is, by our conventions, definable in that structure. (One can of course also consider definability without parameters (also known as
-definability), in which arbitrary constants are not permitted, but we will not do so here.)
We can isolate some special subclasses of definable sets:
- An atomic definable set is a set of the form (1) in which
is an atomic formula (i.e. it does not contain any logical connectives or quantifiers).
- A quantifier-free definable set is a set of the form (1) in which
is quantifier-free (i.e. it can contain logical connectives, but does not contain the quantifiers
).
Example 1 In the theory of a field such as
, an atomic definable set is the same thing as an affine algebraic set (also known as an affine algebraic variety, with the understanding that varieties are not necessarily assumed to be irreducible), and a quantifier-free definable set is known as a constructible set; thus we see that algebraic geometry can be viewed in some sense as a special case of model theory. (Conversely, it can in fact be quite profitable to think of model theory as an abstraction of algebraic geometry; for instance, the concepts of Morley rank and Morley degree in model theory (discussed in this previous blog post) directly generalises the concepts of dimension and degree in algebraic geometry.) Over
, the interval
is a definable set, but not a quantifier-free definable set (and certainly not an atomic definable set); and similarly for the primes over
.
A quantifier-free definable set in is nothing more than a finite boolean combination of atomic definable sets; in other words, the class of quantifier-free definable sets over
is the smallest class that contains the atomic definable sets and is closed under boolean operations such as complementation and union (which generate all the other boolean operations). Similarly, the class of definable sets over
is the smallest class that contains the quantifier-free definable sets, and is also closed under the operation of projection
from
to
for every natural number
, where
is the map
.
Some structures have the property of enjoying quantifier elimination, which means that every definable set is in fact a quantifier-free definable set, or equivalently that the projection of a quantifier-free definable set is again quantifier-free. For instance, an algebraically closed field (with the field operations) has quantifier elimination (i.e. the projection of a constructible set is again constructible); this fact can be proven by the classical tool of resultants, and among other things can be used to give a proof of Hilbert’s nullstellensatz. (Note though that projection does not necessary preserve the property of being atomic; for instance, the projection of the atomic set
is the non-atomic, but still quantifier-free definable, set
.) In the converse direction, it is not difficult to use the nullstellensatz to deduce quantifier elimination. For theory of the real field
, which is not algebraically closed, one does not have quantifier elimination, as one can see from the example of the unit circle (which is a quantifier-free definable set) projecting down to the interval
(which is definable, but not quantifer-free definable). However, if one adds the additional operation of order
to the reals, giving it the language of an ordered field rather than just a field, then quantifier elimination is recovered (the class of quantifier-free definable sets now enlarges to match the class of definable sets, which in this case is also the class of semi-algebraic sets); this is the famous Tarski-Seidenberg theorem.
On the other hand, many important structures do not have quantifier elimination; typically, the projection of a quantifier-free definable set is not, in general, quantifier-free definable. This failure of the projection property also shows up in many contexts outside of model theory; for instance, Lebesgue famously made the error of thinking that the projection of a Borel measurable set remained Borel measurable (it is merely an analytic set instead). Turing’s halting theorem can be viewed as an assertion that the projection of a decidable set (also known as a computable or recursive set) is not necessarily decidable (it is merely semi-decidable (or recursively enumerable) instead). The notorious P=NP problem can also be essentially viewed in this spirit; roughly speaking (and glossing over the placement of some quantifiers), it asks whether the projection of a polynomial-time decidable set is again polynomial-time decidable. And so forth. (See this blog post of Dick Lipton for further discussion of the subtleties of projections.)
Now we consider the status of quantifier elimination for the theory of a finite field . If interpreted naively, quantifier elimination is trivial for a finite field
, since every subset of
is finite and thus quantifier-free definable. However, we can recover an interesting question in one of two (essentially equivalent) ways. One is to work in the asymptotic regime in which the field
is large, but the length of the formulae used to construct one’s definable sets stays bounded uniformly in the size of
(where we view any constant in
as contributing a unit amount to the length of a formula, no matter how large
is). A simple counting argument then shows that only a small number of subsets of
become definable in the asymptotic limit
, since the number of definable sets clearly grows at most polynomially in
for any fixed bound on the formula length, while the number of all subsets of
grows exponentially in
.
Another way to proceed is to work not with a single finite field , or even with a sequence
of finite fields, but with the ultraproduct
of a sequence of finite fields, and to study the properties of definable sets over this ultraproduct. (We will be using the notation of ultraproducts and nonstandard analysis from this previous blog post.) This approach is equivalent to the more finitary approach mentioned in the previous paragraph, at least if one does not care to track of the exact bounds on the length of the formulae involved. Indeed, thanks to Los’s theorem, a definable subset
of
is nothing more than the ultraproduct
of definable subsets
of
for all
sufficiently close to
, with the length of the formulae used to define
uniformly bounded in
. In the language of nonstandard analysis, one can view
as a nonstandard finite field.
The ultraproduct of finite fields is an important example of a pseudo-finite field – a field that obeys all the sentences in the languages of fields that finite fields do, but is not necessarily itself a finite field. The model theory of pseudo-finite fields was first studied systematically by Ax (in the same paper where the Ax-Grothendieck theorem, discussed previously on this blog, was established), with important further contributions by Kiefe, by Fried-Sacerdote, by two papers of Chatzidakis-van den Dries-Macintyre, and many other authors.
As mentioned before, quantifier elimination trivially holds for finite fields. But for infinite pseudo-finite fields, such as the ultraproduct of finite fields with
going to infinity, quantifier elimination fails. For instance, in a finite field
, the set
of quadratic residues is a definable set, with a bounded formula length, and so in the ultraproduct
, the set
of nonstandard quadratic residues is also a definable set. However, in one dimension, we see from the factor theorem that the only atomic definable sets are either finite or the whole field
, and so the only constructible sets (i.e. the only quantifier-free definable sets) are either finite or cofinite in
. Since the quadratic residues have asymptotic density
in a large finite field, they cannot form a quantifier-free definable set, despite being definable.
Nevertheless, there is a very nice almost quantifier elimination result for these fields, in characteristic zero at least, which we phrase here as follows:
Theorem 1 (Almost quantifier elimination) Let
be a nonstandard finite field of characteristic zero, and let
be a definable set over
. Then
is the union of finitely many sets of the form
where
is an atomic definable subset of
(i.e. the
-points of an algebraic variety
defined over
in
) and
is a polynomial.
Results of this type were first obtained essentially due to Catarina Kiefe, although the formulation here is closer to that of Chatzidakis-van den Dries-Macintyre.
Informally, this theorem says that while we cannot quite eliminate all quantifiers from a definable set over a nonstandard finite field, we can eliminate all but one existential quantifier. Note that negation has also been eliminated in this theorem; for instance, the definable set uses a negation, but can also be described using a single existential quantifier as
.) I believe that there are more complicated analogues of this result in positive characteristic, but I have not studied this case in detail (Kiefe’s result does not assume characteristic zero, but her conclusion is slightly different from the one given here). In the one-dimensional case
, the only varieties
are the affine line and finite sets, and we can simplify the above statement, namely that any definable subset of
takes the form
for some polynomial
(i.e. definable sets in
are nothing more than the projections of the
-points of a plane curve).
There is an equivalent formulation of this theorem for standard finite fields, namely that if is a finite field and
is definable using a formula of length at most
, then
can be expressed in the form (2) with the degree of
bounded by some quantity
depending on
and
, assuming that the characteristic of
is sufficiently large depending on
.
The theorem gives quite a satisfactory description of definable sets in either standard or nonstandard finite fields (at least if one does not care about effective bounds in some of the constants, and if one is willing to exclude the small characteristic case); for instance, in conjunction with the Lang-Weil bound discussed in this recent blog post, it shows that any non-empty definable subset of a nonstandard finite field has a nonstandard cardinality of for some positive standard rational
and integer
. Equivalently, any non-empty definable subset of
for some standard finite field
using a formula of length at most
has a standard cardinality of
for some positive rational of height
and some natural number
between
and
. (For instance, in the example of the quadratic residues given above,
is equal to
and
equal to
.) There is a more precise statement to this effect, namely that the Poincaré series of a definable set is rational; see Kiefe’s paper for details.
Below the fold I give a proof of Theorem 1, which relies primarily on the Lang-Weil bound mentioned above.
— 1. Proof of theorem —
Fix a nonstandard finite field of characteristic zero. Let us temporarily define a basic set to be a set (2) for some atomic definable set
and some polynomial
, and a good set to be a finite union of basic sets in some vector space
. Our objective is thus to show that all definable sets are good.
Clearly, every atomic definable set is already a basic set, and is thus certainly good. Given the relationship between atomic definable sets and all definable sets, it thus suffices to show that the class of good sets are closed under boolean operations (e.g. union, intersection, negation) and projection. Closure under unions is automatic by definition; the interesting and difficult closure properties are projection and negation. (Closure under intersection is, strictly speaking, a consequence of closure under unions and negations, thanks to de Morgan’s laws; but it turns out to be convenient to obtain closure under intersection first as a stepping stone to closure under negation.)
We first observe that we may place good sets in a suitable “normal form”. Let be the algebraic closure of
. This is an algebraically closed field that has a nonstandard Frobenius endomorphism
, defined as the ultralimit of the Frobenius maps
on each of the finite fields
. This is a nonstandard automorphism of
whose fixed points are precisely
. An atomic definable set over
is the same thing as the
-points
of an variety
over
which is invariant with resspect to the Frobenius automorphism (this can be established for instance by inspecting a reduced Grobner basis for the associated ideal). Such a variety can be decomposed over
into absolutely irreducible components. Some of these components may remain Frobenius-invariant and will thus continue to be defined over
. If a component is not Frobenius invariant, we may intersect it with all of its Frobenius conjugates without losing any of its
-points; by the Noetherian property, this will reduce that component to an absolutely irreducible variety defined over
. Thus, every variety defined over
can be replaced with a union of finitely many absolutely irreducible varieties defined over
, without losing or gaining any
-points. From this, we see that any good set can be placed in a form in which all of the varieties involved are absolutely irreducible.
Now consider a basic set of the form
with absolutely irreducible. We consider the constant term
of
. If this term vanishes identically on
, then this basic set is all of
. Otherwise, observe that we may factorise
for some polynomial with
, and then
From this, we see that any good set can be expressed as the union of the -points of a variety (i.e. an atomic definable set) and finitely many basic sets which have unit constant term in the sense that they have the form
with absolutely irreducible and
.
This leads us to the following useful lemma:
Lemma 2 (Hypersurface removal) Let
be a good set in
, and let
be a polynomial. Then
is also good.
Proof: By replacing with its norm over the polynomials over
, we may assume that
takes coefficients in
. By the preceding discussion, it suffices to establish this in the case when
is either the
-points
an irreducible variety
, or a basic set on such a variety with unit constant term. In the former case, we simply note that
In the latter case, we observe that
whenever has unit constant term.
This allows us to generalise to be constructible rather than algebraic:
Corollary 3 Any set of the form
, where
is a quantifier-free definable set (i.e. the
-points of a constructible set), and
is a polynomial from
to
, is good.
Proof: Again, we may assume takes coefficients in
. A constructible set can be expressed as the union of finitely many quasiprojective varieties (i.e. the set-theoretic difference of two affine varieties). A quasiprojective variety
can be written as the finite union of sets of the form
for various polynomials
, and the claim then follows from the preceding lemma.
Now we can handle images of constructible sets with -dimensional fibres:
Proposition 4 Let
be constructible sets defined over
with irreducible Zariski closures, and let
be a regular map, also defined over
, with the property that all fibres
,
are zero-dimensional. Then
is good. (In particular, the set of
-points of any constructible set is a good set.)
Proof: By restricting if necessary, we may assume that
is dominant. The field
of
-valued rational functions on
can then be viewed as a field extension of
(using the pullback map by
); as the fibres are zero-dimensional, this extension has transcendence degree zero, i.e. it is algebraic. By the primitive element theorem (and it is here that we are using our hypothesis that
has characteristic zero, to ensure that the field extension is separable),
is generated by
and a rational function
on
, which by clearing denominators one can assume to be integral with respect to the polynomial ring
, thus
for some polynomial with coefficients in
with
monic as a polynomial over
. Each of the coordinate functions
on the affine variety
can then be viewed (again after clearing denominators) as a polynomial (over
) in
, divided by some non-trivial polynomial
, thus
for all and all
in
, and some polynomials
with coefficients in
. From this, we see that for any
with
,
has an
-point if and only if
has a root in
. Thus we have
Using Corollary 3 and an induction hypothesis on the dimension of to deal with the residual term
, we obtain the claim.
Next, we record an important property of infinite pseudofinite fields, known as the pseudo algebraically closed (PAC) axiom:
Lemma 5 (PAC axiom) Let
be an absolutely irreducible variety defined over
. Then
is non-empty. Furthermore, for any Zariski-dense constructible subset
of
,
is non-empty also.
Proof: We write as an ultralimit of varieties
defined over
. As
is absolutely irreducible, we have
absolutely irreducible with dimension
for
close to
(see Lemma 3 and Lemma 8 of this previous blog post), and so by the Lang-Weil bound (see this previous blog post) we have
Since goes to infinity as
, we conclude that
is non-empty for
sufficiently close to
, and so
is non-empty also by Los’s theorem. The same argument works for
, using the Schwarz-Zippel bound from the previous post to handle the set removed from
to form
.
We can now establish closure with respect to projection:
Proof: It suffices to establish that
is a good set whenever is a constructible subset of
. Let
be the projection map from
to
, so our task is to show that
is good. By decomposing
, we may assume that
has irreducible closure, and that all the fibres of
in
have the same dimension, which is either
,
, or
. If the fibres have dimension
, then by the PAC axiom
, and the claim follows from Corollary 3. If the fibres have dimension
, the claim follows instead from Proposition 4. So the only remaining case is when the fibres are one-dimensional curves.
We can work generically in , since the contribution of any subvariety of
of strictly smaller dimension can be handled by an induction hypothesis. Generically, the fibre over a point
is then given by a curve
for some polynomial
defined over
. If
is an
-point of
, then by the PAC axiom, this curve contains an
-point in the plane
if and only if
, viewed as a polynomial of two variables, contains an absolutely irreducible factor defined over
. If one fixes the degree
of this factor, this is a definable constraint on
coefficients, and by Proposition 4, the set of
for which this constraint is obeyed is a good set. Letting
vary from
to
and taking unions, we obtain the claim.
From closure under projection, we can deduce closure under intersection:
Corollary 7 The intersection of two good sets in
is again a good sets in
.
Proof: It suffices to show that the intersection of two basic sets
and
in is good. But observe that
can be obtained from the atomic definable (and hence basic) set
by applying two projections, so the claim follows from two applications of Corollary 6.
The only remaining task is to show closure under negation. To get a hint of how to proceed on this, consider again the example of the quadratic residues in :
One can almost describe the complement of this set by selecting a non-quadratic residue of
and using the set
though this is not quite the complement of and one has to delete the origin, for instance by using Lemma 2. Applying the algorithm implicit in the proof of that lemma, one ends up with
as an explicit representation of the non-quadratic residues as a basic set.
The selection of an arbitrary non-quadratic residue may seem like a rather ad hoc step to take in the above example, which does not initially bode well for generalisation to other definable sets. However, one can make the argument a bit more “canonical” by a bit of Galois theory. Note that while an arbitrary element
of
need not necessarily have a square root
in
, it always has a square root
in the (unique) quadratic extension
of
. Of course,
will also be a square root. Letting
be the Galois conjugation of
over
, we have
and so is equal to either
or
. In the former case,
is a quadratic residue; otherwise, it is a non-quadratic residue. (Let us ignore the situation when
, as Lemma 2 and Corollary 7 give us enough tools to deal with what is going on exceptional sets of lower-dimensional varieties.) Thus, outside of the point
, we can express
as the set of
for which one has
for the two square roots
, and the complement as the set of
for which
(extending
from
to
in some arbitrary manner if desired). The descriptions of
and its (near-)complement in (3), (4) can be viewed as (shadows of) “norms” of the above assertions
,
in some sense. This suggests that more generally, given a basic set
on a variety
, one can use Galois theory to naturally partition
(outside of some exceptional set of smaller dimension) into disjoint components, one of which is
, and all of which can be described as unions of basic sets, which would lead to the desired closure property with respect to negation. (Not coincidentally, a very similar trick was used in Bombieri’s version of Stepanov’s proof of the Hasse-Weil bound discussed in this previous post, in which an upper bound on
for curves
was converted into a matching lower bound.)
Proposition 8 The complement of a good set in
is again a good set.
Proof: In view of Corollary 7 and de Morgan’s laws, it suffices to show that the complement of a basic set in
is a good set, whenever
is absolutely irreducible and defined over
. From Proposition 4, we already know that
is a good set, so it suffices to show that the set
is good. By an induction on the dimension of , we may freely delete any lower-dimensional component of
, thus it will suffice to show that
is good for some generic subset of
.
By an induction on the degree of , and Corollary 7, we may assume that
is absolutely irreducible on
. In particular,
and
have no common factor (here we are again implicitly using the characteristic zero hypothesis), and so for generic
,
and
have no common zero (otherwise their resultant would vanish generically, leading to a common factor). If we write
for some polynomials on
with
generically non-zero on
, we thus see that for generic
,
has
distinct zeroes in
. Indeed, recalling that any finite field
has a unique extension
of degree
,
has a unique extension
of degree
, and all the
zeroes of
lie in this extension.
It is a standard fact from Galois theory that is a Galois extension of
, with the Galois group isomorphic to
and generated by the Frobenius endomorphism
; taking ultralimits, we see that
is also a Galois extension of
, with Galois group again isomorphic to
and generated by the ultralimit
of the standard Frobenius endomorphisms
. For any generic
, the
distinct zeroes
of
are acted upon by the Frobenius map. Thus, there exists a permutation
such that
for all
.
Observe that for generic the statement
holds precisely when
has no fixed points. Thus, it suffices to show that for each permutation
, the set of generic
for which one can enumerate the zeroes of
as
with the property that
for all
, is a good set. But by viewing
as a
-dimensional linear vector space over
and expanding in terms of some arbitrarily chosen basis (cf. the factor
in the quadratic residue discussion), and noting that the Frobenius map is a linear transformation of this vector space, we see that this set is of the form covered by Proposition 4, and the claim follows. (One could also avoid the use of an arbitrarily chosen basis here by working with a more abstract version of Proposition 4, if desired.)
Remark 1 A closer inspection of the above arguments show that there were really only two properties of the nonstandard finite field
that were really used (besides the fact that
was a characteristic zero field). The first was the PAC axiom (Lemma 5). The second was the properties of the Galois group
, namely that this group is isomorphic to the profinite integers
(or equivalently, that
has precisely one field extension of each degree). It turns out that these properties in fact completely characterise infinite pseudo-finite fields, which is the main result of the paper of Ax cited previously.
12 comments
Comments feed for this article
12 September, 2012 at 5:06 pm
Fred Lunnon
“andrelations” -> “and relations”
“carindality” -> “cardinality”
[Corrected, thanks – T.]
13 September, 2012 at 9:31 am
tomsim
The first occurence of “metoynm” looks funny, even to this mathematical layman.
[Corrected, thanks. -T.]
13 September, 2012 at 9:48 am
rlss
R^2 should be R in definition of [1,-1]. [Corrected, thanks – T.]
13 September, 2012 at 5:17 pm
Neat Post Related to Formal Systems « Let Us Figure This Out
[…] Definable subsets over (nonstandard) finite fields, and almost quantifier elimination Share this:FacebookStumbleUponRedditLike this:LikeBe the first to like this. This entry was posted in Uncategorized. […]
15 September, 2012 at 11:02 pm
Uri Andrews
Quick note: A quantifier-free definable set is not also known as a 0-definable set. 0-definable simply means definable without parameters.
[Corrected, thanks -T.]
16 September, 2012 at 3:18 pm
William Newman
another trivial typo: ‘cosntructible’
[Corrected, thanks – T.]
20 September, 2012 at 5:27 am
Emmanuel Kowalski
You mention “Cherlin-van den Dries-Macintyre”, but the paper you link to is the one by Chatzidakis, van den Dries and Macintyre…
[Corrected, thanks – T.]
3 October, 2012 at 4:57 am
Anonymous
what you mean by( the length of the formula is uniformaly bounded in m)…
Which definition of uniformaly bounded did you use here?
3 October, 2012 at 7:54 am
Terence Tao
“uniformly bounded in m” means “bounded by some quantity C that does not depend on m”.
14 November, 2012 at 9:42 am
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[…] out that one can essentially write all definable sets as an intersection of sets of this form; see this previous blog post for more discussion.) To regularise the set , it is convenient to square the adjacency matrix, […]
30 December, 2012 at 8:44 pm
hgutahw
In “a good set to be a finite union of good sets in some vector space”, the second occurrence of “good” should be replaced by “basic”.
[corrected, thanks – T.]
29 October, 2013 at 8:09 pm
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