Let be a quasiprojective variety defined over a finite field , thus for instance could be an affine variety

where is -dimensional affine space and are a finite collection of polynomials with coefficients in . Then one can define the set of -rational points, and more generally the set of -rational points for any , since can be viewed as a field extension of . Thus for instance in the affine case (1) we have

The Weil conjectures are concerned with understanding the number

of -rational points over a variety . The first of these conjectures was proven by Dwork, and can be phrased as follows.

Theorem 1 (Rationality of the zeta function)Let be a quasiprojective variety defined over a finite field , and let be given by (2). Then there exist a finite number of algebraic integers (known ascharacteristic valuesof ), such thatfor all .

After cancelling, we may of course assume that for any and , and then it is easy to see (as we will see below) that the become uniquely determined up to permutations of the and . These values are known as the *characteristic values* of . Since is a rational integer (i.e. an element of ) rather than merely an algebraic integer (i.e. an element of the ring of integers of the algebraic closure of ), we conclude from the above-mentioned uniqueness that the set of characteristic values are invariant with respect to the Galois group . To emphasise this Galois invariance, we will not fix a specific embedding of the algebraic numbers into the complex field , but work with all such embeddings simultaneously. (Thus, for instance, contains three cube roots of , but which of these is assigned to the complex numbers , , will depend on the choice of embedding .)

An equivalent way of phrasing Dwork’s theorem is that the (-form of the) zeta function

associated to (which is well defined as a formal power series in , at least) is equal to a rational function of (with the and being the poles and zeroes of respectively). Here, we use the formal exponential

Equivalently, the (-form of the) zeta-function is a meromorphic function on the complex numbers which is also periodic with period , and which has only finitely many poles and zeroes up to this periodicity.

Dwork’s argument relies primarily on -adic analysis – an analogue of complex analysis, but over an algebraically complete (and metrically complete) extension of the -adic field , rather than over the Archimedean complex numbers . The argument is quite effective, and in particular gives explicit upper bounds for the number of characteristic values in terms of the complexity of the variety ; for instance, in the affine case (1) with of degree , Bombieri used Dwork’s methods (in combination with Deligne’s theorem below) to obtain the bound , and a subsequent paper of Hooley established the slightly weaker bound purely from Dwork’s methods (a similar bound had also been pointed out in unpublished work of Dwork). In particular, one has bounds that are uniform in the field , which is an important fact for many analytic number theory applications.

These -adic arguments stand in contrast with Deligne’s resolution of the last (and deepest) of the Weil conjectures:

Theorem 2 (Riemann hypothesis)Let be a quasiprojective variety defined over a finite field , and let be a characteristic value of . Then there exists a natural number such that for every embedding , where denotes the usual absolute value on the complex numbers . (Informally: and all of its Galois conjugates have complex magnitude .)

To put it another way that closely resembles the classical Riemann hypothesis, all the zeroes and poles of the -form lie on the critical lines for . (See this previous blog post for further comparison of various instantiations of the Riemann hypothesis.) Whereas Dwork uses -adic analysis, Deligne uses the essentially orthogonal technique of ell-adic cohomology to establish his theorem. However, ell-adic methods can be used (via the Grothendieck-Lefschetz trace formula) to establish rationality, and conversely, in this paper of Kedlaya p-adic methods are used to establish the Riemann hypothesis. As pointed out by Kedlaya, the ell-adic methods are tied to the intrinsic geometry of (such as the structure of sheaves and covers over ), while the -adic methods are more tied to the *extrinsic* geometry of (how sits inside its ambient affine or projective space).

In this post, I would like to record my notes on Dwork’s proof of Theorem 1, drawing heavily on the expositions of Serre, Hooley, Koblitz, and others.

The basic strategy is to control the rational integers both in an “Archimedean” sense (embedding the rational integers inside the complex numbers with the usual norm ) as well as in the “-adic” sense, with the characteristic of (embedding the integers now in the “complexification” of the -adic numbers , which is equipped with a norm that we will recall later). (This is in contrast to the methods of ell-adic cohomology, in which one primarily works over an -adic field with .) The Archimedean control is trivial:

Proposition 3 (Archimedean control of )With as above, and any embedding , we havefor all and some independent of .

*Proof:* Since is a rational integer, is just . By decomposing into affine pieces, we may assume that is of the affine form (1), then we trivially have , and the claim follows.

Another way of thinking about this Archimedean control is that it guarantees that the zeta function can be defined holomorphically on the open disk in of radius centred at the origin.

The -adic control is significantly more difficult, and is the main component of Dwork’s argument:

Proposition 4 (-adic control of )With as above, and using an embedding (defined later) with the characteristic of , we can find for any real a finite number of elements such thatfor all .

Another way of thinking about this -adic control is that it guarantees that the zeta function can be defined *meromorphically* on the entire -adic complex field .

Proposition 4 is ostensibly much weaker than Theorem 1 because of (a) the error term of -adic magnitude at most ; (b) the fact that the number of potential characteristic values here may go to infinity as ; and (c) the potential characteristic values only exist inside the complexified -adics , rather than in the algebraic integers . However, it turns out that by combining -adic control on in Proposition 4 with the trivial control on in Proposition 3, one can obtain Theorem 1 by an elementary argument that does not use any further properties of (other than the obvious fact that the are rational integers), with the in Proposition 4 chosen to exceed the in Proposition 3. We give this argument (essentially due to Borel) below the fold.

The proof of Proposition 4 can be split into two pieces. The first piece, which can be viewed as the number-theoretic component of the proof, uses external descriptions of such as (1) to obtain the following decomposition of :

Proposition 5 (Decomposition of )With and as above, we can decompose as a finite linear combination (over the integers) of sequences , such that for each such sequence , the zeta functionsare entire in , by which we mean that

as .

This proposition will ultimately be a consequence of the properties of the Teichmuller lifting .

The second piece, which can be viewed as the “-adic complex analytic” component of the proof, relates the -adic entire nature of a zeta function with control on the associated sequence , and can be interpreted (after some manipulation) as a -adic version of the Weierstrass preparation theorem:

Proposition 6 (-adic Weierstrass preparation theorem)Let be a sequence in , such that the zeta functionis entire in . Then for any real , there exist a finite number of elements such that

for all and some .

Clearly, the combination of Proposition 5 and Proposition 6 (and the non-Archimedean nature of the norm) imply Proposition 4.

** — 1. Constructing the complex -adics — **

Given a field , a *norm* on that field is defined to be a map obeying the following axioms for :

- (Non-degeneracy) if and only if .
- (Multiplicativity) .
- (Triangle inequality) .

If the triangle inequality can be improved to the ultra-triangle inequality

then we say that the norm is *non-Archimedean*. The pair will be referred to as a *normed field*.

The most familiar example of a norm is the usual (Archimedean) absolute value on the complex numbers , and thus also on its subfields and . For a given prime , we also have the -adic norm defined initially on the rationals by the formula

for any rational , where is the number of times divides an integer (with the conventions and ). Thus for instance for any integer , which is of course inverse to the Archimedean norm . (More generally, the fundamental theorem of arithmetic can be elegantly rephrased as the identity for all non-zero rationals , where ranges over all places (i.e. over all the rational primes , together with .) It is easy to see that is indeed a non-Archimedean norm. A classical theorem of Ostrowski asserts that all norms on are equivalent to either the Archimedean norm or one of the -adic norms , although we will not need this result here.

A norm on a field defines a metric , and then one can define the metric completion of this field in the usual manner (as equivalence classes of Cauchy sequences in with respect to this metric). It is easy to see that the resulting completion is again a field, and that the norm on extends continuously to a norm on the metric completion .

The metric closure of a non-Archimedean normed field is again a non-Archimedean normed field. Once one has metric completeness, one can form infinite series of elements of the field in the usual manner; but the non-Archimedean setting is somewhat better behaved than the Archimedean setting. In particular, it is easy to see that if is a non-Archimedean metrically complete normed vector field, then an infinite series is convergent if and only if it obeys the zero test , and furthermore that convergent series are automatically unconditionally convergent. (The notion of absolute convergence is not particularly relevant in non-Archimedean fields.) Thus we can talk about a countable series in a non-Archimedean metrically complete normed vector field being convergent without having to be concerned about the ordering of the series.

As key examples of metric completion, we recall that using the Archimedean norm , the metric completion of the rationals is the reals , whereas using a -adic norm , the metric completion of the rationals is instead the -adic field .

Note that the metric notion of completeness (convergence of every Cauchy sequence) is distinct from the algebraic notion of completeness (solvability of every non-constant polynomial equation, also known as being algebraically closed). For instance, the fields and are metrically complete, but not algebraically complete. However, the two notions of completeness are related to each other in a number of ways. Firstly, the metric completion of an algebraically complete field remains algebraically complete:

Lemma 7Let be a normed field which is algebraically closed. Then the metric completion is also algebraically closed.

*Proof:* Let be a monic polynomial of some degree with coefficients in . We need to show that has at least one root in . By construction of , we can view as the limit of polynomials with coefficients , where the convergence is in the sense that each coefficient converges to as for . As is already algebraically closed, each has roots (possibly with repetition). Because the are bounded, it is easy to see from the equation that the roots are uniformly bounded in . Among other things, this implies that converges to zero as , since and the coefficients of converge to zero. Writing , we conclude that the distance between and the zero set goes to zero as . From this one can easily extract a Cauchy sequence with , which then converges to a limit which can be seen to be a zero of , giving the claim.

In the other direction, in the case of the -adics at least, it is possible to extend a norm on a field to the algebraic closure of that field:

Lemma 8For any in the algebraic closure of , define the norm of by the formulawhere are the Galois conjugates of in (so in particular ). Then becomes a non-Archimedean normed field with this norm.

The situation is much more complicated in the Archimedean case, as there is no canonical way to extend the norm in this case. For instance, if one wishes to extend the Archimedean norm from to , one can do so by choosing an embedding and using the Archimedean norm on , but this is not a Galois-invariant definition. For instance, one of the two roots of the equation will have a larger norm than the other (one norm being the golden ratio, and the other being its reciprocal), but the choice of root that has the larger norm depends on the choice of embedding . Note that the definition (4) fails to be a norm in the Archimedean case; for instance, in , (4) would require and to have norm , while their sum would have norm , violating the triangle inequality.

*Proof:* The only difficult task to show is the ultra-triangle inequality (3). It suffices to show that for every Galois extension of and every , one has

We view as a finite-dimensional vector space over of some dimension , and identify each with the multiplication operator defined by . These can be viewed as an element of , the space of -linear maps from to itself, and the determinant of has norm by construction. We pick some arbitrary -basis of and use this to define a non-Archimedean “norm” on by the formula

for , and then define a “norm” on by

It is easy to see that the space is then a closed linear subspace of . In particular, since is locally compact, we see that for any compact interval , the set is compact. On the other hand, as all the are invertible, is non-zero on this compact set. Thus, for any , there exists a constant such that

for all . Since , we then see from a rescaling argument that there is a constant such that

for all . Since , we conclude the spectral radius formula

Now we can prove the ultra-triangle inequality via the tensor power trick. If , then from (5) we have

as ; from this and the easy bounds and binomial expansion we also conclude that

as . A second application of (5) then gives , and the ultra-triangle inequality follows.

Combining the two lemmas, we see that if we define

to be the metric completion of the algebraic completion of the -adic field , then this is a non-Archimedean normed field which is both metrically complete and algebraically complete, and serves as the analogue of the complex field . Note that comes with an embedding , since may clearly be embedded into . Also, the norm on induced from this embedding is clearly Galois-invariant and thus independent of the choice of embedding. Finally, we remark from construction that every non-zero element of has a norm which is a rational power of , so on taking limits (and using the ultra-triangle inequality) we see that the same is true for non-zero elements of .

Remark 1In the Archimedean case, the analogue of is the reals , and in this case the algebraic completion is a finite extension of (in fact it is just a quadratic extension) and is thus already metrically complete. However, in the -adic case, it turns out that is an infinite extension of (for instance, it contains roots of for every ), and is no longer metrically complete, requiring the additional application of Lemma 7 to recover metric completeness.

** — 2. From meromorphicity to rationality — **

We now show how Proposition 3 and Proposition 4 imply Theorem 1. The basic idea is to exploit the fact that a non-zero rational integer cannot be simultaneously small in the Archimedean sense and in the -adic sense, and in particular that we have an “uncertainty principle”

which is immediate from the fundamental theorem of arithmetic. We would like to use this uncertainty principle to eliminate the error term in Proposition 4, but run into the issue that many of the quantities involved here are not rational integers, but instead merely lie in . To get around this, we have to work with expressions that are guaranteed to be rational integers, such as polynomial combinations of the with integer coefficients. To this end, we introduce the following classical lemma:

Lemma 9 (Rationality criterion)Let be a sequence in a field , with the property that there exists a natural number such that the determinantsvanish for all sufficiently large . Then there exist , not all zero, such that we have the linear recurrence

for all sufficiently large . (Equivalently, the formal power series is a rational function of .)

Note that in the converse direction, row operations show that if one has the recurrence (8), then (7) vanishes.

*Proof:* We may assume that the determinants

are non-vanishing for infinitely many (this is a vacuous condition if ), since otherwise we can replace by in the hypotheses and conclusion.

Let be large enough that (7) vanishes, and suppose that the determinant (9) vanishes for this value of . We claim that the determinant

also vanishes; induction then shows that (9) vanishes for all sufficiently large , a contradiction.

To see why (10) vanishes, we argue as follows. As (9) vanishes, there is a non-trivial linear dependence among the rows of the matrix in (9). If this dependence does not involve the first row, then it also creates a non-trivial dependence among the first rows of the matrix (10), and we are done. Thus we may assume that the first row in (9) is a linear combination of the next rows. As a consequence, the first row in (7) is a linear combination of the next rows, plus a vector of the form for some . If is non-zero, then the row operations and cofactor expansion show that the determinant (7) is plus or minus times the determinant (10), giving the claim. If is instead zero, then the first rows of the matrix in (7) have a non-trivial linear dependence, which on deleting the first column shows that the rows of the matrix in (10) also have a non-trivial linear dependence, giving the claim.

We thus conclude that (9) does not vanish for all sufficiently large . In particular, the matrix in (7) always has rank . An easy induction then shows that the row span of the matrix in (7) is a hyperplane in (spanned by either the first rows or the last rows), which is independent of . Writing this hyperplane as , we obtain the claim.

Now let be as in Proposition 3 and Proposition 4, let be a large natural number to be chosen later, and consider the determinant (7). This is clearly a rational integer. On the one hand, from Proposition 3 we have the upper bound

for all and some , with independent of . On the other hand, from Proposition 4 we can write each row in the matrix in (7) (after applying the embedding ) as the linear combination of at most vectors of the form for various , plus an error vector whose coefficients all have norm at most (say), where is independent of . Taking determinants, we conclude that

for sufficiently large and some . Inserting the two bounds (11), (12) into the uncertainty principle (6), we conclude the vanishing

for all sufficiently large . Applying Lemma 9, we conclude that there exists a natural number and rational coefficients , not all zero, such that

for all sufficiently large . By clearing denominators, we may assume that the are all rational integers. By deleting zero terms, we may assume that and are non-zero.

We can use this recurrence to improve the conclusions of Proposition 4. Observe from that proposition, after collecting like terms and absorbing any characteristic value with into the error term, that for any we can find a finite number of distinct characteristic values with for , as well as non-zero integers , such that

for all . Applying (13) to eliminate the , we conclude that

for all sufficiently large , where is the characteristic polynomial

If one of the is not a root of , then by applying difference operators

to eliminate all the other characteristic values, we eventually conclude that

for all sufficiently large and some independent of , contradicting the hypothesis . Thus all the are zeroes of and in particular lie in . If we then let be an enumeration of the distinct zeroes of (which are all non-zero by the non-vanishing of ), and choose such that for all , we conclude that for each , there exist integers for such that

for all . The coefficients ostensibly depend on , but a repetition of the above arguments show that they are in fact independent of , since given two with , we see from the triangle inequality that

for all , and then by applying difference operators to isolate a single , we see that for all . (Note that this argument also gives the uniqueness of the characteristic values that was asserted in the introduction.) As the are independent of , we may send in (14), and conclude that

We are now nearly done, except that the are algebraic *numbers* rather than algebraic integers. However, as the are rational integers, we have for all and , and applying difference operators to (15) to isolate we conclude that for all and all . As the characteristic values are closed under the absolute Galois group of , we conclude that all Galois conjugates of also have -adic norm at most one, so the minimal polynomial of in has coefficients that are rational and have -adic norm at most one for every , and are thus rational integers, so that is an algebraic integer as required. Theorem 1 follows.

Remark 2The argument above is has been slightly rearranged from the standard argument in the literature, in which one establishes rationality of the zeta function directly, rather than first establishing rationality of the generating function (which is essentially the logarithmic derivative of the zeta function). The reason I did so was to highlight the fact that transcendental operations such as exponentiation do not play a role in this portion of the argument, in contrast to Propositions 5 and 6, which crucially exploit the properties of the exponential function.

** — 3. The -adic Weierstrass preparation theorem — **

Now we prove Proposition 6. We begin with a theorem somewhat analogous to Rouche’s theorem in complex analysis, which approximately locates a zero of an entire function that is dominated by a monomial .

Lemma 10 (Rouche-type theorem)Letbe an entire function on , thus and as . Suppose that for some and . Then there exists a root of (thus ) with .

*Proof:* We first consider the polynomials

for some . As is algebraically closed, there must be a factorisation

for some , thus is plus or minus the symmetric polynomials of the . Since , we conclude from the non-archimedean nature of the norm that for at least one . Similarly, given any , if there are exactly for which , then by computing the symmetric polynomial we conclude that . Since goes to zero, we conclude that for any , the number of for which is bounded uniformly in ; the same argument shows that the are uniformly bounded away from zero.

Now we run an argument somewhat similar to the proof of Lemma 7. Let be large natural numbers, and let be such that . We have

and hence (since )

as ; thus

On the other hand, since , and since for all but a bounded number of , we see from the non-archimedean nature of the norm that for all but a bounded number of . Since the are also uniformly bounded away from zero, we conclude that

as . From this, we can form a Cauchy sequence such that and ; taking limits, we obtain with such that , giving the claim.

One can refine the methods in this proof to read off the -adic magnitudes of all the zeroes of in terms of the Newton polytope of (yielding a -adic analogue of Jensen’s formula from complex analysis), but we will not need to do so here.

By iteratively removing the zeroes generated by the above lemma, we have

Proposition 11 (-adic Weierstrass preparation theorem, alternate form)Letbe an entire function on , and let . Then there exists a factorisation

where and

is an entire function such that for all .

*Proof:* We can make a rational power of , and then by rescaling we may normalise .

Since , the function goes to zero as , and so there exists a natural number such that

for all , with the convention that , and with strict inequality if . We now induct on . If then we are already done (setting ). Now suppose that , and that the claim has already been proven for . From the strict form of (16) with we have , so by Lemma 10 we can find with such that . We can then factor

where

and

Since , we also have

Using (16) and the non-archimedean property, one easily verifies that

for all , with strict inequality if . Applying the induction hypothesis to , we obtain the claim.

Now we prove Proposition 6. Let be as in that proposition, and let be arbitrary. By Proposition 11 we have

for some and some entire function

with for all . Now we apply formal logarithms

to both sides. Clearly for any formal power series that actually converges in ; comparing coefficients, we conclude that the formal identity holds in any characteristic zero field. For similar reasons we have for any formal power series with with coefficients in a characteristic zero field. We conclude that

But by working out the power series, we see that

where the coefficients obey the bounds

for some constant independent of . The claim then follows after increasing as necessary.

** — 4. Factorising the zeta function — **

Now we establish Proposition 5, which is the most “number-theoretical” component of Dwork’s argument.

First observe that by covering the quasiprojective variety into affine pieces, and using an induction on the dimension of to take care of any double-counted terms, we may reduce to the case when is an affine variety (1), thus

We can view as a sum over the affine variety . The next step is Fourier expansion in order to “complete” the sum into exponential sums over an ambient affine space. Write . For any define the trace map by the formula

This is a linear map over .

Let be a primitive root of unity in . Then from Fourier analysis we see that for any , the sum

is equal to if and equal to zero otherwise. Hence

In view of this (replacing by , and rescaling the zeta function), it suffices to show that for any polynomial defined over , one can decompose the sequence

as a finite linear combination over of sequences with entire.

It is convenient to remove the coordinate hyperplanes. Note that splits as the space plus some lower-dimensional spaces, where . By an induction on dimension, it thus suffices to show that the sequence

decomposes as a finite linear combination over of sequences with entire.

To prove this, we will establish the following trace formula:

Theorem 12 (Trace formula)There exists a formal power seriesin variables with coefficients in , or more compactly

for all and some (where ), such that one has the trace formula

for all , where is the -linear map on the (infinite-dimensional) vector space of all formal power series in variables defined by

and the trace on is computed using the monomial basis of , thus if

then

one can easily verify that this sum is convergent. (We will not address the subtle issue as to whether trace is a basis-independent concept in infinite dimensions.)

Let us assume this trace formula for the moment and conclude the proof of Theorem 5. Expanding out the factor in (18) and arguing as before, it will suffice to show that the zeta function

is entire; thus if is the coefficient of this zeta function, our task is to show that

Note from the formal identity

(which is true for small complex , and is thus also true for formal power series in characteristic zero) and the Jordan normal form that

on the level of formal power series in for any finite-dimensional matrix in characteristic zero. In particular, for any natural number , the coefficient of is given by the formula

where the sum ranges over distinct elements of , and over permutations of . This is a universal polynomial identity in characteristic zero, and so we conclude that the coefficient of the zeta function (20) is given by the formula

where now range over distinct natural numbers, and ranges over permutations of ; again, one can check that this sum is convergent. By the non-archimedean nature of the metric, it thus suffices to show that

as .

Now from (17) and construction of , we have

for any , and so (as is a permutation)

But because there are only a finite number of elements of of a given length , we see that grows superlinearly in (in fact it must grow by ), and the claim follows.

It remains to establish the trace formula (18). We first write the trace in a more tractable form. For any natural number , let denote the group of roots of unity.

Lemma 13Ifis a power series with for some and all , then for any we have

where we use the notation

Note that the power series for converges at all roots of unity.

*Proof:* Observe that

for any . Iterating this using (19), we conclude the identity

and so to prove the lemma it suffices to do so in the case, that is to say

The right-hand side expands as

From Fourier analysis we see that equals when is a multiple of and zero otherwise, so the sum simplifies to

The claim follows.

In view of this lemma, it will suffices to obtain an identity of the form

with for some and all , since the claim will then follow with replaced by the power series

This will be deduced from the following basic fact in -adic analysis, namely the existence of a canonical multiplicative embedding of the algebraic closure of inside .

Lemma 14 (Teichmuller lifting)Let be the algebraic closure of . Then there exists a map , known as the Teichmuller lift, with the following properties:

- (Homomorphism) One has
- (Bijection) For each , is a bijection (and hence group isomorphism, by (23)) between and . In particular, for all .
- (Description of trace) There exists a power series

Let us see how the lemma implies an identity of the form (22). Writing

for some finite set of multi-indices and coefficients , we have from the linearity of trace that

and hence by (24) we have

where

and . Note that the required decay of the coefficients of follows from that of , since the have unit -norm. The claim now follows from the bijective nature of the Teichmuller lift.

The only remaining task is to establish Lemma 14; here I will follow the exposition of Koblitz. We begin by constructing the Teichmuller lift for a given . Let be a primitive element of , thus

(The existence of such a primitive element can be seen by counting how many elements of have order strictly less than .) The minimal polynomial of over thus has degree , that is to say it is of the form

for some . We arbitrarily lift this to the -adic integers as

where reduce to modulo . Since the minimal polynomial is irreducible in , the lift is irreducible in and hence also in (here we use Lemma 10 to reach a contradiction if a monic factor of has a coefficient of -norm greater than ). Thus, if we let be a root of , then and is a degree extension of . In this field, we define the *valuation ring* and its *maximal ideal* . Then is a field generated over by , which is a root of , thus is a degree extension of and may be identified with .

We claim that the field extension is *unramified* in the sense that all of the non-zero elements of have norms that are integer powers of , and in particular that . Suppose this were not the case, then there exists an element of with . If one lets be a linear basis of over , and let be representatives of this basis in , one can then show that are linearly independent over , contradicting the fact that is a degree extension.

Let be an element of , then . As discussed earlier, we can view as an element of . By applying (a slight variant of) Hensel’s lemma, we can find a lift of such that . This gives an injective map from to , which on comparing cardinalities must be a bijection. Since the quotient map from to is a homomorphism, we see that is a homomorphism. One can check that the maps for different are compatible, and glue together to form a single map obeying the homomorphism and bijection properties.

Now we have to construct . We first give a heuristic discussion. From the construction of , we morally have for all , where we are deliberately vague as to what “” means. Since the map should morally be periodic modulo , we thus expect

and so one is led to the initial guess

for . To make this heuristic discussion rigorous, we have to formally define what means as a power series in . We write , thus and

so that

thus the Galois conjugates of multiply to , and so . We can then define by formal binomial expansion as

This is well-defined (over ) as a formal power series in . However, the convergence properties are bad, because of the denominator . Indeed, a standard computation shows that

where is the sum of the digits of the base expansion of , and so

The sequence does not go to infinity as , and so the power series does not converge for . This is a problem, since we want to apply to norm one quantities such as , and furthermore we are claiming a slightly larger radius of convergence in Lemma 14, namely (almost) .

It turns out that there is a way to tweak the series to significantly improve the -adic convergence behaviour. Namely, we (formally) define the corrected function

where is the formal power series in two variables defined by

one can verify that this is well-defined as a formal power series, and for fixed , is a formal power series in . By telescoping series, we have

as a formal power series, whenever . In particular,

for . If we can show that makes sense as a formal power series with for all , we thus have

since are the Galois conjugates of over , one can verify that are the Galois conjugates of over , and so lies in ; since has norm , this quantity in fact lies in . Quotienting out by the maximal ideal , we conclude that

and (24) follows.

So we are at last reduced to showing that with for all . From (25) we have the identity

On the other hand, we have , and hence

for some formal power series with coefficients in , and hence

We can expand as , where is a formal power series with coefficients in and no constant term, hence

As , the coefficient of this identity lets us express the coefficient of as a polynomial combination (over ) of lower degree coefficients of (as well as coefficients of ), which by induction shows that all coefficients of lie in . Replacing by , the desired claim follows.

Remark 3The function can also be defined aswhere is the Artin-Hasse exponential

and is a root of the power series .

Remark 4A small modification of Dwork’s argument also establishes rationality (the zeta function associated to) exponential sums such asfor some polynomial defined over , and some multiplicative character .

## 13 comments

Comments feed for this article

13 May, 2014 at 10:11 pm

AnonymousThere is an extraneous “sup” at the start of the post.

[Corrected, thanks – T.]14 May, 2014 at 3:04 am

KCdIn the paragraph after Theorem 2, should be . For example, if then and with converges for , as you’d want.

[Corrected, thanks – T.]14 May, 2014 at 10:43 am

Anonymous“as we wil seesup below” What does that mean?

[Corrected, thanks – T.]14 May, 2014 at 1:19 pm

Gergely HarcosVery interesting post!

I have trouble following the argument below (14). Let us consider the polynomial . It is irreducible over but reducible over . In fact, in , the two roots have valuations . This means that we cannot just say that , because will depend on the embedding we use. Perhaps is meant under a suitable embedding? This should follow from the fact that if is not an algebraic integer, then for some finite place of . Am I missing something?

Further small remarks and suggestions:

1. In the display below (1), “” is missing from the chain of equations.

2. “seesup” should be “see”.

3. should be .

4. In the proof of Proposition 3, should be .

5. In Proposition 6, the words “there exist” are missing.

6. The example “cube root of 2″ after Lemma 8 is not the best one, as this algebraic number has the same norm in every embedding into .

7. In Section 2, “Proposition 4″ and “Proposition 5″ should really be “Proposition 3″ and “Proposition 4″ at several occurrences (e.g. in the beginning, or before and after (10), or after (12)).

8. Before (13), the zeros of should be labeled as , because the degree of equals .

[Corrected, thanks – T.]14 May, 2014 at 4:46 pm

David SpeyerI thought I’d advertise an old blog post of mine http://sbseminar.wordpress.com/2011/12/12/rationality-of-the-zeta-function-mod-p/ where I show is given by the trace of the -th power of a matrix, acting on monomials in a manner similar to the infinite operator in this proof. My vague intuition for Dwork’s argument is that he figured out how to make this construction work to count solutions modulo any power of , not just the first power.

14 May, 2014 at 7:29 pm

Terence TaoThis is an interesting variant of Dwork’s argument! It’s slightly different though. In your argument, your starting point is the factorisation

(1)

for all , where is a polynomial that does not depend on k. In Dwork’s argument, the analogous factorisation is

(2)

in for all , where is a root of unity, is the Teichmuller lift of x, and is a certain formal power series independent of k with coefficients in that converges at the Teichmuller lifts. In your case, is a polynomial, which makes your matrices finite dimensional, but in Dwork’s case is transcendental, which creates the infinite-dimensional matrices.

My understanding of the Teichmuller lift and associated objects (e.g. the Artin-Hasse exponential) is not firm enough to see if there is any substantial connection between the easy factorisation (1) and the harder factorisation (2). Part of the problem is that the root of unity does not live in the p-adic field , instead being of the form for some of norm . The coefficients of G also live in some finite extension of that I don’t understand very well. In principle one can try to reduce (2) mod p to get something more closely resembling (1), but I don’t see any precise connection. (The left-hand sides of (1), (2) are of course related through the Fourier identity .)

p.s. as far as I can tell, your argument does not use the smoothness or irreducibility of , which bothers me a little bit because I’m pretty sure one can have nontrivial even and odd cohomology in the non-smooth or reducible case.

15 May, 2014 at 7:59 am

David SpeyerYour PS is bothering me too. The easiest way I can see that the story can be consistent is if those terms vanish modulo . That’s what happens in the simplest case: Take . Then the complex curve defined by is three spheres glued into a circle, so with Betti numbers . The corresponding action of Frobenius is by , and ; the argument I give is computing that the second of these is , and the reason that the third isn’t causing a problem is that it also vanishes modulo . A more interesting example is . On complex points, this is the same as the previous one (note that ) but the Frobenius action now depends on . However, it still seems to be true that the action on vanishes modulo .

15 May, 2014 at 9:32 pm

Will SawinThe Lefschetz hyperplane theorem still works for the low cohomology degrees, so the issue is large degree eigenvalues. These tend to vanish modulo p – I can see an l-adic proof that they always do for smooth compact varieties using Poincare duality + the Weil conjectures, but I don’t see a similar argument for smooth hypersurface complements.

20 May, 2014 at 7:58 am

aquazorcarsonAre there bounds for S_n that use the number of defining polynomial k? Also this k seems to collide with the k for the number of characteristic values. The results of Bombieri and Hooley also remind me of work by J. Milnor that bounds the number of intersection points of two complex varieties of complementary dimension, using Morse theory.

[Notation changed, and dependence on number of defining polynomials clarified – T.]20 May, 2014 at 5:40 pm

aquazorcarsonThanks for the quick update. With the dependency on m, is Hooley’s bound strictly weaker than Bombieri’s?

20 May, 2014 at 7:31 pm

Terence TaoIn practice yes. Technically, the bound stated in Hooley’s paper may occasionally be stronger than that stated in Bombieri’s, if the variety is not a complete intersection, but I think that if one pushes the methods in Bombieri’s paper to the limit (beyond what is stated as the main result in his paper), it will give slightly stronger results than the corresponding limit of Hooley’s methods.

5 July, 2014 at 6:57 am

Angel FernandezTheorem 2 (hypothesis Riemann).

In its indication, the similarity with the classical Riemann hypothesis.

I have te inform you that it is shown by (Albana Diez) than no-trivial zeros of the zeta function are on the straight line for ( a = 1/2).

See in: UniversaĆ±l Journal of Applied Mathematics Vol 1 (3);2013

title: Convergents series for Riemann hypothesis.

Regards.

Angel

20 March, 2015 at 7:05 am

A p-adic proof that pi is transcendental | Matt Baker's Math Blog[…] following well-known characterization of rational functions, whose proof we omit (see Lemma 9 in this blog post by Terry […]