 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. 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. 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.

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. 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.

## Ivan Fesenko - Research in texts

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. Below the fold I give a proof of Theorem 1 , which relies primarily on the Lang-Weil bound mentioned above.

CV , math. MG , math. One of the most well known problems from ancient Greek mathematics was that of trisecting an angle by straightedge and compass , which was eventually proven impossible in by Pierre Wantzel, using methods from Galois theory. Formally, one can set up the problem as follows. Define a configuration to be a finite collection of points, lines, and circles in the Euclidean plane. Define a construction step to be one of the following operations to enlarge the collection :.

We say that a point, line, or circle is constructible by straightedge and compass from a configuration if it can be obtained from after applying a finite number of construction steps. Problem 1 Angle trisection Let be distinct points in the plane. Is it always possible to construct by straightedge and compass from a line through that trisects the angle , in the sense that the angle between and is one third of the angle of?

On the other hand, some special angles can certainly be trisected by straightedge and compass, such as a right angle. Also, one can certainly trisect generic angles using other methods than straightedge and compass; see the Wikipedia page on angle trisection for some examples of this. The impossibility of angle trisection stands in sharp contrast to the easy construction of angle bisection via straightedge and compass, which we briefly review as follows:.

The key difference between angle trisection and angle bisection ultimately boils down to the following trivial number-theoretic fact:.

Lemma 2 There is no power of that is evenly divisible by. Proof: Obvious by modular arithmetic, by induction, or by the fundamental theorem of arithmetic. In contrast, there are of course plenty of powers of that are evenly divisible by , and this is ultimately why angle bisection is easy while angle trisection is hard. The standard way in which Lemma 2 is used to demonstrate the impossibility of angle trisection is via Galois theory. The implication is quite short if one knows this theory, but quite opaque otherwise.

We briefly sketch the proof of this implication here, though we will not need it in the rest of the discussion. Firstly, Lemma 2 implies the following fact about field extensions. Corollary 3 Let be a field, and let be an extension of that can be constructed out of by a finite sequence of quadratic extensions.

Then does not contain any cubic extensions of. Proof: If contained a cubic extension of , then the dimension of over would be a multiple of three. On the other hand, if is obtained from by a tower of quadratic extensions, then the dimension of over is a power of two. The claim then follows from Lemma 2.

To conclude the proof, one then notes that any point, line, or circle that can be constructed from a configuration is definable in a field obtained from the coefficients of all the objects in after taking a finite number of quadratic extensions, whereas a trisection of an angle will generically only be definable in a cubic extension of the field generated by the coordinates of. The Galois theory method also allows one to obtain many other impossibility results of this type, most famously the Abel-Ruffini theorem on the insolvability of the quintic equation by radicals.

However, this argument has the drawback that it requires one to first understand Galois theory or at least field theory , which is usually not presented until an advanced undergraduate algebra or number theory course, whilst the angle trisection problem requires only high-school level mathematics to formulate. In this post I would therefore like to present a different proof or perhaps more accurately, a disguised version of the standard proof of the impossibility of angle trisection by straightedge and compass, that avoids explicit mention of Galois theory though it is never far beneath the surface.

To describe the intuitive idea of the proof, let us return to the angle bisection construction, that takes a triple of points as input and returns a bisecting line as output. We iterate the construction to create a quadrisecting line , via the following sequence of steps that extend the original bisection construction:. Let us fix the points and , but not , and view as well as intermediate objects such as , , , , , , as a function of. Let us now do the following: we begin rotating counterclockwise around , which drags around the other objects , , , , , , that were constructed by accordingly.

For instance, here is an early stage of this rotation process, when the angle has become obtuse:. Now for the slightly tricky bit. We are going to keep rotating beyond a half-rotation of , so that now becomes a reflex angle. At this point, a singularity occurs; the point collides into , and so there is an instant in which the line is not well-defined.

### Number Theory Seminar

However, this turns out to be a removable singularity and the easiest way to demonstrate this will be to tap the power of complex analysis, as complex numbers can easily route around such a singularity , and we can blast through it to the other side, giving a picture like this:. Note that we have now deviated from the original construction in that and are no longer on the same side of ; we are thus now working in a continuation of that construction rather than with the construction itself.

Nevertheless, we can still work with this continuation much as, say, one works with analytic continuations of infinite series such as beyond their original domain of definition. We now keep rotating around. Here, is approaching a full rotation of :. When reaches a full rotation, a different singularity occurs: and coincide. Nevertheless, this is also a removable singularity, and we blast through to beyond a full rotation:. And now is back where it started, as are , , , and … but the point has moved, from one intersection point of to the other.

As a consequence, , , and have also changed, with being at right angles to where it was before. In the jargon of modern mathematics, the quadrisection construction has a non-trivial monodromy. But nothing stops us from rotating some more. If we continue this procedure, we see that after two full rotations of around , all points, lines, and circles constructed from have returned to their original positions. Because of this, we shall say that the quadrisection construction described above is periodic with period.

Similarly, if one performs an octisection of the angle by bisecting the quadrisection, one can verify that this octisection is periodic with period ; it takes four full rotations of around before the configuration returns to where it started. More generally, one can show. Proposition 4 Any construction of straightedge and compass from the points is periodic with period equal to a power of. The reason for this, ultimately, is because any two circles or lines will intersect each other in at most two points, and so at each step of a straightedge-and-compass construction there is an ambiguity of at most.

Each rotation of around can potentially flip one of these points to the other, but then if one rotates again, the point returns to its original position, and then one can analyse the next point in the construction in the same fashion until one obtains the proposition. But now consider a putative trisection operation, that starts with an arbitrary angle and somehow uses some sequence of straightedge and compass constructions to end up with a trisecting line :.

What is the period of this construction? If we continuously rotate around , we observe that a full rotations of only causes the trisecting line to rotate by a third of a full rotation i. Because of this, we see that the period of any construction that contains must be a multiple of. But this contradicts Proposition 4 and Lemma 2. Below the fold, I will make the above proof rigorous. Unfortunately, in doing so, I had to again leave the world of high-school mathematics, as one needs a little bit of algebraic geometry and complex analysis to resolve the issues with singularities that we saw in the above sketch.

Still, I feel that at an intuitive level at least, this argument is more geometric and accessible than the Galois-theoretic argument though anyone familiar with Galois theory will note that there is really not that much difference between the proofs, ultimately, as one has simply replaced the Galois group with a closely related monodromy group instead. The GIMPS approach to finding Mersenne primes relies of course on modern computing power, parallelisation, and efficient programming, but the number-theoretic heart of it — aside from some basic optimisation tricks such as fast multiplication and preliminary sieving to eliminate some obviously non-prime Mersenne number candidates — is the Lucas-Lehmer primality test for Mersenne numbers, which is much faster for this special type of number than any known general-purpose deterministic primality test such as, say, the AKS test.

This test is easy enough to describe, and I will do so later in this post, and also has some short elementary proofs of correctness; but the proofs are sometimes presented in a way that involves pulling a lot of rabbits out of hats, giving the argument a magical feel rather than a natural one. In this post, I will try to explain the basic ideas that make the primality test work, seeking a proof which is perhaps less elementary and a little longer than some of the proofs in the literature, but is perhaps a bit better motivated.

### Bibliography

Create a free website or blog at WordPress. Ben Eastaugh and Chris Sternal-Johnson. Subscribe to feed. What's new Updates on my research and expository papers, discussion of open problems, and other maths-related topics. Tag Archive. The fundamental theorem of Galois theory then gives a one-to-one correspondence also known as the Galois correspondence between the intermediate extensions between and and the subgroups of : Theorem 1 Fundamental theorem of Galois theory Let be a Galois extension of.

In that case, is isomorphic to the quotient group. We then have the analogous Galois correspondence: Theorem 4 Fundamental theorem of covering spaces Let be a Galois covering. 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.

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. A quantifier-free definable set is a set of the form 1 in which is quantifier-free i. 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. Define a construction step to be one of the following operations to enlarge the collection : Straightedge Given two distinct points in , form the line that connects and , and add it to.

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2. mathematics and statistics online.
3. On the Asymptotic Values and Paths of Certain Integral and Meromorphic Functions!
4. Complex Analysis and Riemann Surfaces.
6. Riemann zeta function - Wikipedia.

Compass Given two distinct points in , and given a third point in which may or may not equal or , form the circle with centre and radius equal to the length of the line segment joining and , and add it to. Intersection Given two distinct curves in thus is either a line or a circle in , and similarly for , select a point that is common to both and there are at most two such points , and add it to.

The impossibility of angle trisection stands in sharp contrast to the easy construction of angle bisection via straightedge and compass, which we briefly review as follows: Start with three points. Form the circle with centre and radius , and intersect it with the line. Let be the point in this intersection that lies on the same side of as.

Form the circle with centre and radius , and the circle with centre and radius. Let be the point of intersection of and that is not. The line will then bisect the angle. The key difference between angle trisection and angle bisection ultimately boils down to the following trivial number-theoretic fact: Lemma 2 There is no power of that is evenly divisible by.

What Is Asymptotic Analysis? And Why Does It Matter? A Deeper Understanding of Asymptotic Bounding.

We iterate the construction to create a quadrisecting line , via the following sequence of steps that extend the original bisection construction: Start with three points. Let be the point on the line which lies on , and is on the same side of as. Form the circle with centre and radius. The line will then quadrisect the angle. For instance, here is an early stage of this rotation process, when the angle has become obtuse: Now for the slightly tricky bit. However, this turns out to be a removable singularity and the easiest way to demonstrate this will be to tap the power of complex analysis, as complex numbers can easily route around such a singularity , and we can blast through it to the other side, giving a picture like this: Note that we have now deviated from the original construction in that and are no longer on the same side of ; we are thus now working in a continuation of that construction rather than with the construction itself.

Here, is approaching a full rotation of : When reaches a full rotation, a different singularity occurs: and coincide. Nevertheless, this is also a removable singularity, and we blast through to beyond a full rotation: And now is back where it started, as are , , , and … but the point has moved, from one intersection point of to the other. More generally, one can show Proposition 4 Any construction of straightedge and compass from the points is periodic with period equal to a power of.

But now consider a putative trisection operation, that starts with an arbitrary angle and somehow uses some sequence of straightedge and compass constructions to end up with a trisecting line : What is the period of this construction? What is good mathematics? Why global regularity for Navier-Stokes is hard. Top Posts Eigenvectors from eigenvalues Almost all Collatz orbits attain almost bounded values Career advice Books Does one have to be a genius to do maths? On writing The Euler-Maclaurin formula, Bernoulli numbers, the zeta function, and real-variable analytic continuation About The Collatz conjecture, Littlewood-Offord theory, and powers of 2 and 3 Work hard.

Categories expository tricks 10 guest blog 10 Mathematics math. AC 8 math. AG 41 math. AP math. AT 17 math. CA math. CO math. CT 7 math. CV 27 math. This has led to further work of A. Levin, P. Autissier, M. Ru, G. Heier, and others. In particular, Ru has defined a number, Nev D , that concisely describes the best diophantine approximation obtained by this method, where D is an effective Cartier divisor on a projective variety X. In this talk, I will give an overview of variants of these constants as developed by Ru and myself, and indicate how an example of Faltings can be derived using these constants. Abstract: Let f and g be algebraically independent entire functions. We first give an estimate of the Nevanlinna counting function for the common zeros of f n -1 and g n -1 for sufficiently large n.

## Forum Mathematicum

We then apply this estimate to study divisible sequences in the sense that f n -1 is divisible by g n -1 for infinitely many n under the assumption that f and g are multiplicatively independent. This is joint work with Ji Guo. Abstract: On the triaxial ellipsoid there are various classical dynamical systems.

In the talk we shall report on joint work of Ronald Garcia on curvature lines and geodesics. We are dealing with the question under which conditions these curves are closed or not. These are classical problems in the theory of dynamical systems and there were almost no results in this direction. We also discuss the cases of ellipsoids of revolution and ellipsoids in Minkowski space. The proofs make basic use of the analytic subspace theorem in the case of elliptic periods, and periods on abelian surfaces come up naturally. In the elliptic case we solve an extended problem of Th.

Schneider dealing with periods of differentials of the third kind. Toggle navigation.