**(h)** is *henselian*, i.e. admits a nontrivial henselian valuation,

**(eh)** is *elementarily henselian*, i.e. every is henselian,

**(def) ** admits a definable nontrivial henselian valuation, and

**(-def) ** admits a -definable nontrivial henselian valuation.

By `definable’ we mean `definable in the language of rings’. Our paper is the latest work on this topic, which began with the work of Prestel and Ziegler who show, in [PZ78], that there exist t-henselian, nonhenselian fields. Subsequently, several more recent papers explore this topic and related issues of definable valuations; examples include [JK15] and [FJ15]. Our work builds on these three papers, and others. A more complete survey of the literature is available in our paper.

There are some trivial implications between the above properties:

and

Franziska and I investigate whether any other implications hold between these properties across the class of all fields of characteristic zero, and across the smaller classes

and,

for a prime;

where denotes the canonical henselian valuation on . Our main theorem gives the `complete picture’ for each of these classes

Main Theorem (Theorem 1.1, [AJ15])

- In the class the complete picture is

- For each prime , in the class the complete picture is

- Consequently, in the class the complete picture is simply given by (1) and (2), above.

Given the `trivial’ implications, (a) and (b) above, the proof of Theorem 1.1 boils down to showing the following:

- In the class the implication holds.
- For each prime , in the class the implications and hold.
- No implications hold other than the ones listed above and the `trivial’ ones.

The proof of (4) has two stages: first we show that using Theorem B from [JK15]; then we employ the Omitting Types Theorem to remove the parameter.

The proof of (5) is more tricky. The implication follows from basic properties of the canonical henselian valuation combined with some model-theoretic trickery. Franziska and I call these arguments `Sirince tricks’ because we wrote them down when we were on a research visit to the wonderful Nesin Mathematics Village, near Sirince, Turkey! The proof of the implication depends on some delicate arguments involving uniform definitions of canonical -henselian valuations.

Finally, the proof of (6) is a set of four counterexamples. The first (given by Prestel-Ziegler, [PZ78]) simply shows the existence (in equicharacteristic zero) of t-henselian nonhenselian fields. The second and third examples were given in [JK15] and are based on the construction in [PZ78]. Finally, the fourth example is based on a more sophisticated construction give in [FJ15] of t-henselian nonhenselian fields with various properties – we extend it to defectless fields of positive characteristic.

There are several key open questions. In my opinion, perhaps the most prominent is: can our results be extended to equicharacteristic ?

**[AJ15]** Sylvy Anscombe and Franziska Jahnke. *Henselianity in the language of rings*. Manuscript, 2015. arXiv

**[FJ15]** Arno Fehm and Franziska Jahnke. *On the quantifier complexity of definable canonical henselian valuations*. Mathematical Logic Quarterly, 61(4-5):347-361, 2015.

**[JK15]** Franziska Jahnke and Jochen Koenigsmann. *Defining coarsenings of valuations*. To appear in the Proceedings of the Edinburgh Mathematical Society, 2015. arXiv

**[PZ78]** Alexander Prestel and Martin Ziegler. *Model-theoretic methods in the theory of topological fields*. Journal für die reine und angewandte Mathematik, 299(300):318-341, 1978.

Filed under: general, research Tagged: model theory, valued fields ]]>

In this post I want to discuss the solution to Puzzle #4. Let’s recall the puzzle: it was to find an uncountable chain in .

First we note that and have the same cardinality, i.e. . There are several well-known bijections that are often used to show this, but as I was writing this post I was reminded that none of the bijections that I’m familiar with are *explicit* in the sense of being given by some sort of `formula’. Here is a stackexchange discussion of this point (in particular, see the answer by Thomas Andrews).

Anyway, for the sake of this puzzle, we just need some bijection .

Now, for each , let be the preimage (under ) of the set of rational numbers in the interval .

**Proposition ** is an uncountable chain

This doesn’t need too much proof. Let be real numbers. If is such that then also . Thus .

Furthermore, there exists a rational number such that [why?]. Therefore , and so . This shows that , as required.

Filed under: Sylvy's puzzle corner Tagged: fun maths ]]>

Here is my next puzzle! Sorry that it’s been greatly delayed.

I mentioned previously the problem of constructing the midpoint of a given circle, using only ruler and compass. One student solved it nearly immediately (congratulations!), so here I offer a more general problem.

Given a parabola in the plane, construct its vertex using straightedge-and-compass construction (sometimes called `ruler-and-compass’ construction).

I want to give a rough description of what it means to perform a `straightedge-and-compass’ construction. We work in the Euclidean plane . You may be familiar with using a ruler and pair of compasses to construct geometrical drawings. For example, if you are given two points you can use the compasses to draw a circle with centre and radius .

Further, you can then construct a regular hexagon inside that circle such that the edge-length of the hexagon is equal to the radius of the circle. If you’ve not done it before, try it! It’s quite fun!

The issue arises: what shapes can one draw in this way? And which points in the plane can arise as intersections of circles/lines that have been constructed in this way?

We start with two points , as above. What can we do with ? We can draw a unique straight line which passes through , we can draw a circle with centre and radius , and we can draw a circle with centre and radius .

These three `circle-lines’ intersect in various points: draw them! The two circles intersect in two points; the line intersects in two points, one of which is ; and the line intersects in two points, one of which is .

Anyway, we get four new points. Using these new points, together with in various ways, we can construct other circle-lines. Then we can consider the intersections of *those*. And so on. All of these points that we obtain are said to be *constructible*. The key point is that there must be some *finite* sequence of points and circle-line constructions to get from to .

For simplicity (and so that we don’t have to worry about our choice of and ), we usually set to be the origin and set to be .

**Deadline: **11am Thursday 3rd of December

(you can either send your solution by email or put a hand-written solution into the folder on my office door)

**Prize: **something chocolatey? I’m open to suggestions…

**Solution: **I’ll post the solution soon after the deadline.

The webpage for these puzzles is http://anscombe.sdf.org/puzzle.html

Have fun!

Filed under: general, Sylvy's puzzle corner Tagged: fun maths ]]>

In this post I want to carry on discussing the solution to puzzle #3. Recall that denotes the world that Sonic explores in the `Special Stage’. We are thinking of as a *topological* space, upto topological equivalence.

We have already taken a look at the cases of the sphere and the torus, and we had shown that *cannot be* a sphere and that *could be* a torus. In fact we saw the `map’ of the level: so it really IS a torus! But the question is: `what can we say about using only the topological information that sonic can gather?’

The final part of puzzle #3 asked us to see if could be a Klein Bottle. A Klein bottle is the topological space obtained by taking a square and gluing together the opposite edges, as in the following diagram on the left. Note that the blue pair are glued together with a twist, while the red pair are glued `straight’.

Well, the answer is `yes’. Let be such that is even and is odd. Let be a rectangle which is divided into an chessboard, coloured as usual so that white squares and black squares alternate. I think of the `bottom’ or `horizontal’ edges as being -squares long, and the `side’ or `vertical’ edge as being -squares long. Below on the right is a picture for the case .

Gluing the edges of this chessboard together in the way described above results in a Klein bottle which is tiled in the usual `alternating black and white’ pattern that Sonic sees all around him. This shows that could be a Klein bottle.

Note that we’ve only taken into account topological information. In later posts I want to tackle the `geometric’ question.

Filed under: Sylvy's puzzle corner Tagged: fun maths ]]>

It seems that my first-year students (who form a large part of the target audience for my puzzles) are unfamiliar with the `countable’ and `uncountable’ terminology that I used in puzzle #4. …oops!

Instead of introducing it now, I’ll give a completely different puzzle which was originally shown to me by a friend (more on this later).

**Sylvy’s weekly puzzle #4a**

Show by means of a `geometrical drawing’ that

.

**deadline:** 11am Monday 9th of November

**how to hand-in: **send typed solution by email or put hand-written solution into the folder on my office door

**prize:** I’m very pleased to announce that the winner of this puzzle will get a book of mathematical puzzles!!

**webpage:** http://anscombe.sdf.org/puzzle.html

Have fun!

Filed under: Sylvy's puzzle corner Tagged: fun maths ]]>

This is an old one, but good fun.

**Sylvy’s weekly puzzle #4**

Let denote the powerset of the natural numbers. For clarity, we *will* consider 0 to be a natural number. Then the pair

is a *partial order*: the (weak) ordering is reflexive, antisymmetric, and transitive. But it is very definitely **not** a *total order*: given we may have neither nor . A *chain* is a subset of on which **is** a total order. For example:

is a chain, but note that is countable.

This week’s problem is to find an *uncountable* chain, that is a subset of which is totally ordered by and uncountable.

**Deadline: **10am Thursday 5th of November

(you can either send your solution by email or put a hand-written solution into the folder on my office door)

**Prize: **I’m very please to anounce that the winner will get a book of mathematical puzzles!!

**Solution class:** there will be a class at 10am on Thursday 5th of November in my office to go over the solution to this puzzle.

The webpage for these puzzles is http://anscombe.sdf.org/puzzle.html

Have fun!

Filed under: Sylvy's puzzle corner Tagged: fun maths ]]>

- this which talks about completeness, compactness, zeroth-order logic, and Skolemisation;
- and this which talks about nonstandard analysis, notions of `elementary convergence’ and `elementary completion’, countable saturation, compactness and saturation re-written from an analytical point-of-view, and the Szemerédi regularity lemma (something I always want to know more about).

I don’t have time to write anything more now, but later I will get back to it.

Filed under: thinking about mathematics Tagged: model theory ]]>

I’m really pleased at the interest the third puzzle has generated! Or maybe I just haven’t been able to stop talking about it… Anyway, here is the **first part** of the solution:

The question asked us to investigate the world about which Sonic runs around in a certain special level. Let’s call that world . I want to describe as a *topological suface. *If you’ve not come across this notion before, you might like to think of a few key examples:

- the (real) plane ,
- the open unit disk , and
- the sphere .

Slightly more exotic examples:

- the Möbius band,
- the Torus,
- (variations on the theme…) multi-handled tori, and
- the Klein bottle.

Perhaps now isn’t the time for a complete exposition of topological surfaces (okay, I *may* have got part-way through drafting one…!), that may come later. For the present I just want to say a few words about the *Euler characteristic* (which is denoted ) of a surface, and the formula that goes with it:

.

The first time I ever heard about this it was in the context of the Platonic solids, as follows.

- Let’s begin by thinking of a cube. A cube has 8 corners (we’re going to call these
*vertices*), 12*edges*, and 6*faces*. If denotes the number of vertices, the number of edges, and the number of faces; then we have .

- Next, let’s think of a tetrahedron. In this case there are 4 vertices, 6 edges, and 4 faces; thus we have .

(Can you see where this is going?)

- Now let’s think of an octahedron. Here there are 6 vertices, 12 edges, and 8 faces; we have .

You can check that the same formula

holds for the other Platonic solids – but it doesn’t stop there. What about other solids, for example the Archimedean solid the rhombicuboctahedron?

(I picked this one just because it has such a beautiful name.) The formula holds here too: we have . Therefore holds for this solid:

.

In fact:

**The formula holds for all polyhedra that are topologically equivalent to the sphere .**

Although I won’t properly define `topological equivalence’ here, let me just say that – roughly speaking – two shapes and are topologically equivalent if one can be transformed into the other by continuous shrinking, stretching, bending, twisting, etc. (discontinuous transformations such as tearing or joining are not allowed). Each of the polyhedra discussed above is topologically equivalent to the sphere.

Let’s now think about the surface that sonic explores. It is given to us equipped with a vertices/edges/faces structure so we can calculate . On , each face has four edges and each vertex is the endpoint of four edges. We don’t seem to know how many faces there are, so let be the number of faces. Then there are edges and vertices. [Why?] Thus $V-E+F=0$. This shows that is *not* topologically equivalent to a sphere. This is a negative answer to the first part of the puzzle.

On the other hand, could be a torus, as shown by the following map for one of these levels:

[Map of a Special Stage in Sonic 3, copyright Sega (fair use of image for educational purposes)]

How does this show that could be a Torus? You can see quite easily that the programmers have joined the left edge to the right edge and joined the top edge to the bottom edge. Thus, if Sonic walks off the left edge of the map, he will just pass round to the right edge (without noticing!). This shape is quite easily seen to be a Torus. This is a positive answer to the second part of the puzzle.

In the third part, I asked about whether could be a Klein bottle, and I’ll write a second post with the answer.

[**Edit:** perhaps it’s unfair to use the map of the level that I found online to argue that *might be* a Torus. In fact the map (plus the description of how the edges are `obviously’ joined together) shows that *definitely is* a Torus. If you didn’t have the map of the level then I think it’s reasonable to simply argue that a Torus *can *admit a vertex/edges/faces structure as seen in the level (i.e. with `square’ faces joined in fours at vertices). We’ll address the question of whether *must be* a Torus in the second part of this answer, when I’ll also discuss the Klein bottle.]

Filed under: Sylvy's puzzle corner Tagged: fun maths ]]>

My students already know that I am setting `weekly’ fun maths puzzles – although they’re not quite weekly! The first two can be found here, and I may well elaborate on them in later posts. For now, I want to discuss puzzle number 3. Remember: this is aimed at undergraduates in the first term of their first year, so please: no spoilers from more knowledgeable folks.

**Sylvy’s Weekly Puzzle #3**

This one is a little more `open ended’ than the previous puzzles. The winner will be the best attempt at an `investigative’ solution.

I tried to describe one of my favourite levels on the computer game Sonic to you (in fact I think it was on Sonic 3 – my bad), here is a screenshot:

[Special stage on Sonic 3 Copyright Sega (fair use of image for educational purposes)]

The game is played on a certain kind of `grid’. It’s an infinite two-dimensional grid, like a chessboard but infinite in all directions. No matter where Sonic goes, his world looks like this. My question is:

*What can you say about the topology of his world?*

Of course, that’s really an unfair question, because it should be asked to students that have studied a bit of topology or graph theory. Let me rephrase it more simply: could the world Sonic inhabits in the bonus level be:

(i) a sphere?

(ii) a torus?

or (harder)

(iii) a Klein Bottle?

Key things to investigate here are: Euler’s formula and the classification of closed surfaces.

**Deadline: **10am Thursday 22nd October (either by email or hand-written solutions in the folder on the outside of my office door)

**Prize: **something edible (several of the prizes from Puzzle #4 onwards will include a book of Mathematical Puzzles!)

**Don’t forget**: Class at 10am THIS THURSDAY in my office to go over the previous puzzle.

The webpage for these puzzles is http://anscombe.sdf.org/puzzle.html

Have fun!

Filed under: Sylvy's puzzle corner Tagged: fun maths ]]>

Filed under: general ]]>